rugged basalt
But im lost
#linear-algebra
quasi vale
@rugged basalt So the equation of the line is r = <1,0,2> + t<-2,1,0>. The direction vector of the line is <-2,1,0> and the plane contains this line. So the normal of the plane will be perpendicular to <-2,1,0>. We just need <a,b,c> dot <-2,1,0> = 0. Just take any values of a,b,c that fulfils the condition.
Or, you now have two points on the plane. One is (3,0,-1) and the other that we got from the equation of the line, (1,0,2).
You can find another vector the plane contains, which means you'll have two.
Then you know how to find the normal.
main nacelle
can someone explain this and @ me if you do
tulip basalt
A square real matrix. A is positive definite iff its symmetric part is positive definite. How do I show this?
wintry sphinx
@main nacelle basically you have to list 5 linear combinations of v1 and v2
main nacelle
so just do any constant * the vectors and add them?
wintry sphinx
yes, then write out the constants used to generate it and list out the results
@tulip basalt $\mathbf x^\top A \mathbf x = \frac{1}{2}\mathbf x^\top(A + A^\top)\mathbf x + \frac{1}{2} \mathbf x^\top(A-A^\top) \mathbf x$
stoic pythonBOT
Saccharine:
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wintry sphinx
You can show that $\mathbf x^\top A \mathbf x = \mathbf x^\top A^\top \mathbf x$
stoic pythonBOT
Saccharine:
winged bluff
I've tried looking at the eigenvectors of A and could find a sort of general form to it. I also found that scaling each row i of the vector with x_i would give an eigenvector of the eigenvalue k+1, where k is the current eigenvalue. But i don't know how to proceed
tulip basalt
@tulip basalt $\mathbf x^\top A \mathbf x = \frac{1}{2}\mathbf x^\top(A + A^\top)\mathbf x + \frac{1}{2} \mathbf x^\top(A-A^\top) \mathbf x$
@wintry sphinx thank you for this!
stoic pythonBOT
emphatic_wax:
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fallen karma
wintry sphinx
is that not the definition of affine subset?
fallen karma
The definition of an affine subset of $V$ given in this book is a subset of $V$ of form $v+U$ for some $v \in V$ and some subspace $U$ of $V$.
stoic pythonBOT
seth.delacroix:
dusky epoch
(v1 + U1) \cap (v2 + U2) = w + (U1 \cap U2) i believe, where w is some function of the v's and U's
assuming it's not empty of course
i think?? not sure ngl
hard coral
I'm pretty sure we had to prove or disprove that one in class...
But I also dont remember which way it was :'D
fallen karma
I'll see if I can show that
ionic lynx
Could somebody explain what N(A) and {0} means? It's not mentioned in the chapter of this book.
native rampart
N(A) is x such that Ax=0,ig
ionic lynx
Sort of. The book is quite difficult in explaining it
linear combination of v1, v2, vn
native rampart
{0} means the subspace is spanned by 0 alone
Basically subspace with only 0 in it
ionic lynx
So the answer to question a should be that A is 0 or that x is 0, to meet the condition of Ax=c, correct?
native rampart
It means Ax=0 means x=0 is the only solution
ionic lynx
I don't understand this question. What is the relation between y and Ax=b?
marble lance
y is a solution to Ax=b iff Ay=b
tacit flicker
Hi, Im new here and I had a question. I am trying to prove that if the det(A)=0 , then A*<v1,v2> =0 where A is a 2x2 matrix and V is non-zero . I understand that ad=cb ; i don't really know how to start Thank you
wintry steppe
What do you mean by A*<v1,v2> @tacit flicker
tacit flicker
a 2x2 matrix A multiplied by a non-zero vector
wintry steppe
Can you show the entire question
elfin schooner
ooo determinants
tacit flicker
i already proved that if Ax=0 then ad-bc=0
but i dont know how to prove it the other way around
i tried many times
native rampart
Take (x,y) such that ax+by=0
tacit flicker
well, it is an if and only if proof. So i have to prove 1) assume Ax=0, prove ad-bc=0 and 2) assume ad-bc=0, prove Ax=0
elfin schooner
yes assume ad-bc=0
wintry steppe
The matrix
1 1
1 1
Multiplied by (1,1) does not = 0
wintry steppe
Can you show the entire question
tacit flicker
this is the entire question
elfin schooner
@wintry steppe can i take over?
elfin schooner
@tacit flicker so you assumed ad-bc=0, what next
tacit flicker
thank you; then ad=bc and i am trying to prove that Ax=0
elfin schooner
okay, lets prove that Ax=0
wintry steppe
Do you know what the intuition behind a determinant is? It might be easier to explain in terms of that @elfin schooner
tacit flicker
assuming x=<v1,v2> then I have that Ax= <av1+bv2 ; cv1+dv2>
native rampart
Do you know what the intuition behind a determinant is? It might be easier to explain in terms of that @elfin schooner
Absolutely not
wintry steppe
lol
elfin schooner
why use determinants here
assuming x=<v1,v2> then I have that Ax= <av1+bv2 ; cv1+dv2>
@tacit flicker
we gotta prove that Ax=0. how do we do that?
tacit flicker
maybe proof by contradiction; im really not sure
wintry steppe
Well, if you want an intuitive method a determinant of a transformation (in 2 dimensions) tells you how areas change. If it's 0 all space is squished into a single line.
zealous vine
wintry steppe
Perhaps think about which "lines" will be mapped onto the origin, or alternatively, consider what type of vectors have a dot product of 0
tacit flicker
perpendicular vectors?
@elfin schooner do i assume that while doing my proof by contradiction
elfin schooner
yep
wintry steppe
perpendicular vectors?
@tacit flicker exactly. So try to find one
tacit flicker
i am still stuck; i am so sorry
for my proof by contradiction; i am assuming that av1+bv2=0=cv1+dv2
tacit flicker
the dot product=0; but im not sure how to do it with a matrix
wintry steppe
For a proof by contradiction youd assume the opposite of that
Well if the determinant is 0 for a 2x2 matrix, it means one row is a multiple of the other
tacit sonnet
Can someone explain in a simple terms what is meant by a Hilbert Space?
wintry steppe
So try finding the perpendicular to the first row, and see what happens after you multiply and simplify with the 2nd
tacit flicker
ill try that ; thank you
would the perpendicular to the 1st row be av1+bv2=0 so that av1= -bv2
wintry steppe
That's one way to put it
Can you think of the simplest values for v1,v2 that would satisfy this?
In terms of elements of the matrix
tacit flicker
v1/v2= - b/a ?
wintry steppe
Yep
tacit flicker
if i also assume that cv1+dv2=0, then v1/v2=-d/c
wintry steppe
Don't worry about that yet
But yes
Once you find a vector perpendicular to one row it's guaranteed to work for the other
tacit flicker
so from v1/v2= -b/a, we find that v1= -b/a v2. Then i replace v1 in the 2nd equation which is cv1+dv2=0 and we get cb=ad which=0
wintry steppe
Well, you're over complicating it a little. The simplest way to find a perpendicular of a vector x,y is to swap the coordinates and flip the sign on one: -y,x. So in your case, the simplest way is to use the vector -b, a as V
Your answer is good too tho
tacit flicker
but if i consider my answer i have no idea how i proved that Ax=0 since i just assumed it equal to 0
pallid tusk
Can anyone answer an easy af question, about jugs on uh #math-discussion
tacit flicker
yep , i just tried it both ways and they both give me that bc=ad
wintry steppe
Yeah, if you picture what the transformation is doing, it might be easier to get an idea of where to start - at least for me it is. What you're doing is finding the null space (also called kernel)
Which is the space perpendicular to all linearly dependent vectors in a matrix
tacit flicker
i have no idea what a kernel is. I know that we will learn it later
wintry steppe
Well you've just found one
tacit flicker
true... but i can't answer using kernels
@wintry steppe thank you; i think i figure it out and now i know what a kernel is
@elfin schooner thanks again; your hints were super helpful
wintry steppe
👌
ionic lynx
y is a solution to Ax=b iff Ay=b
@marble lance That's quite a short answer. If I'm correct, Z is in the vector space of A, so any multiplication of Z + x1 can be a solution for Ax=b. Correct?
marble lance
I just explained what "y is a solution to Ax=b" means. It's not a solution to your question
And I can't remember what it was
bold ivy
So I was trying to compute eigen vector, With eigen value 2 with the matrice
-1-lambda 3
-1 3-lambda
and I got -3x_1+3x_2 , trynna solve x.
but for somereason when I check the solutions it turns out to be t(1,1) whaaat?
why isn't it t(3,3)
dire arrow
If you format your question better you'll probably have a better response
bold ivy
ok
so I had matrice A = -1 3
-1 3 .
and I got two eigen values
2 and 0. When I trynna get eigen vector for eigenvalue 2. I get -3x_1+3x_2.
dire arrow
$A = \begin{matrix} -1 & 3 \ -1 & 3 \end{matrix}$
stoic pythonBOT
wOne:
dire arrow
yea
bold ivy
but v should be identity matrice?
dire arrow
Sorry I realized the expression was not very well formatted
stoic pythonBOT
wOne:
Compile Error! Click the reaction for details. (You may edit your message)
stoic pythonBOT
wOne:
bold ivy
i thought you solved eigen vector, by solving the linear equation you get from
det(A-ILambda)
dire arrow
since I listed it as a row vector. You need it to be a horizontal vector.
You do that to get the eigenvalues
Remember eigenvectors are the vectors satisfying $A \underline{v} = \lambda \underline{v}$
stoic pythonBOT
wOne:
bold ivy
det (A- i*Lambda) =
-1-h 3
-1 3-h
^ I thought you put eigen value in that matrice and then you solve the linear equation.
dire arrow
That's to find the eigenvalue
bold ivy
Yes
dire arrow
You want to use the eigenvalues to find the eigenvectors
bold ivy
but I thought that if you got the eigen values, you would just plug in
these value in lambda (h)
dire arrow
Yes but you're doing it to get a series of equations
bold ivy
wdym
dire arrow
If you have $A \underline{v} = \lambda \underline{v}$, you also have$A \underline{v} -\lambda \underline{v} = 0$ correct?
stoic pythonBOT
wOne:
dire arrow
Then you have
bold ivy
yes
dire arrow
$(A -\lambda) \underline{v} = 0$
stoic pythonBOT
wOne:
bold ivy
yes
dire arrow
And then you just put your value of lambda in
bold ivy
but then I have to do invers
bold ivy
?
dire arrow
You just multiply the matrix with v
Using your example for eigenvalue 2
$[\begin{matrix} -1 - (2)& 3 \ -1 & 3- (2) \end{matrix}] [v_1, v_2]^T = 0$
stoic pythonBOT
wOne:
stoic pythonBOT
wOne:
bold ivy
but you get ?
dire arrow
It's just matrix multiplication?
bold ivy
it is linear equation tho?
dire arrow
Do you know how to multiply a mxn matrix with a nx1 column vector?
bold ivy
wdym
dire arrow
Do you know how matrix multiplication works in general
bold ivy
yes
of ccourse
3v1 3v2
-1v1 1v1
which is a linear equation
to be honest, doing gaussian elimination on a 2x2 is prob the difficult
dire arrow
Okay I reread your question
t(3, 3) = t(1, 1)
i didnt get that part at first
since t is presumably a constant which can veary
your vectors are pointing in the same direction
bold ivy
yeah
bold ivy
just solve it
split sundial
so I set up the augmented matrix
split sundial
then I row reduced and got stuck
split sundial
so set up the matrix with those three then get the determinant?
split sundial
so s = {(1,2,1), (-2,-3,-1), (a,b,c)} then find determinant
bold ivy
do what I did
bold ivy
I already did that
split sundial
sorry im kinda new to linalg
bold ivy
think that determinant, is calculating the area between these lines or planes. So when determinant of an linear equation system is 0, then that means all these vectors has an area of 0 which means they are on top of each other aka an line. A line for an obvious reasons has an infinity of vectors. aka it is dependent.
now You want to solve a,b,c when det is not equal to 0.
bold ivy
in determinant form
bold ivy
@dire arrow I still don't get this
t(3, 3) = t(1, 1)
dire arrow
@dire arrow I still don't get this
t(3, 3) = t(1, 1)
@bold ivy Your vectors are in the same direction
(3, 3) is just (1, 1) scaled up
and the way that eigenvectors work is that they're just extended or shrunk without changing direction, right?
So since 3, 3 is just a scaled up version of 1, 1 then they are the same eigenvector
Conventionally we put the eigenvector in simplest terms so we'd say the eigenvector is k(1, 1) where k is a real number
bold ivy
1 -2 a
2 -3 b
1 -1 c
if det of this system is 0 then that means they are linearly dependent, since the area between theese area is 0. which they on top of each other. so you want to check when det is not 0, what values you get
but isn't the eigen value 3 then @dire arrow
dire arrow
eigenvalues are how much they're scaled by
meaning that if you put 1, 1 into the system you'd get 2, 2
split sundial
im going to send my work
dire arrow
2, 2 is also an eigenvector
bold ivy
oh
split sundial
I don't see where those are coming from
dire arrow
Yep
I was stupid too
I didn't understand your answer and thought you were going about the wrong way entirely
bold ivy
so, there are infinite of eigen vector if that makes sense
split sundial
dire arrow
if that makes more sense
split sundial
is what im doing so far right
bold ivy
ye
split sundial
how do i get from my part to ur part
bold ivy
augemented matrix
bold ivy
lol
bold ivy
skip zero column, you just gonna do determinant of the system
you are not going to solve the system of equation, only the constants
bold ivy
@dire arrow but I guess you would get t(3,3) if we do it in your way. but it feels like it's often you get dependent linear equation when you have 2x2
yes do determinant of it @split sundial
dire arrow
@dire arrow but I guess you would get t(3,3) if we do it in your way. but it feels like it's often you get dependent linear equation when you have 2x2
@bold ivy I don't know about the frequency. But the underlying concept still holds
bold ivy
yeah
bold ivy
one of the subject I find most difficult regarding linear algebra is prob, distance between non-pararell planes /lines
split sundial
does not equal zero?
bold ivy
you could just do equal to zero for now
dire arrow
one of the subject I find most difficult regarding linear algebra is prob, distance between non-pararell planes /lines
@bold ivy non parallel planes is always 0 in R^3
Skew lines in R^3 is probably the hardest
bold ivy
I mean skew lines
dire arrow
in R^2 non parallel lines always meet
bold ivy
I simply dont get the equation of distance of skewlines in R^3
dire arrow
What's the equation for that?
bold ivy
dire arrow
P1P2 are?
dire arrow
looks like just projection?
bold ivy
it is in R^3
well yeah it kinda is. but it is difficult to grasp it visually, for me atleast.
dire arrow
It's probably the same as distance from plane to line and distance from parallel lines if you think about it
bold ivy
I just
apperantly lines is always on a uniqe plane in R^3, but they are on the side of the plane apperantly. but in some sketch I see the lines in the plane.
like
dire arrow
planes don't have sides since they don't have thickness
dire arrow
Yeah
bold ivy
@dusky epoch didn't you say that was not the case?
dire arrow
It's just using the planes to illustrate that the distance between skew lines is really the distance between planes, just slightly restricted
bold ivy
eitherway it makes more since if a line is in a unique plane, since R^2 is a plane itself
dire arrow
yes
bold ivy
but that means you want to know the normal vector, from plane P, to plane A. and when you check when the normal vector interject the plane A, you can know the distance I guess.
dire arrow
Parallel planes share a common normal vector
mild igloo
dire arrow
Yep, so you can see how the distance between skew and parallel lines can really just be reduced to distance between planes?
mild igloo
anyone mind taking a look at this and checking my answers
I got undefined on all but b
dire arrow
I got undefined on all but b
@mild igloo check d and e
mild igloo
❤️
bold ivy
@split sundial did you get it right?
bold ivy
did you get the correct answers tho?
?
@dire arrow I dont understand why they take unit vector of normal vector ? in the dot product?
dire arrow
Probably because you're doing projection
Projection without a normal vector scales your result and you don't want that is the simple answer
To get a more detailed answer you'll want to look at the derivation
bold ivy
you mean projection without unit vector of normal vector... ?
thorn lichen
um, not sure how to incorporate the reflection part
pretty sure the rotation is [cosphi sinphi] and [-sinphi cosphi]
thorn lichen
how would you reflect it then?
bold ivy
check when it is 45 degrees in unit circle, then you just solve x,y and x,y for the second solution.
native rampart
Can you show this projection theorem?
molten fable
Yes sure
bold ivy
aka you solve cos sin, cos sin, for rotation. I think
I'm not sure.
but test it.
check 45 degrees in exact value
for cos and sin
cunning arch
i'm good with a-c, but absolutely stuck on d-g 
i know what i need to do, just not quite sure how to get started
native rampart
You know V is direct sum of W and U for some U?
molten fable
Go on
native rampart
Basically (Take all vectors not in W and form a subspace U)
molten fable
Yes
native rampart
Form an orthonomal basis on W and extend to an orthonomal basis on V and manipulate
bold ivy
@cunning arch just create some matrice that has these properties they are in question for.
molten fable
Ohhhh okayy gotcha
bold ivy
and check for your self
cunning arch
ik that's what they're asking for... i just don't know how to create some matrix that works for it
bold ivy
I think they say that two pairs are not linear combination of each other?
I can be wrong here.
@molten fable can't read
dire arrow
If you could solve and send
@molten fable If you have the picture sending it would be better than sending a screenshot of the picture....
molten fable
Oops okayy
bold ivy
if the columns of A are such that no two pairs are multiples of each other, then Ax = b has a solution for any vector b
i think they mean that assume there areno linear combination.
but it also assume that Ax = b is linear dependent
cunning arch
how do i type matrices in here lol. my top row is 1 0 0 and 0 1 0, plus we haven't learned about linear dependent/independent in class yet, so i'm not 100% sure
i don't even know where i'm going with that matrix though
bold ivy
are you sure? have you done gaussian elimination?
bold ivy
yes
molten fable
bold ivy
if you've done gaussian elimination, you should now what linear dependent is and independent. Eitherway, independent is when you get one unique solution for a given system, and dependent is when you get infinite of solutions.
molten fable
I hope it can be read now
dire arrow
how do i type matrices in here lol. my top row is 1 0 0 and 0 1 0, plus we haven't learned about linear dependent/independent in class yet, so i'm not 100% sure
@cunning arch if you know latex this will be easier
cunning arch
i don't know how to LaTeX, or else I would
native rampart
Just use the definition of orthogonal complement
dire arrow
$\begin{matrix} A& B & C \ D & E & F \ G & H & I \end{matrix}$
stoic pythonBOT
wOne:
native rampart
For 1) let (a,b) be an element of orthogonal complement , then by definition a(x)+b(0)=0
For all x
cunning arch
so i just need a matrix that linearly dependent?
native rampart
Implying a is 0 and b is an arbitrary real number
bold ivy
I intepreted as that but if you haven't gone through it then idk.
molten fable
For 1) let (a,b) be an element of orthogonal complement , then by definition a(x)+b(0)=0
@native rampart
Is fhis for me?
native rampart
Ye
molten fable
Okayy
native rampart
You get (0,y) to be the orthogonal complement set
bold ivy
but
cunning arch
idek anymore, my prof is all over the place. basically, i need a linearly dependent vector and show that there is no solution for Ax=b?
bold ivy
@cunning arch it has a condition
if it is not linearly combination is it linearly dependent
cunning arch
yeah it is, i just need to prove it
bold ivy
if it is false
oh
yeah
it is because linearly dependent is literally linearl combination
cunning arch
i'm lost on how being a linear combination/linearly dependent means that there is no solution for Ax=b
bold ivy
-linearlycombination -----> linearly dependent.
all you have to do is to prove that linear combination in a matrice will give you linearly dependent.
i'm lost on how being a linear combination/linearly dependent means that there is no solution for Ax=b
if the columns of A are such that no two pairs are multiples of each other, then Ax = b has a solution for any vector b
It asks when Ax has infinity solutions to B
cunning arch
we're trying to prove that ax=b does not have a solution
it's what the problem is asking for
bold ivy
what
cunning arch
to prove that then 'then" statement is false
to find a matrix that is true for the "if" statement but false for the "then" statement lol
bold ivy
you prove this statement false
-linearlycombination -----> linearly dependent.
just find a matrice with linearly combination and check if you get linearly dependent
wintry steppe
can someone help
cunning arch
i think i'm getting more confused lol, ty for your help.
wintry steppe
cunning arch
yes. This is what we need to do: Find a matrix where no two pairs are multiples of each other, but Ax=b does not have a solution for any vector b
dire arrow
@wintry steppe do you know what the span is?
wintry steppe
I'm a be honest I'm completely lost
bold ivy
@cunning arch oh
@cunning arch i thought it was
This is what we need to do: Find a matrix where no two pairs are multiples of each other, but Ax=b does
have a solution for any vector b
no, we're trying to prove the "then" statement false, i thought i said that?
bold ivy
wait it is on d
cunning arch
wait what on d?
bold ivy
wintry steppe
someone told me it means something like a vector v is ins span {v1,v2} if and only if there are real numbers a and b such that v=av1+bv2
but not sure what that means
cunning arch
yes, we're trying to prove the "then" statement false @bold ivy , so that it is false that Ax=b has a solution for any vector b
bold ivy
move to questions
dire arrow
someone told me it means something like a vector v is ins span {v1,v2} if and only if there are real numbers a and b such that v=av1+bv2
@wintry steppe That means that if you can find coefficients for v1 and v2 such that v3 = av1 + bv2, v3 is in the span of v1 and v2.
So play around with simple values (0, -1, 1, 2, -2, etc) and see if you can find a way to make them equal the question vector
wintry steppe
heh, that's similar to the question someone else had earlier. You're looking for the "null space" (or "kernel") of A, which is the space orthogonal (perpendicular) to all linearly independent vectors of A.
(by "linearly independent vectors of A" I mean rows, since x and y are column vectors presumably)
You need to find some linearly independent vectors that are perpendicular to (1, 2, 3) @split sundial
@deep kite PLEASE
native rampart
Why do you have to spam?
wintry steppe
That's what the other person said too. Technically you don't need to know the names for most "things", but it makes it much easier to find online resources and help if you do... and practically everything has a name. This is from a self taught POV tho, and I suspect that teachers intentionally don't tell you, so you have to use their method, and not just something you find. (take this with a grain a salt tho, I wear a tinfoil hat)
cunning arch
It is possible for Ax=0 to have nontrivial solutions, but no solution for Ax=b for any b?
Before for it to have a nontrivial solution, there must be at least 1 free variable, right? And if there's 1 free variable, doesn't that mean that there's always a solution for vector b?
dusky epoch
if you meant "for any b != 0" then sure just take A as the zero matrix
cunning arch
is that valid 
lol i didn’t even think of that, ty
bold ivy
is division part of definitin of an vector space?
wintry steppe
No
bold ivy
oh ok
I simply dont understand why polynomials is a vector space.
isn't just polynomials just R
maaybe not
bold ivy
but polynomial vector space has X,Y given dimension?
wintry steppe
Might be a little abstract but any objects, (fish, furniture, bricks) might satisfy these with correct definition of "addition" and "multiplication"
Polynomials have N of the dimensions equal to the number of different powers of X
bold ivy
yeah but that is because vector spaces is just about mathematical objects with vectors in it.
I kinda think of vector spaces as domain ?
wintry steppe
Wym?
bold ivy
but do you agree with me that vector spaces is consists of one type of mathematical objects?
wintry steppe
Again, little abstract, but any objects work if they have a defined addition and multiplication that follows the above
bold ivy
for an example , polynomial, consists of polonomial vectors. Polynomial is just an abstract predicate that you apply to mathematical objects.
i guess
wintry steppe
"type" of object isn't a very well defined term, but you could in fact you can use the above to define it
bold ivy
Like for an example,
animals is a domain,
and lion is part of that domain and so on.
wintry steppe
i.e. object are of the same type if you can add and multiply them somehow and satisfy those axioms
bold ivy
likewise, polyonomial is a domain and poloniomial vector is part of that domain.
vector space is abstract but not the members in it
Vector space is just a set, and vectors are members in it.
wintry steppe
"Domain" has a different meaning tho
bold ivy
?
I think people have difficulties, to differentiate vectorspace and vectors in it.
vectorspace is not an object while vectors in it is an object. but idk
it all ends up to set theory.
wintry steppe
I don't think domain used in the context of mappings has much to do with vector spaces
wintry steppe
The idea of a vector space is independent of that
bold ivy
wikipedia is perhaps not the best choice of information but
A vector space (also called a linear space) is a collection of objects called vectors
https://en.wikipedia.org/wiki/Vector_space
they are pretty much saying vectorspace is a set, and vectors are just members
gray dust
a domain is only talked of in the context of a function. you use domain in a very odd way as if you're not sure what a domain is
bold ivy
I know
but I was thinking you could perhaps think of vector space, as a set
like
animal is a set
while
lion is a member of that set
polynomial
vector polynomials is a set of that polynomial
gray dust
a vector space IS a set. it also has structure, namely a sense in which you can add & scale its elements, where those operations of adding & scaling satisfy the vector space axioms
bold ivy
yeah
every vector space have a different way of adding each other and scaling.
like matrices.
?
bold ivy
no I mean that different vector space, has different way of adding each other.
fallen karma
you mean vector addition has to be defined independently for each?
bold ivy
adding a matrice with each other, is not the samething as adding two numbers. My question is, do you have define how you add and scale in vector space? like adding two matrices with each other?
fallen karma
yes you do
bold ivy
I see
but also
a matrice isn't a vector space tho? it consists of plenty of vector space?
or maybe a matrice is just a subspace
fallen karma
the set of matrices of the same dimension is itself a vector space
bold ivy
but R^1 is a vector space and R^2 is a different vector space
fallen karma
yes
bold ivy
but then matrices is a subspace?
gray dust
@fallen karma set of fibonacci sequences as an R-vector space is sorta fun
fallen karma
so span{(1,0),(0,1)} is what we think of R2
fallen karma
and a subspace of that would be span{1,0}} which is a little different from R^1
but p much the same thing
bold ivy
but R^2 and matrice are not distinct from each other?
fallen karma
The set of 2x1 matrices with real numbers as components you can think of as R^2
gray dust
but p much the same thing
i find this misleading to say to someone not very strong in linear algebra without backing up what you mean
fallen karma
theyre isomorphic
gray dust
yes they are but epic doesn't know what that means
bold ivy
so matrices and R^2 ismorphic? don't know what it is yet.
gray dust
no. span{(1,0)} (as a subspace of R^2) and R, both taken as real vector spaces with the usual rules of adding/scaling, are isomorphic, which you can vaguely read as saying they have the same structure
fallen karma
Two vector spaces are isomorphic if there is a one-to-one and onto linear map from one into the other and is vaguely read as rokabe said
bold ivy
so R and R^2 are ismoprhic?
bold ivy
but what is it that is ismophic? matrix and R^n?
fallen karma
or rather there is no onto linear map from R into R^2
R^n is the set of all lists of length n with components that are real numbers, and this is isomorphic to the set of 1Xn matrices with real number components, which is isomorphic to the set of all nx1 matrices with real number components
bold ivy
yeyeyyeyeyeyeyeyyeyey
structeres as in predicate logic
<....>
?
maybe it is better if we go through homomorophism,
it is when they have different structure, but they are equal with each other?
but they have one to mapping
gray dust
homomorphism is a word you'll more often hear in an abstract algebra class, not linear algebra
generally a homomorphism is a function that respects some operation in the domain & codomain
bold ivy
but I've heard homorophism is like ismopmorophism
gray dust
a homomorphism on vector spaces respects 2 pairs of operations in the domain & codomain, namely vector addition & scalar multiplication
gray dust
is your keyboard broken
bold ivy
my brain is broken
gray dust
an isomorphism is a bijective homomorphism but i think the introduction of this into the conversation as well as your fixation on it has led you astray from your original question
bold ivy
my original question was if, matrice is a vector space and R^n is a vectorspace, what relations does these two vector space have. I thought that matrice was a subspace, since a matrice is in R^n. Now you guys said that set of matrices and R^n, has the same structures, ie. <+-.......>
gray dust
matrice is a vector space
matrice was a subspace
matrice is in R^n
these make 0 sense, explain them, and try to be exact in wording
bold ivy
oh
my original question was if, matrice is a vector space and R^n is a vectorspace, what relations does these two vector space have. I thought that matrice was a subspace, since a matrice is in R^n. Now you guys said that set of matrices and R^n, has the same structures, ie. <+-.......>
Since a subspace is when a vectorspace is in another vectorspace. And the same thought of line, I thought the matrice was a subspace of R^n.
since matrice is a vector space.
golden drum
The set of nxn matrices isn't a subspace of R^n
gray dust
again most of these make 0 sense, i have very vague ideas of what you're trying to say but none of it comes out right
bold ivy
ok I see
golden drum
The set of matrices is "larger"
golden drum
Yes
bold ivy
ok
gray dust
no don't encourage that wording @golden drum
bold ivy
wdym
gray dust
the SET of all n by n matrices with real entries can be taken as a real vector space
golden drum
Yes, sorry, I was going to talk about Isomorphism
Hahaha
The definition of a vector space is independent of R^n
bold ivy
oki oki oki
golden drum
R^n is just an example of vector space
bold ivy
what relational does set of all n by n matrice have with R^n
golden drum
You can multiply a matrix with a vector of R^n
Ax = b
And get another vector of R^n
bold ivy
can you really do operation between different vector space :S
fallen karma
@bold ivy Do you know about linear transformations
golden drum
bold ivy
my professor never went throught it
golden drum
Read a book that covers Vector Spaces
bold ivy
but isn't a vector just a matrice
golden drum
No
fallen karma
a lot of things can be vectors
bold ivy
you just represent vectors in matrice
golden drum
By definition, the set of continuos functions in [0,1] are a vector space
You have to see the Definition
,w Vector Space
See the Definition
bold ivy
I understand the definition of a vector space and the concept of such. I see no reason why a vector is not a matrice.
fallen karma
The thing is that each finite dimensional vector space over a field $F$ (such as $\mathbb{R}$) of dimension $n$ is isomorphic to to $F^{n}$ . That might be what you're trying to say @bold ivy
stoic pythonBOT
seth.delacroix:
fallen karma
but it's not always useful to think of a finite dimensional vector space of dimension n as F^n
golden drum
I understand the definition of a vector space and the concept of such. I see no reason why a vector is not a matrice.
@bold ivy
Give me an example
fallen karma
yes
bold ivy
a vector consists of [0,n] this is a matrice tho?
fallen karma
that looks like a vector in R^2
fallen karma
but the polynomial nx + 0 is also a vector in the vector space of polynomials of degree 1 or less
bold ivy
then it seems to me R^2 is just vector space matrice but in dimension 2
fallen karma
but you see how you can map the polynomial i gave to the vector you gave
bold ivy
yes
wide grail
I'm confused what part b is asking exactly
Should I be looking to see if v1 and v2 are multiples of each other or any combination of the three vectors?
gray dust
use the definition of linear dependence
bold ivy
@wide grail just do determinant 3x3 and just check when it is 0
gray dust
see if a nontrivial linear combo of v1,v2,v3 produces the 0 vector
wide grail
@bold ivy Yeah I did that for part a
hard coral
Basically b want's you to write down 2 ways of how those 3 vectors are dependent
bold ivy
@fallen karma but what is your point with your example with matrices and vector??
gray dust
for b, find 2 such nontrivial linear combos that produce the 0 vector
wide grail
idk what that means
bold ivy
I think you should you just multiply with a constant but idk
golden drum
@bold ivy I think you can see vectors in R^n as matrices with columms > 1 equal to 0
fallen karma
because it sounds like what youre trying to say is that matrices are the only vectors that there are
@bold ivy
bold ivy
huh i've never thought about that
when you say that, it makes sense but it is obviously false, if you say so.
gray dust
@wide grail what's the definition of linear dependence?
bold ivy
@golden drum columms > 1 equal to 0, what do you mean by this ?
golden drum
@bold ivy I think you can see vectors in R^n as matrices with columms > 1 equal to 0
@golden drum
This is an Isomorphism, I think
But for example, the set of continuous Functions on [0,1] are a vector space
And that isn't a set of matrices
wide grail
@gray dust They're dependent if they're a multiple of each other
sacred turret
wide grail
any two of them
golden drum
@bold ivy All the columms after the first one have 0
bold ivy
@golden drum I simply don't understand this notation [0,1]
sacred turret
<@&286206848099549185>
golden drum
That's a closed interval
sacred turret
I need help on this.
sacred turret
<@&286206848099549185>
gray dust
stop pinging helpers
wide grail
bruh
gray dust
@wide grail that works sometimes but we have a precise definition, do you know it?
bold ivy
but why in [0,1], can't have matrices?
sacred turret
I did. And now I am asking @golden drum
bold ivy
is it because matrice defined in R?
golden drum
Channel is overload, let's wait Epic
bold ivy
oki
gray dust
@wide grail does your book have a precise definition
wide grail
hold on im looking through my notebook maybe i can find the def. you are referring to
bold ivy
I must say, I like abstract thinking. Maybe that is because I'm a philosophy major.
wide grail
@gray dust So they're linearly dependent if theres more than one solution to Ax=0?
is that correct?
gray dust
@wide grail what's A, what's x
golden drum
there are infinite combination, in linear combination 😄
@bold ivy
If the set is infinite
There exists finite fielea
wide grail
A would be the matrix formed by our vectors v1 v2 v3
golden drum
Fields
bold ivy
wait what
wide grail
the columns
bold ivy
is that what you are trying to say ?
cloud coral
Just to clarify, i have to wait until he's finished with this explanation to ask another question? I'm sorry for interrupting, im new to this server:(
golden drum
Just to clarify, i have to wait until he's finished with this explanation to ask another question? I'm sorry for interrupting, im new to this server:(
@cloud coral
Yes, please
cloud coral
Okay thank you
gray dust
i don't see the need to rephrase the defn in terms of a matrix eqn just yet
wide grail
Oh so I just have to set up the matrix equation with two of the vectors? @gray dust
gray dust
the defn should involve some linear combo
@wide grail no i want the defn not rephrashed as matrix eqn
wide grail
Hmm not sure I thought the matrix eq was the def
and if we get infinite solutions to Ax=0 then its linearly dependent
gray dust
it's a compact way to rephrase the defn but i don't want you using that if you've not seen the defn as presented in every other linalg book
A set of vectors $\brc{v_1,\ldots,v_n}$ is said to be linearly dependent if there exist scalars $c_1,\ldots,c_n$, where at least one of these scalars is nonzero, such that
$$c_1v_1+\ldots+c_nv_n=0$$
stoic pythonBOT
RokettoJanpu:
wide grail
Oh I see
So i can make a matrix equation with two of the vectors and try to find a linear dependence between two of them
and then try another pair
gray dust
again there shouldn't be an obsession with writing everything as a matrix eqn
wide grail
and that should be two different linear dependence relations
gray dust
matrices can be used to compactly rewrite problems but you should learn to not always jump to doing that
wide grail
but is using pairs to find dependence relations what the question is asking?
gray dust
apply the definition i wrote to a). you show {V1,V2,V3} is linearly dependent by finding scalars c1,c2,c3, where at least one of the scalars is nonzero, such that c1V1+c2V2+c3V3=0
for b) use the definition again, find 2 different sets of scalars that satisfy
where at least one of the scalars is nonzero, such that c1V1+c2V2+c3V3=0
wide grail
Yeah I already did part a with Ax=0
gray dust
again i don't want you saying everything in terms of matrices. as long as you're doing hw on linear dependence, just use the definition i wrote above
wide grail
Mmmk so I just used that def and did the matrix thing to make solving easier and did the pairs
The first pair I chose is independent but I’m wondering if the second pair is dependent
slate quartz
o
gaunt field
I think this has something to do with projection matrix but idk what to put for v and vT
steep wraith
@rugged basalt use A, B and the equation to find a point on the sphere
get the radius vector, make a a value such that the normal vector and the radius vector are orthogonal
tacit flicker
can someone please help me for that question
Since it is an "if and only if " proof ;i need to show 1) assume Ax=0, prove ad-bc=0 and 2) assume ad-bc=0 , prove Ax=0
I already proved 1) but i don't know how to prove 2)
gray dust
take a few secs to make an example showing that’s false
id_R is linear. f:R->R, f(x)=x^2 isn’t. id_R(1)=1=f(1)
R as an R vector space is 1dim. so what?
id_R^2 is linear. let f(x,y)=(x^2,y^2). id(i)=f(i) & id(j)=f(j)
this is your q
Say I had a basis B for a vector space V over a field F. If I mapped every basis element into a arbitrary point v in the space, it is not necessarily true that the mapping is linear right?
i’m giving examples to show defining T on a basis isn’t enough to make T linear
honestly making T nonlinear is less work than what i just did
soft burrow
I already proved 1) but i don't know how to prove 2)
@tacit flicker think of it as a system of equations ax+by=cx+dy=0, where X=(x,y). particularly (x,y)=||(d,-c)|| works
gray dust
define T on a basis. then define T everywhere else such that T isn’t linear
dusk hemlock
right, i know that it is not enough to make T linear, but I was wondering if there was any possible way to define a linear map in such a way
steep wraith
I don't think the basis vectors will be linearly independent then
dusk hemlock
Right, that was what I thought as well
does this still hold if we have a infinite dimension basis?
gray dust
Say I had a basis B for a vector space V over a field F. If I mapped every basis element into a arbitrary point v in the space, it is not necessarily true that the mapping is linear right?
is this still your q
tacit flicker
@tacit flicker think of it as a system of equations ax+by=cx+dy=0, where X=(x,y). particularly (x,y)=||(d,-c)|| works
@soft burrow if a do that , then i am assuming that Ax=0 and that's actually what i want to prove
soft burrow
no, you're only assuming that ad-bc=0. Then ad+b(-c)=0 and cd+d(-c) is always 0, so there exists a vector X=(d,-c) such that AX=0
steep wraith
are there real additive maps that are not linear @dusk hemlock
dusk hemlock
@steep wraith Yes, for example https://math.stackexchange.com/questions/2377638/additive-function-t-mathbbr-rightarrow-mathbbr-that-is-not-linear
tacit flicker
@soft burrow thank you very much
ocean glade
how do we know that an orthogonal transformation without real eigenvalues must be a rotation
i mean it makes sense in the sense that rotations have complex eigenvalues only
but not sure how to prove
steep wraith
Bro @dusk hemlock what if u did like T(x, y, z) to T(x, y, 0)
Resulting vectors could be dependent
nvm I don't think that answers your original question
wintry sphinx
you want to find a polynomial such that the inner product of it with p and q are both 0
halcyon pollen
hey guys i kinda learning linear algebra where do i need to pratice the sums ?
kinda important for me tho
novel sedge
Ehi guys, I've a problem understanding how can i get the tensor component (so a matrix) in a base (x1,y1,z1) if I have the tensor matrix in a base (x,y,z) and i know every angle between (x,y,z) and (x1,y1,z1)
this should be the "solution" but I don't understand it
what i mean when i say " i have every angle" i mean i have something like this:
novel sedge
<@&286206848099549185>
gritty linden
Hey guys! im stuck on this question. i need to solve function of f(x)=k*a^2, the line passes through (0.3) and (3,8)
i need to solve f(1)
dire arrow
Your function looks slightly off, also this isn't a linear algebra question
wintry sphinx
@novel sedge I don't know if this is enough information; the angle loses important information. The end goal is to write the new basis as a linear transformation of the old basis, and I think with only the angles, this isn't unique
novel sedge
it should be enough... i have the tensor component (the matrix), i know the last base, the standard e1 e2 e3, and i know that the new coordinate system e'1 e'2 e'3 has the axis inclined in respect to the base e1 e2 e3 by those angles... i know how i can answer that question if i had a vector and not a tensor, but don't know how to do it with the tensor
@wintry sphinx
wintry sphinx
think about it in 2D
let's say the black vectors are your original basis
consider the original basis vector e_1 to be the one pointing rightward
if I tell you that the new basis's first vector is 30 degrees from that
it could be either of those two
hmm though the other angles might constrain it
I suppose if you're talking about bases for the entire space
maybe
idk
gritty linden
Where should i ask it?
novel sedge
yeah they constrain it
actually, there is right there the solution on the first image.. but i don't know how to do it 😦
U for example in this case, is a 3x3 matrix and the components are just row1: cos90, cos 45, cos 134 row2: cos 45 etc etc
F= the tensor matrix is a 3x3 (2,-2,0)(-2, 6,0)(0,0,4)
i just don't know how to use those formulas
@wintry sphinx
wintry sphinx
U, as written in there, is the change of basis matrix
and I'm not sure that it's just the cosines of the various things
novel sedge
yeah i have U and F
at least for my prof it is... but let's say it is
if i have U and F
how do you use that formula?
wintry sphinx
wait actually it is
U gives you the dot products of the ei and ej
what's F? the tensor?
novel sedge
yeah
wintry sphinx
are you familiar with einstein summation notation?
seems pretty straightforward to compute
novel sedge
just learnt today
stoic pythonBOT
Saccharine:
novel sedge
so, using last formula F'ij=Uik Ujl Fkl
what is k and l
i can understand ij like he's going through each element of F'
but how do you know what are k and l?
wary moss
Can someone give me a hint on this? So far I've just been trying to prove the determinant of the wronskian isn't 0 for like an hour 😔
I also tried to look at it from a proof by contradiction for a while but it seems like that's just the same thing
i doubt this is important either but i proved any two of the elements will be linearly independent of each other
and three
using the wronskian thingy
wintry steppe
Given some matrix A, show that there exists a d > 0, such that for all 0 < lambda < d, A + I*(lambda) is invertible
how do I even start
and we're not allowed to use properties of the characteristic polynomial
@wary moss perhaps you're overthinking it, think about the definition of linear independence
soft burrow
@wary moss one way is to suppose there's a nontrivial linear combination, differentiate n-1 times to obtain n systems of equations and the resulting matrix will be a Vandermonde matrix. Another very nice way is to consider that exponentials are eigenvectors of the differential operator in the space of differentiable functions, as explained by Sheldon Axler himself (who I certainly wasn't expecting to find in math.SE) here https://math.stackexchange.com/a/1451686
@wintry steppe is A invertible to begin?
wary moss
hm i did look at like trying to see if there was a nontrivial linear combination but it just seemed like a dead end
soft burrow
one way to do it might be with some multivariable calc/basic topology. The determinant of a matrix is a continuous (in fact, polynomial) function of the entries of the matrix. (edit: this was directed at n3therite's question.)
wary moss
also i haven't learned what a vandermonde matrix is so idk if I can use that
ill take another look at the nontrivial linear combiination way
thank you by the way
both of u
🥺
wintry steppe
@wintry steppe is A invertible to begin?
@soft burrow no you're not given anything
just A is V to V
novel sedge
no wait, what you mean? let's say we are using F'ij= Uik Ujl Fkl
so to calculate F'23
we do F'23= U2k U3l Fkl
what are those k and l? @wintry sphinx
wintry steppe
@wary moss have you tried showing it purely algebraically? I don't think you need anything too complicated if you consider how a linear combination of b_i * e^(a_i x) looks ||at the extrema||
||from there you can construct some simple inequalities||
wary moss
what do you mean by extrema?
i've been looking at it algebraically this whole time but ig just not in the right way
at least for the last half hour of me trying to solve
wintry steppe
Well, do you mind if I make the hints a little more explicit?
Idk how to explain otherwise
wintry steppe
By extrema I just mean as x gets big or small
wary moss
mk thanks
wintry steppe
||The first thing you ought to guarantee is that all the coefficients b_i add up to 1.|| ||Then think about which terms dominate - at negative x it's those with the smallest a_i, and at large x it's those with the biggest a_i.|| ||So for the decomposition of a given e^(a_j x), denoting the biggest a_i as A (and the respective b_i as B), this leaves you with the inequality: sum of b_i * e^(a_i x) > B e^(A x) and at a large x: B e^(A x) > e (a_j x) iff A > a_j||
how would i do this
wary moss
I think I'm making progress actually off the first hint which is rly cool.
Thanks for the second one I still might need it
wintry steppe
@wintry steppe if you enumerate the dimensions, think about which would have to belong to W1 and to W2 in order to make them unequal.
what do you mean? like do dim(V) = Dim(W1+W2)?
Both W1 and W2 have (n - 1) dimensions...
And they're unequal
I see
i'm not sure how to use that fact to conclude equality of V
You only need a single dimension to get from W1 to V
||And the condition that they're unequal guarantees it's in W2||
ah, so since they're both (n-1) the one element difference can tell us that the missing element can be in the other subspace?
hence the addition is n
thanks
soft burrow
@soft burrow no you're not given anything
@wintry steppe oh okay. Can you assume that $\ker (A-\lambda I)$ is non-trivial iff $\lambda$ is an eigenvalue of $A$? That doesn't involve the characteristic polynomial and it's easy to see from the fact that $Av=\lambda v = \lambda Iv$ if $v$ is an eigenvector for the eigenvalue $\lambda$.
stoic pythonBOT
bastian.uwu:
soft burrow
if not, then I'm not sure how you're expected to solve it
wintry steppe
@soft burrow that seems like a good way to do it
and yeah we're allowed to assume that
although, this problem has A + \lambda I
soft burrow
yep but it's close. Here the matrix $A$ represents a finite-dimensional linear transformation so it has a finite amount of eigenvalues. Consider the set $N$ of all real eigenvalues of $A$ that are negative. Since this set is finite it has a maximum, which we denote by $-d$. Then for all $-d<-\lambda<0$, $\ker(A+\lambda I)={0}$ since $-\lambda$ is never an eigenvalue of $A$ in this case.
stoic pythonBOT
bastian.uwu:
soft burrow
(if there are e.g. no real eigenvalues or none that are negative reals this in fact shows that A+\lambda I is invertible for any positive \lambda)
gray dust
recall defn of inverse matrix & properties of det
inner oxide
im just not sure how you would move towards a value
gray dust
2 givens, det(A)>0 & A^T=A^-1. write what A^T=A^-1 means
inner oxide
im slow in the head
this was a longer homework and i did all the harder problems and got stuck on this
thanks for the help @gray dust
gray dust
you're welcome
cloud bloom
@inner oxide easiest example is probably identity matrix
half ice
I would assume it's more like "prove that detA will be 1"
wintry steppe
with wolframalpha there is
cloud bloom
@vast thicket if you are really quick at doing co factor expansion there is lol
vast thicket
oh
wintry steppe
When doing cofactor expansion, would I be able to choose the row in red? or is it only limited to the outer rows and columns?
gray dust
ANY row/col
soft burrow
sure, in fact you generally choose the one with as many 0s as possible b/c it's easier
half forge
can someone help me in latex
im not sure why code is weird
3.3 Problem 19\\
Show that S =\{0,4,8,12,16,20,24\} is a subring of $\mathbb{Z}_{28}.$ Then prove that the map $f:\mathbb{Z}_{7} \rightarrow S $ given by $f([x]_{7}) = [8x]_{28}$
\end{problem}```
limber sierra
mathmode doesnt always play well with line breaks; you can try inserting a forced line break \\ after the \rightarrow to see if that looks better
half forge
okie it worked
how can i create a table?
i tried creating mine by 8 x 8
but it didnt work
We need to show that $S$ is in $S$. Now, we will create $\mathbb{Z}_{28}$ table.
\begin{center}
\renewcommand\arraystretch{1.3}
\setlength\doublerulesep{0pt}
\begin{tabular}{r||*{4}{2|}}
-\mathbb{Z}_{28} & 0 & 4 & 8 & 12 & 16 & 20 & 24\\
\hline\hline
0 & 0 & 24 & 20 & 16 & 12 & 8 & 4 \\
\hline
1 & 1 & 2 & 3 & 10 \\
\hline
2 & 2 & 3 & 10 & 11 \\
\hline
3 & 3 & 10 & 11 & 12 \\
\hline
\end{tabular}
\end{center}```
radiant topaz
hi, i dont know much about PCA and i have a couple questions that i cant figure out the answer to
first is this, i was under the impression that the eigenvectors of the covariance matrix that correspond with the largest eigenvalues are the principal components
and the only eigenvector for this matrix out of the three i found that are in the solutions was (0,1,2)
but there are multiple answers expected (i think?)
and for this last one,
when running it through symbolab to find the eigenvectors, (1/6, 1/3) isnt even there
i think im missing something but dont know what, any help or pointers in the right direction would be greatly appreciated
<@&286206848099549185>
gray dust
@radiant topaz you’re missing the defn of eigenvector. a nonzero vector x is said to be an eigenvector of C with eigenvalue L if Cx=Lx
let x=given vector in hw. compute Cx & note the factor by which x gets scaled. that’s its eigenvalue
radiant topaz
thanks
tired garden
These vectors are linearly dependent. If they were independent then they would be orthogonal. Correct?
The dot product of all these vectors should be 0 if they were orthogonal
dusky epoch
If they were independent then they would be orthogonal.
no.
a set of vectors can be LI without being orthogonal.
tired garden
I see
violet flume
saying ||FUCK|| is now banned (WARNING PRESS SPOILER AT OWN RISK)
native rampart
Because diagonalisability has nothing to do with invertibility
copper pulsar
Yeah A={{1,1},{0,1}}
stuck stratus
Prove that for invertible matrices A and B it holds that AB and BA are similar
How do I do this?
spiral star
that's super simple. just write down the definition of similarity and then use either that A or B is invertible
its a one-liner
strange dove
Can someone help me with how I'd approach this question
do I just go u(t) dot v(t) = 0 and solve for a?
spiral star
i would say u' and v' should be orthogonal
but my english isnt good enough to fully understand what they mean by trace. i would interpret it the trace as the vector u(t) - u(t_0)
so (a,3,1) dot (-2,1,0) = 0 and solve for a
strange dove
oh wow, when you say it like that it makes way more sense
yea i believe (a,3,1)dot(-2,1,0)=0 is correct, ima do that
thank you @spiral star
drowsy harness
howdy howdy - one of the steps in this homework problem im doing is just find the eigenvalues/eigenvectors of a matrix, and the matrix is all zeros except for the off-diagonal
i was like, boy, sure would be nice if i could just flip this matrix upside down and make my life way easier, since it would already be diagonalized
turns out this is what one of my classmates did, and apparently this is completely valid? why would this be the case?
native rampart
What was your matrix?
dusky epoch
this has 1 and -1 as eigenvalues doesn't it
drowsy harness
i believe so
i just dont get why you can move around rows and it wont change the eigenvalues/eigenvectors
like, i get that this whole deal is just systems of linear equations, but eigenvalues/eigenvectors were explained to me as values/vectors that can diagonalize a matrix, so it doesnt make sense to now hear that it doesnt have to be diagonal
and order doesnt matter
slate haven
its certainly not true in general, since
$\begin{pmatrix}
0 && -1\
1 && 0
\end{pmatrix}$
has i and -i as eigenvalues and the flipped one has 1 and -1
stoic pythonBOT
chrisply:
drowsy harness
ah! thats a good point. i am much happier knowing this is not true in general - cheers for the example!
trim galleon
What's a hyperplane?
strange dove
a plane in n-space that contains a set of solutions for some vector?
cobalt vale
this matrix isnt 1-1
its also not onto right ?
since it deosnt have a pivot in every row (third row has no pivot)
slow scroll
correct
cobalt vale
thank you!
slow scroll
np
1-1 if and only if every column has a pivot and
onto if and only if every row has a pivot
wintry steppe
i don't understand linear combinations, bases, and linear dependence
slow scroll
u need to understand linear combinations first. Then you need to understand linear independence/dependence, and then you can understand bases
cobalt vale
2 is true
wait no. false
p>n. this means number of entries is greater than number of vectors. this is false because linearly dependent means the number of vectors > number if entries
can someone help me with 1 tho
gray dust
recall T linear implies T(0)=0. its contrapositive will help
gray dust
do you know/have you proven that if T is a linear map then T maps the 0 vector of the domain to the 0 vector of the codomain?
cobalt vale
nah we havent talked about that
gray dust
you can try proving it now or take it for granted