Since we have three variables but only two pivots, we set one to be free. Let z = t. Then, x = -t
#linear-algebra
native ore
yes I understand now, thankyou
half ice
Np. Feel free to ask if you need anything else
native ore
Actually there is one more thing
Im not sure how to espress the system as a vector eqn like they did
Normally when Im asked to express a line as a vector equation I would find 2 points, take the difference of the points as a direction vector and use one point as position.
but now there a system and im kind of confused
I dont understand how they expressed it in vector form, can you even solve that system.
half ice
You're finding the nullspace of
1 2 -1 0
1 0 0 6
When I say "nullspace", I mean all of the vectors you'd multiply the matrix by, in order to get the zero vector as an output.
@native ore
In order to do this, you'd normally want to rref first. They chose not to. Instead, they assumed there's two pivots, which is fair because the rows are pretty clearly independent
Both questions are the same question - we're finding the nullspace.
- Find rref
- Difference between columns and pivots is the number of free variables. Set them to something like s, t...
- Solve each equation for x1, x2, x3... In terms of s, t
gray dust
no, work em individually
dusky epoch
the spans
north sierra
For a matrix to be invertible it has to nxn right?
dusky epoch
it has to be a square matrix, yes.
native ore
@half ice what do you mean by set the number of frew vars to s,t
In my first question it made sense how we got the one direction vector from 2 equations
How did we end up with 2 direction vectors here
north sierra
Thx ann
half ice
@native ore
There's two columns without pivots. Call them "free". Let x3 = t, x4 = s
native ore
oh I see
slender yarrow
well you're told how a projection is defined, ie you have P²=P
that lets you simplify I^2-2P+P^2 quite a bit
(+ how could you simplify I² also ?)
@wintry steppe
you screwed it up a bit
-2P+P isn't -3P
it happens 
pliant thistle
How would I do this?
Since the lines aren’t parallel but they intersect what would I have to do to find the equation of the plane
silent finch
what can you psot a picture of the whole task
clever cedar
I don't quite understand what the dot product represents
I understand that I can find the included angle with it, but independant of that
what does the dot product do
half ice
Yes, as well as x3 = t, x4 = s.
Those four equations allow you to get each in terms of s and t
clever cedar
if we are given two vectors and it asks use to find the included angle of those two vectors
then do those two vectors have the same tail
?
native ore
the same tail? doesn't that just mean they intersect
clever cedar
i mean the tip of their tail touching
native ore
oh you mean does it mean if they originate from the same point
native ore
Yea it does
when you find the angle of 2 vectors you just use the direction vectors
so they originate from the origin
the location of a vector doesnt matter when you are finding the angle between vectors
native ore
np
clever cedar
wait quick question
other than direction vector
which has magnitude & direction, what other kinds of vectors are there
native ore
When I said direction vector there can also be a position vector
clever cedar
Oh, im lost then
native ore
like t[1,0]+[5,6]
clever cedar
in my head a vector is just a set of coordinates [1,2,3] that start from a point
and end at a that point
native ore
that would just be the vector in the direction [1,0] starting and intersects the point [1,0]
native ore
this way we can express vectors that dont go through the origin
clever cedar
oh so thats what a parametric equation is
native ore
yea, because whenever we just say a direction vector by itself
we assume that its going through the origin
if we dont want it to go through the origin and specify a different location
we say a point that it goes through
clever cedar
and to clarify, we have to specify a point that it goes through because
thatll be the direction?
native ore
No the point isnt the direction
the point tell you where it is in a space
the direction vector is the direction
Here if you're give two points, (3,2) and (4,5)
find the vector that goes through those two points
to find the direction vector you would do B-A
And this would give us direction vector AB
native ore
we could just express that as t[1,3]
but that would also be assuming it goes through the origin
half ice
@native ore
You have x2 - 1/2 t - 3s = 0
Can you express x2 in terms of s and t?
native ore
and would be incorrect
so we say t[1,3] + [3,2]
(which is one of the given points because the question tells us it goes through that point)
@half ice yes like that?
half ice
If the work is right, yeah! That's the idea
This gives all of the vectors that your matrix will map to 0. Just throw in an s and t
half ice
They actually wrote it in terms of the span, which you can too with a bit of algebra
clever cedar
Ok sorry with my annoying questions. But a position vector is just a general vector for two points and a direction vector is just a vector for anywhere in the space
Am i correct
native ore
Ayyyyyyyy
@clever cedar
The position vector is a point p in an space Rn
e.g [5,6]
clever cedar
😦
native ore
that why its called position
native ore
mike
I see the confusion, the position vector is associated with the direction vector
the magnitude of a vector depends on the direction vector
t[1,0]
has no magnitude
because we havent given it a value for t yet
think of the direction vector and position vector together as a formula
(which it basically is)
clever cedar
ah okay, so they kinda related
native ore
you dont get a vector until you tell them what the value of t is
so there is no magnitude until you plug a t into the equation
because there is no vector
t[1,0] is nothing but a span of the x axis
but 5[1,0] is the vector that is [5,0] starting from the origin
clever cedar
ohhh
native ore
5[1,0] + [5,6]
is the vector [5,0] going through the point [5,6]
so now its no longer on the x axis
and we couldnt show that before without giving it position vector [5,6]
clever cedar
ok ty
clever cedar
oh
so a position vector is just a vector starting at origin
and direction vector is the position vector (kind of)
that is anywhere
as long as it has same direction
and magnitude
native ore
The position vector is the vector that is defined by positions.
e.g 2 points on a plane
the position vector is the vector connecting those 2 points
A direction vector is a vector that states the direction of a vector but it has no magnitude unless its defined by a scalar value.
A direction vector will always come out the origin unless stated otherwise.
as seen in the image
My bad mike, calling the point a position vector wasn't right. I meant that point is the position of the direction vector.
The position vector is like I said the resultant vector given 2 points in a space.
@clever cedar
heres a simpler more organized explanation
It might seem confusing because you might not have gotten to vector equations yet
But a vector equation consists of a direction vector + a position vector
plush mural
Hey all, I have the general equation of a plane x+2z=-1. I want to fint the vector equation, usually it is simple enough but there not being an y in the equation is throwing me off and I've been on 3d graphing software for well over an hour trying different stuff without progress ^^
Like my idea was first something like the parameters x=-1-2t_1 and z=t_1 but what happens to the second vector. It can't be all zeroes because that is another plane
slender yarrow
@wintry steppe did you get the rest?
jagged saffron
How would i show any real matrix over R is similar to a matrix in rational canonical form over R?
toxic pendant
Is there any situation other than two planes being parallel where there would be no solution?
half ice
No solution of what?
toxic pendant
a line
How I'm visualizing it is that two planes will always intersect in a line if they are not parallel, if they are parallel they can have no solution or infinite solutions. I was wondering if my thoughts were correct
lone cove
yeah thats the only solutions
toxic pendant
good to know I'm not crazy
clever cedar
"a vector n is normal vector to a plane if the dot product is 0 for every vector v in the plane"
what does every vector in the plane mean
how do i visualize that
also what does perpendicular mean in R^3
clever cedar
o
toxic pendant
Think of a plane as an infinitely large sheet of paper
clever cedar
yh that was another issue
my textbook spoke of plane like i know what that is
never even explained it to me
half ice
No, "every vector in the plane" means every vector from some point in the plane, that ends at some point of the plane
n is the normal vector
Mb that's what you meant nvm
clever cedar
xd
young pasture
I am not sure if this is a trick question but my homework asks me to show that I + A is invertible
I here is the identity matrix
it seems easy to just say yeah and the inverse is: (I + A)^-1 is the inverse
but I feel like there is more and this is just a trick question
quartz compass
you need more information about A
suppose A = -I then clearly I+A is not invertible
young pasture
Let I be the n×n identity matrix and let O be the n×n zero matrix . Suppose A is an n×n matrix such that A3 = O. Show that I+A is invertible
All we know is that A is nilpotent since its power results in 0
That's all that we are given
but this info is more relevant to the second part of the q
quartz compass
good, the trick for nilpotent matrices like this is to basically use the geometric series
quartz compass
it's just an algebra trick, sure
are you familiar with this series from calculus
$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$
stoic pythonBOT
Merosity:
quartz compass
hmm well how about this factorization then:
$\frac{x^n-1}{x-1} = 1+x+x^2+\cdots+ x^{n-1}$
stoic pythonBOT
Merosity:
young pasture
Nope not covered in the course.
quartz compass
could have seen that from highschool
well at any rate this is essentially the trick you need
young pasture
Why can't I just write that the inverse of I + A is the inverse
I feel like this might be noobish but it is a possible solution
no?
quartz compass
because you're just saying that it exists without showing anything
putting a little -1 doesn't make it exist
otherwise O^{-1} would be the inverse of the all zeros matrix yeah?
young pasture
Yeah but all the calc and factorization formulas above are not covered in the course.
I would be marked wrong if I put it down that way
quartz compass
well you have to come up with the inverse
young pasture
I need to explain why it is invertible only using those formulas
quartz compass
you can start multiplying things by it
(I+A)(I-A+A^2) = I
that's the answer
by showing you have a matrix that multiplies I+A to make the identity matrix
you have to get it somehow by playing around, distribute it out and check that the multiplication does work out
young pasture
Ight thanks I will try that
quartz compass
the geometric series is one of the key ingredients in many problems in math
here it is in a simpler form 10000-1 =9*1111
slim narwhal
Having problems with matricies
I'm attempting to solve for Lambda in the equation Ax = (lambda)Bx
All of which are matrices, and I am given the matrices M & K
half ice
x is a matrix too? That's very non-standard
slim narwhal
yes
half ice
What's M and K? How do they relate?
slim narwhal
would a picture of the problem help?
stoic pythonBOT
half ice
Does your class have a definition for "positive definite"?
slim narwhal
"A symmetric matrix with positive pivots is a positive definate matrix"
In addition, we recently learned a theorem which let me conclude that matrix K has real eigenvalues which are positive
though I do not know how to apply that
unless the lambda in front of M are its eigenvalues
half ice
I'm seeing on Wiki that a positive definite matrix A satisfies x'Ax > 0
for x' being the transpose of x
Not a property I know a ton about though D:
half ice
However, that may work. What happens if we multiply both sides by x' on the left?
slim narwhal
Not entirely sure (?) could you then regard it as xx'?
I thought you could maybe multiply both sides by x^-1 on the right and get rid of x entirely
slim narwhal
x inverse
slim narwhal
yes
slim narwhal
ah
dusky epoch
what
dusky epoch
are you saying it didn't occur to you that "inverse of a vector" doesn't make sense
slim narwhal
i wasn't thinking
dusky epoch
oh that explains it then
slim narwhal
yes unfortunately I have a rudimentary understanding of matrices and chose to leave the two hardest problems of my homework for when my mind was the most dead
👍
half ice
I meant to get
x'Kx = λ x'Mx
dusky epoch
ok wait what's the problem again
stoic pythonBOT
serene axle
so i tried doing this problem. the scalar lamda is indeed the eigen value of matrix A. However I'm having trouble believing that v is the eigen vector for this matrix.
this is what i've done so far:
i reduced the matrix so that only 1 equation remains in the system since the numbers worked out that way for me. however i'm not getting a second equation as shown when i wrote the linearly independent equations as a basis.
chegg, and slader seems to agree that v is the correct eigenvector but i just don't believe it from the work i've done here.
can anyone clarify and point me in the right direction? thanks.
slim narwhal
o7
gray dust
spark finch
ok
serene axle
when im asking for help and people roasting me instead
plush mural
Guys, anyone that could help me with my question? Was a long time ago but how can I get the vector equation of the plane x+2z=-1? There not being a y-value is throwing me off
plush mural
No, the vector equation of the form (x,y,z)=(x_0,y_0,z_0) + t_1(v_1) +t_2(v_2)
where v_1 and v_2 are vectors parallel to the plane
if y is 0, then one of the vectors should be the zero vector but that does not make so much sense to me and when graphin it it is wrong 😦
dusky epoch
if y is 0, then one of the vectors should be the zero vector
first off what do you mean by "y is 0" and second why does that mean one of the vectors should be the zero vector
plush mural
Ok sorry I'm really bad at explaining haha
You can describe a plane with a point on it and two vectors parallel to it, right?
My book calls that the vector equation of the plane
if we had for example x+y+z=2
dusky epoch
yes
plush mural
they would do: x=2-y-z
let y=t1 and x=t2
(x,y,z)=(2,0,0)+t1(-1,1,0)+t2(-1,0,1)
where t1, t2 are scalars that scale the two vectors and (2,0,0) is a point on the plane that translates it
Now, when I have the equation I wrote first, there is no way I can get it right because I don't know what to put for the secon vector. You can only get colinear points from that equation
or obviously you can't because it is a plane but I'm confused
dusky epoch
x = -1 - 2z
y = t1, z = t2
(x,y,z) = (-1, 0, 0) + t1(0, 1, 0) + t2(-2, 0, 1)
plush mural
Ok so now to my second problem then, I got the exact same equation as you wrote now (in one of my attempts) and when graphing they do not seem to be the same plane
It says that the vector equation you wrote is the plane x-y+x=-1
dusky epoch
z
plush mural
Oh you know what
I input the plane in the wrong way in the calculator
your solution is correct
thank you a lot for your help 😄
dusky epoch
i was about to say just that.
rich hornet
I don't get Example 2...
why should I subtract A from B?
which is B-A
Why should I do this?..
empty copper
stoic pythonBOT
Rijinaru:
slow scroll
basically, what you want are two equations that are not linear combinations of each other.
serene axle
@wintry steppe what happens when you try reducing the matrix?
winter moat
What do you mean?
clever cedar
like my book introduced this idea of normal vectors and says
"a normal vector is a vector that is orthogonal to every vector on a plane"
but idk what a plane is
it never explained it
winter moat
Do you know what a line is?
clever cedar
yea :p
winter moat
If you extend a line in a 3D space in any direction, it would become a plane.
Does this make sense to you?
clever cedar
so its just a line with a z axis?
winter moat
There can be infinitely many planes passing from z axis
winter moat
Yes, they extend to infinity.
clever cedar
ooh thats cool
winter moat
You can imagine a page from your noteook is a plane
and you can draw different arrows in your notebook, in different directions
They are vectors of that plane
clever cedar
oh i see
so if i take a pencil and stand it up on my peice of paper
the pencil (representing a vector for instance)
is orthogonal to the plane
winter moat
Yes
winter moat
🙂
clever cedar
thank you for ur time
winter moat
You're welcome!
finite sleet
guys
ill show u one nice thing about determinants
$|D| = \begin{vmatrix}
A & 0 \
X & B
\end{vmatrix}$
stoic pythonBOT
Ordy Chan:
finite sleet
where A and B are square matrices , 0 whatever kind of 0 matrix and X whatever kind of matrix
the determinant of D is just
$|D| = |A| \cdot |B|$
stoic pythonBOT
Ordy Chan:
finite sleet
very cool.
steady fiber
Multiply the matrix and the vector
Find how much the vector was scaled by
That is the eigenvalue
That's literally how eigenvalue is defined
gray dust
@wintry steppe the product of A and one of its eigenvectors xi is just xi scaled up by some constant lambda
$A\xi=\lambda \xi$
stoic pythonBOT
RokettoJanpu:
wintry steppe
Got tyvm
arctic fox
Just a quick question, are there any scenarios where multiplying two non-zero matrices produce a zero matrix?
dense dock
yes in fact there are cases where multiplying a non zero matrix by itself turns into zero
steady fiber
there's a lot of cases like that
arctic fox
hmm.. then how do I show that I+A is invertible? https://gyazo.com/7486cb52109a47e5c0d79b26dc4508f1
uneven bloom
Look at eigenvalues
subtle badge
@arctic fox Use the fact that I-A^3 = I
uneven bloom
That works too, but the eigenvalue method gives you the determinant identically
arctic fox
I haven't learned eigenvalue yet so I tried some stuff with I-A^3=I but I'm not sure about introduce the inverse of A without even knowing that it's invertible
quartz compass
honestly I would just prove it by computing (I+A)(I-A+A^2)
uneven bloom
Kei’s idea is using multiplicativity of the determinant
arctic fox
Our prof didn't teach us about determinant for nxn matrices yet, so far we only dealt with determinants for 2x2 squares using the formula ad-bc
dense dock
matrix multiplication is distributive, so a quick way to go about is doing I^3 + A^3 = I and using the sum of cubes
uneven bloom
Actually, showing that I+A and I-A+A^2 commute is enough
arctic fox
like how Merosity suggested?
uneven bloom
You can show that they commute (both equalling the identity)
So yeah
Eigenvalues are powerful because they allow you to find the determinant of I+A
Which is better than proving it’s nonzero
arctic fox
Sorry, I'm a bit confused. Can you elaborate on showing how they commute?
dense dock
AB = BA
uneven bloom
(I+A)(I-A+A^2)=I^3+A^3=I and (I-A+A^2)(I+A)=I^3+A^3=I, so they commute
Since we found a B such that (I+A)B=B(I+A)=I, I+A is invertible by definition
charred stirrup
to find the inverse of any invertible matrix, you need to turn the coefficient matrix into the identity matrix?
dense dock
yea thats the algorithm
charred stirrup
is there any other check to see if it is invertible in the first place?
other than finding its determinant
mint sentinel
where does the -6 come from?
dusky epoch
sounds like something that was given in the problem to which this is a solution
@mint sentinel please post the original problem exactly as stated
mint sentinel
lol, nvm, i didnt read the start of the problem where they said that the determinant of the original matrix is -6...
just blind
blissful vault
Every nonempty subset of a basis in R3 is a basis of R3
i don't understand this statement
what's a nonempty subset of a basis in r3?
sonic osprey
Do you know what a subset of a set is
blissful vault
please correct me then
trim sapphire
Is this a true or false q? A subset of another set is a set with some of the elements of the other set and no elements outside of the other set.
It could also have all of the elements of the other set & be an improper subset denoted by the line under the ring.
blissful vault
em yes it's a prove true or give example if false
sonic osprey
So if you take a set like {1,2,3}
A subset is just another set containing only those elements
steady fiber
oh it makes sense if it's true of false
sonic osprey
So a subset would be {1,2} or {1} or even {1,2,3}
But {3,4} would not be a subset, since 4 isn't an element of the original set
A proper subset is when the subset doesn't contain all the elements
So {1,2,3} is not a proper subset, but the rest are proper subsets
blissful vault
what about basis and subspace?
trim sapphire
The basis of a vector space is a minimal set of vectors which can be linearly combined with some set of scalars to yield any vector in the space. I don't really know much linear algebra so correct me if I am wrong. Do you understand this statement? A subspace of a space is just a subset of a space. I think I have heard that subspaces do not need to satisfy as many properties algebraically as spaces.
steady fiber
that's correct for basis
the basis is the smallest set of vectors that span a vector space
for R^3, conveniently enough a basis has 3 vectors
and that is the fewest number of vectors needed to span R^3
if you can find a subset of a basis of R^3 that has fewer than 3 vectors (but more than 0, as there is a non-zero condition)
then you have disproved the statement
blissful vault
ok so that's easy
for ex
(100)(010)(001)
that's r3
then subset is (100)(010)
and that doesn't span r3
half ice
@blissful vault
A set is a collection that has no rules. Think of it as a jar that has some things. A set of vectors is just a set, that contains vectors.
A vector space is a set of vectors that follow some special rules. You should be able to add vectors together, and scalar multiply them.
A subspace is a vector space inside your vector space.
A span of a set of vectors is the set of all linear combinations of those vectors. This set also happens to be a vector space, and is a subspace of the original vector space.
In a linearly independent set, you can't build one of the vectors using the span of the others.
A linearly independent set is called a "basis" for the set that it spans. Very important theorem in linear algebra, the size of the basis is always the same for the vector space you're working on. We call this number the dimension of the space.
trim sapphire
Yeah, that is right. Because the z-coordinate is fixed in that situation, it does not span R^3.
half ice
"the space"? A basis is a set of lin ind vectors
blissful vault
the subspace must be L.I right
half ice
L.I is a property that belongs to a set of vectors, not a vector space
No vector space itself works as an LI set
brittle juniper
steady fiber
yes
scarlet hamlet
was i looked at a map
blissful vault
and there can be basis for the space (Rn) and the subspace too
steady fiber
you can have a spanning set that's not linearly independent
scarlet hamlet
theres a basis for any space
like think of a basis as an east-west street and a north-south street? and by scaling them, you can get to anywhere on the map
thats how i understood it lol
scarlet hamlet
yeah
blissful vault
so basically the whole map
scarlet hamlet
yeah, so with those 2 basis 'vectors', u can go anywhere in that space
by scaling them
half ice
That's an orthogonal basis. We don't need a basis to follow compass directions
scarlet hamlet
ok well i mean an orthog basis is still a basis
half ice
(1,1) and (0,1) is a basis on R²
blissful vault
yes
half ice
We're not disagreeing lol.
scarlet hamlet
i just major relate to not really getting the 'basis' so im just sharing how it clicked for me
half ice
I see what you mean
But then
(1,1) (0,1) (1,0)
Is not a basis. It does span R² but isn't LI
Despite it not being LI, the set it spans is a vector space
But we no longer call these three vectors a basis
half ice
It's a spanning set for R² yeah
lyric dawn
let x and y in Rn, A a matrix and <.,.> a scalar product. Do we have <x,Ay> = <A^Tx, y> ??? (A^T is the transposed matrix of A)
half ice
Finally, because there's a basis for R² that contains two vectors. We know that ANY basis for R² contains two vectors. We say R² is a 2 dimensional space
blissful vault
ye yes yes
scarlet hamlet
@lyric dawn dot product?
use definition of scalar product and go from one side to another lol
lyric dawn
yea but i always mess up with matrix lmao
but i see why this is true now, thanks
that's trivial
scarlet hamlet
lmfao deadass nothing is trivial for me
golden zealot
can someone remind me what this symbol is?
dusky epoch
this is most likely the characteristic polynomial of A, evaluated at λ (or written in terms of λ)
golden zealot
oh ok thanks
hardy blaze
for this one
i know its false
but could it be false bc R2 cant be a subspace of R3... 
sonic osprey
Why can't R^2 be a subspace of R^3?
hardy blaze
i cant remember why but we talked about it in class 😖
it had to do smth with i hat j hat k hat which is just the standard basis
but he said it in the i hat j hat k hat terms to make it easier for us
😔
o
half ice
@hardy blaze
You're right, saying "R² is a subspace of R³" is technically false.
However, a plane is not R²
steady fiber
go through the vector space axioms
and see which axiom(s) might fail for a random plane in R^n
that's the easiest way possible to do it
and while it's technically false, abuse of words lets people often say R^2 is a subspace of R^3
when they mean R^2 is isomorphic to a subspace of R^3
ripe dagger
hi guys quick question on linear algebra . what does linearly independent mean?
half ice
Have you looked up the definition? Is there an issue with parsing it?
glad comet
So I am working on a problem requiring me to find all values of h such that the matrix is the augmented matrix of a consistent linear system. I'm confused as to the notation they are using for this matrix
gray dust
this matrix looks nothing out of the ordinary
if this is supposedly an augmented matrix representing a linear system, let x_1 and x_2 be some variables to solve
$1x_1+hx_2=4\\3x_1+15x_2=8$
stoic pythonBOT
RokettoJanpu:
gray dust
note how you can go back and forth between equations and matrix by just using the coefficients of each term or the entries in the matrix
glad comet
Ahh okay, so the matrix can be rewritten
yea i was in the middle of typing out something yes I see this now
sorry if it's super basic knowledge I haven't worked with matrices in years and I am just now getting back to them in linear algebra
gray dust
don't sweat it. btw it's actually preferable to rewrite systems as matrices when working with large numbers of equations and/or variables
celest bridge
correct me if I'm wrong, but regarding the R^2 not being a subspace of R^3 thing:
R^2 has elements of the form (x, y), while R^3 has elements of the form (a, b, c). Sure you could have an elememt in R^3 that is of the form (a, b, 0), but that's not the same as (x, y).
right?
half ice
Very true, and that's where particular definitions can conflict with common wording
celest bridge
It's a bit unintuitive
half ice
R² is not a subspace of R³
However there are subspaces of R³ that are trivially isomorphic to R²
celest bridge
R^2 is a subset of C^2 though, right?
half ice
Yes
terse mirage
However there are subspaces of R³ that are trivially isomorphic to R²
such as all elems in R³ with a fixed zero coordinate correct?
or rather
all elems with a single coordinate fixed to 0
half ice
Yeah. Any plane through (0,0,0) is actually isomorphic to R² though, so EHH
terse mirage
yeah ok that makes sense
half ice
But that's the one I was getting at. It's "R² like" but it's not R²
silent dune
How do I do part b using the results from a
terse mirage
If you have A^-1 you can multiply it by the b vector to get the unique sol to the equation
silent dune
So just take the [1 2 3 ] and multiply by my A^ -1 answer from a
I was confused cause I thought I had to actually solve for a x1 x2 x3
terse mirage
the result of [1,2,3] times A^-1 will be the soluitions of x1,x2,x3
ripe dagger
how do i know if I am given vectors in row echelon form, whether they span r3 or not?
i got
1 5 0
0 -25 3
0 0 8
is that even in ref yet?
scarlet hamlet
row reduce it!
half ice
@ripe dagger
Not fully reduced yet, no. This will reduce to the identity matrix, so the vectors are LI
serene axle
@ripe dagger the way i like to think that lets me know if something is in row echelon form is to reduce everything that's possible to reduce
so the matrix you've shown me seems like it can use a bit more tidy up.
you should get
1 0 0
0 -25 0
0 0 8
someone says the sys of eq is LI which is true as you can see here.
remote spire
Ques: If Av is in the null-space of (A^T), v lies in the null-space of A. True or False.
Proof: (My approach)Av lies in the column space of A, and Av lies in the null-space of (A^T) which is the space orthogonal to column space of A.
But from this can I say that v has to lie in the null-space of A
Oh I got it, I think I can since both those spaces are orthogonal and only null vector is contained in both. So, Av has to be zero. Thus v lies in the null-space of A.
dreamy depot
What does the square symbol mean ?
remote spire
They have defined in the question. It means to take max of each element
dreamy depot
@remote spire ahh okay but now i'm trying to figure out how to show that there's no zero vector
stoic pythonBOT
Zophike1:
dreamy depot
^ I don't feel like this is right anyone care to comment
pallid swallow
@dreamy depot the question should be a pushover
the addition rule should fall apart quickly
but it seems like you assumed that the zero vector was [a, a, a]
how about say the zero vector is [a1, a2, a3]?
dreamy depot
I think I got it to show that there's indeed no such vector $a \in \mathbb{R^{3}}$ such that $ a , \square x = x , \square a = x$ for every $x \in \mathbb{R}^{3}$. Suppose for a second that
$$ \big( \big( \max(a_{1},x_{1} )\cdot \cdot \cdot \max(a_{3},x_{3} ) \big) = \big( \big( \max(x_{1},a_{1} )\cdot \cdot \cdot \max(x_{3},x_{3} ) \big) = \big( \max(x_{1}) \cdot \cdot \cdot \max(x_{3})\big).$$
Indeed the addition law should fail since for every $x \in \mathbb{R}^{3} $ since $a + x = x + a \neq x$
@pallid swallow ^
half ice
What's +?
pallid swallow
[a1, a2, a3]+[b1, b2, b3]=[max(a1, b1), max(a2, b2), max(a3, b3)] for this problem
@half ice
dreamy depot
Is my attempted proof correct or is their a gap in my reasoning ?
stoic pythonBOT
Zophike1:
pallid swallow
I think it's not sufficient
What that says is ai<=xi for all xi
you need to spell that out
and say that that is weird
half ice
It's not true that, for every x, a + x ≠ x
dreamy depot
Is it the last line is where it's not enough ?
pallid swallow
before the last line, you probably need more explanation
dreamy depot
ahhhh ok there's a gap essentially
To show that there's indeed no such vector $a \in \mathbb{R^{3}}$ such that $ a , \square x = x , \square a = x$ for every $x \in \mathbb{R}^{3}$. Suppose for a second that
$$ \big( a_{1}, a_{2}, a_{3} \big) + \big(b_{1}, b_{2}, b_{3} \big) = \big( \big( \max(a{1},x{1} )\cdot \cdot \cdot \max(a{3},x{3} ) \big) = \big( \big( \max(x{1},a{1} )\cdot \cdot \cdot \max(x{3},x{3} ) \big) = \big( \max(x{1}) \cdot \cdot \cdot \max(x{3})\big).$$
It's not true for every $x,a+x \neq x$ since,
$$ \big( a_{1}, a_{2}, a_{3} \big) + \big( \max(x_{1} \cdot \cdot \cdot \max(x_{3}) \big) = \big( a_{1} + \max(x_{1}) \cdot \cdot \cdot a_{3} + \max(x_{3}) = \big( \max(x_{1} + a_{1} \cdot \cdot \cdot \max(x_{3}) + a_{3} \neq \big( \max(x_{1} \cdot \cdot \cdot \cdot \max(x_{3})\big) $$
stoic pythonBOT
Zophike1:
dreamy depot
I think I addressed the graps withen the solution @pallid swallow ^
pallid swallow
erm, format the solution properly?
leaden ermine
Hey Element, can I please ask you a question if you have a minute?
pallid swallow
@dreamy depot cleanest way is let x=(a1-1, a2-1, a3-1)
@leaden ermine Try a #❓how-to-get-help channel, you can ping me there
leaden ermine
Ok! Thanks
dreamy depot
ahhh okay so what I did was unnecessary
😦
But what i did is still techncially correct right @pallid swallow ?
quartz compass
just found something funny by accident playing with the characteristic equation maybe someone can tell me more about this
$\lim_{h \to 0} \frac{\det(I+hA)-1}{h} = \tr(A)$
stoic pythonBOT
Merosity:
quartz compass
it's like the directional derivative of the determinant in the direction of A evaluated at the identity is the trace
anywho if anyone has any fun tidbits to add just @ me I guess, back to doing what I was doing
dusky epoch
$\det(I + hA) = h^{-n} \chi_A(-h^{-1})$
stoic pythonBOT
Ann:
feral mountain
shouldn't it be (-h)^~~-~~n? 1 sec
stoic pythonBOT
Ann:
feral mountain
$$\begin{align}\det(I+hA)&=\det(-h(-h^{-1}I-A))
&=(-h)^n\det(-h^{-1}I-A)\
&=(-h)^n\chi_A(-h^{-1})
\end{align}$$
stoic pythonBOT
EpicGuy4227:
Compile Error! Click the 
reaction for details. (You may edit your message)
dusky epoch
oh god you're one of THOSE people.
feral mountain
do you mean that I try to find mistakes in others? is my working correct?
dusky epoch
no i mean
you're one of those people who define the charpoly as det(λI-A) instead of det(A-λI)
feral mountain
ah I see
wikipedia and the class I'm taking has it as det(λI-A) where did you see the other? I haven't read much
quartz compass
haha I specifically alluded to the characteristic equation to avoid the controversy but alas
I like both ways, but I like Ann's way better unless I need a monic polynomial
feral mountain
My take: Let $B={b_1,\ldots,b_n}$ be a basis for $V$. Then define $a_i:=f(b_i)$ and consider ${a_1,\ldots,a_n}$. If $a_i=0$ redefine $a_i$ such that this set forms a basis for $V$. Call this new set $B'$. Then $[f]{B',B}[b_i]B=[f(b_i)]{B'}=[a_i\text{ or }0]{B'}$ so $[f]_{B',B}$ is diagonal and only contains 1's and 0's.
Is this correct?
stoic pythonBOT
EpicGuy4227:
random onyx
Hey all,
Could someone please help me with understanding what analysing a dataset (matrix A) by factorizing it to QR (orthogonal and upper triangular matrices) means in practice? Specifically, if each row is a species of some bacteria and columns represent temperature at some time. How do Q and R now represent this dataset? Thanks 🙂
blissful vault
is it me or instead of A in the middle it should be I
if B is inverse of A, AB = I
half ice
@blissful vault
If A and B are inverses, then AB = BA = I
steady fiber
Yes it should be I there
steady fiber
find distance between A and C, minimize it
so you minimize $d_{AC}=\sqrt{\left(1-t\right)^{2}+\left(2-t\right)^{2}+\left(3-1\right)^{2}}$, and minimizing the square root of something is the same as minimizing the thing, so we can instead minimize $d_{AC}^{2}=\left(1-t\right)^{2}+\left(2-t\right)^{2}+\left(3-1\right)^{2}$
stoic pythonBOT
PorosInMyAshe:
steady fiber
and there's a myriad of ways to find the vertex of a parabola (which is where the minimum would be here)
ocean flame
Can I do the distance formula between AC less than distance between BC?
steady fiber
why would you do that
(according to you) the questions asks to minimize it so that C is closest to A
ocean flame
And solve for t
steady fiber
you don't even need B
steady fiber
yes
ocean flame
I might've misinterpreted it
steady fiber
why do you even care about B
who cares if it is closer to B
you are looking for when C is closest to A
ocean flame
Oh so that it is the closest it can be?
steady fiber
unless the question is "find values of t where C is closer to A than it is to B"
but it just says find where C is closest to A
slow scroll
the question doesn't even make any sense to me 
Like wtf does it mean to "find t such that C is closest to A?"
a recursive formula is given. It wants you to find x_k as a function of k
ocean flame
@steady fiber Wouldn't they say calculate the minimal distance between A and C?
steady fiber
no?
the question asks nothing about the value of the smallest distance
just for what t it occurs
ocean flame
nvm nvm xD
Okay so, your solution
You used the distance formula, and then?
You set it equal to what
slow scroll
is the question to find $t$ such that $$\lVert C - t \rVert \leq \lVert C - x \rVert$$ for all $x$ an element of the subspace spanned by $A$? Is $t$ not just the projection of $C$ onto $A$?
stoic pythonBOT
kxrider:
main jacinth
Just checking, the steps for orthogonal diagonalization are: \
Find Eigenvalues and Eigenvectors \
Orthogonalize Eigenvectors w/ G-S process \
Normalize Orthogonal Eigenvectors \
Throw Orthonormal Eigenvectors in a matrix Q \
Throw Eigenvalues in a diagonal matrix D \
Orthogonal Diagonalization is given by $$QDQ^T$$
Right?
stoic pythonBOT
Natsu:
feral grove
Use vandermonde determinants to show that a polynomial of degree less than n cannot have n distinct roots.
so i get the idea i think but i'm not sure it's clear that every polynomial can be written in the form of vandermonde determinants
feral grove
<@&286206848099549185>
blissful vault
first if r is 0 and t is 0, 000 is in the space
second we take vector x (rrt) and vector y (000) which both exist in U, and x+y =x which is inside U
and third
it must be closed under scalar mult
if c = 0,
0(rrt) = (000) which is also inside U
so since the 3 tests worked U is a subspace of R3?
I was pretty sure you coudln't repeat variables since that would make them dependant
slow scroll
second we take vector x (rrt) and vector y (000) which both exist in U, >and x+y =x which is inside U
no. The way you prove a statement for all vectors v,w in U, v+w is in U is by taking arbitrary vectors and showing that their sum is in U. You can't just cherry pick the vectors to test it on xd
blissful vault
ohh
blissful vault
so second test wouldn't work becayse if (abc) + (def) = (a+d, b+e, c+f) ,
a+d = r
b+e = r
so it wouldn't work? is that enough for proof?
and for multiplication same thing
(cA, cB, cC), cA = cB = r
slow scroll
no, every vector in U has the form (r, r, t) so if you have (a,b,c) in U, then you better have a = b. imo, you should just do (r1, r1, t1) in U and (r2, r2, t1) in U
blissful vault
ooo okok thanks
slow scroll
np
blissful vault
so the proof is just that since a has to =b, not every vector is in U right
slow scroll
no. U has the property that every vector in U looks like (r, r, t). You have to test if two vectors which look like (r,r,t) sum to get another vector that looks like (r,r,t)
slow scroll
ok, but (rrt) and (rrt) are the same vector. Ik i didn't distinguish in the message above cuz im lazy. You need two different arbitrary vectors for the vector addition test and an arbitrary scalar and vector for the scalar multiplication test
blissful vault
so i should have done (rrt) + (abc) is in U
and the only way this can happen is that a = b
slow scroll
youre making it more confusing than it has to be.
By definition of $U$, we can let $(r_1, r_1, t_1), (r_2, r_2, t_2) \in U$. Their sum is $(r_1 + r_2, r_1 + r_2, t_1 + t_2)$. The question is if this is an element of $U$.
stoic pythonBOT
kxrider:
slow scroll
@swift terrace im working on your explanation xd
@swift terrace are you familiar with linear transformations and stuff?
blissful vault
@slow scroll so the answer is yes, since (r1+r2) = (r1+r2)
slow scroll
yes
blissful vault
ok many thanks 😄
slow scroll
np
its easier to think about it in terms of the linear transformations. Let $T: K^n \to K^m$ and an invertible transformation $W: K^n \to K^n $. Now the question is if the image of $TW$ is equal to the image of $T$. Let $v$ be an element of the image of $T$. Then, there exists $w \in K^n$ such that $T(w) = v$. Since $W$ is invertible, there exists $w_2 \in K^n$ such that $W(w_2) = w$. Thus we have that $v = T(W(w_2))$. Therefore $v \in \im(TW)$ and $T \subseteq TW$. Im pretty sure the other inclusion is trivial @swift terrace
stoic pythonBOT
kxrider:
verbal tree
can anyone help me figure out if I did this correctly?
there's no answer in the textbook so im not sure
vast torrent
Looks wrong to me
verbal tree
i needed to find the distance between a point and a line
this is what i did, is this correct?
P being the point, and the d vector being the line
<@&286206848099549185>
wintry steppe
Dude
U know u could write it down and take a pic and upload it
Write it down much more eloquently than writing using a mouse in paint
If i were u
First of all i would write down the equation of rhe line
noble crest
Is there easy way to see why eigenvalues of AB and BA are equal for positive semi-definite A,B?
verbal tree
oh sorry its gud now! @wintry steppe
wintry steppe
Oh cool
verbal tree
it was right
clever cedar
are questions involving finding a point on a plane closest to a point not on a plane generic
like if i just memorize the algorithm for solving them am i good
topaz plover
why is this Linear INdependent
I know there is no scalar multiples
but the determinant is 0
why does that rule not apply to this
slow scroll
if by determinant you mean the determinant of the matrix with columns u1 and u2 then that "rule" doesn't apply because the determinant is not defined/is always zero for non-square matrices. In fact, you can only use the determinant to check if a collection of vectors form a basis for the space they span. @topaz plover
wintry steppe
Hello, can someone explain to me how I solve something like this? 
dusky epoch
by using the definitions
$F^{-1}({0})$ is the set of all $f \in X$ such that $f(A)+f(B)+f(C)=0$
stoic pythonBOT
Ann:
wintry steppe
Yeah that’s what the script of the professor has in it as well but I don’t quite understand how to use that formula like I’ve tried it and I don’t know if this is correct but that’s what I’ve got for (a) F^-1({0})={(f(A)=0, f(B)=0, f(C)=0)} is that how it’s done?
dusky epoch
well. yeah i guess that's ok to specify a function in X
undone garnet
for A is mxn matrix, B is nxm matrix (m > n)
then
det(AB+xI) = x^(m-n).det(BA+xI)
right?
proven magnet
do all vectors in a basis have to be linearly independent?
slow scroll
yes. linearly independent and spanning implies unique representation of every vector in the space
proven magnet
"if any vector v \in V"
slow scroll
also, is that linear algebra done wrong you're reading?
proven magnet
yes, LADW
slow scroll
i learned from that book 🙂
proven magnet
I did 18.06 a few months ago, I'm now doing a 2nd pass but theoretically
slow scroll
ah ok good idea
slow scroll
nah. I jumped around books for a bit (including axler's ladr) but I pretty much settled on LADW. I got different perspectives mostly by asking for help online
proven magnet
nice 🙂
slow scroll
(i do think complementing with ladr is probably a good idea tho)
scarlet hamlet
how do you guys
remember the coefficient for gram schmidt
& projection
i swear i keep getting those mixed up
keen thorn
I am solving a min LP problem with simplex iterations
The RHS is all positive
the row 0 (z row) contains only < 0 numbers on the first step
is that normal?
z | -20 -10 0 0 0 | 0
y3 | 1 1 1 0 0 | 2
y4 | 3 5 0 1 0 | 4
y5 | 2 0 0 0 1 | 3
this is at the very first step
When I do the ratio test against which column should I divide each entry in the RHS here?
the column with min coeff in row 0 or the max coeff in row 0?
by min I mean not by magnitude but really close to -inf
and max I mean close to +inf
clever cedar
if i have a scalar equation of a plane, how do i pick a random point that is on the plane
2x+y+2z = 2,
do i just choose any x,y,z that satisfy the equation?
sonic osprey
yes
clever cedar
thank you
forest onyx
slow scroll
@forest onyx Note that rank of a linear map cannot ever increase after being composed with another function.
pallid swallow
@unkempt robin any use of x hat?
mint sentinel
i have vector u = (a_1, 1, 0) and v = (a_2, 1, 0), why isn't u+v (= a_1 + a_2, 2, 0) a subspace in R^3 ?
dusky epoch
because u+v is a vector and not a subset of R^3
also i'm suspecting that you asked your question improperly
mint sentinel
well, im having a hard time understanding this
dusky epoch
post the exact statement of the problem
mint sentinel
the question is basically if whether all of vectors of the form (a,1,0) is a subspace or R^3
dusky epoch
${ (a, 1, 0) \mid a \in \bR}$
stoic pythonBOT
Ann:
dusky epoch
you're given this set and are asked to check whether it is a subspace of R^3?
mint sentinel
idk, it just says "all vectors of the form (a,1,0)" and determine whether it's a subspace or R^3
dusky epoch
can you post
the exact statement
of the problem
as i asked you to do about 8 messages up
dusky epoch
and what is this "Theorem 4.2.1" that you are expected to use?
mint sentinel
dusky epoch
...that's a theorem and not a definition? weird.
whatever.
call your subspace $U$ for ease of reference.
$(0,1,0) \in U \ (1, 1, 0) \in U \ (0,1,0) + (1,1,0) = (1, 2, 0) \ (1,2,0) \notin U$
stoic pythonBOT
Ann:
mint sentinel
so, if it were vectors of the form (a,2,0), would (a_something, 4, 0 ) be a subspace in R^3?
dusky epoch
what
...
what
no, the set consisting of all vectors of the form (a, 2, 0) is not a subspace of R^3 either.
mint sentinel
i just dont get why (1,2,0) isnt a subspace of R^3
dusky epoch
i think you're failing to understand the difference between a vector and a subspace
or between a vector and a set of vectors
dusky epoch
if you truly understood that you would not be asking why (1,2,0) isn't a subspace of R^3
mint sentinel
but how do i determine all the vectors in a subspace is what i dont get
dusky epoch
because you would then understand that (1,2,0) is a VECTOR, while a subspace of R^3 would have to be a SET OF VECTORS, and a VECTOR is NOT the same as a SET OF VECTORS.
what do you mean, "determine all the vectors in a subspace"
you're either getting ahead of yourself or babbling incoherently right now.
look.
you're given this set
which for ease of reference i'm calling U
$U = { (a,1,0) \mid a \in \bR }$
stoic pythonBOT
Ann:
mint sentinel
okay
dusky epoch
and you are asked to check whether $U$ is a subspace of $\bR^3$.
stoic pythonBOT
Ann:
dusky epoch
if that much is even remotely less than CRYSTAL CLEAR, this is your opportunity to clear up any doubts about the problem statement itself before i begin walking you through a possible solution of it.
dusky epoch
if it is crystal clear, then please confirm that to me by saying "yes, i understand the problem statement clearly".
dusky epoch
okay.
so my claim is that U is in fact NOT a subspace of R^3. and one of the reasons why it is not a subspace of R^3 is that it fails to be closed under addition.
dusky epoch
there really are a lot of disconnects in your knowledge.
mint sentinel
i know
dusky epoch
2 and (0,2,0) are not the same object.
mint sentinel
obviously not, when i said 2 i didnt mean 2 per se
mint sentinel
o-okay ~~
dusky epoch
you NEVER say something without meaning that exact thing.
mint sentinel
sowi
dusky epoch
you NEVER "say 2 without meaning 2 per se".
with that out of the way, this is one of the many possible examples to show that $U$ fails to be closed under addition:
$(0,1,0) \in U \ (1, 1, 0) \in U \ (0,1,0) + (1,1,0) = (1, 2, 0) \ (1,2,0) \notin U$
stoic pythonBOT
Ann:
mint sentinel
yes, so (1000, 1, 0) would be in U right?
dusky epoch
yes, (1000, 1, 0) is in U.
but that is beside the point.
do you understand what i have written, and do you understand why it shows that U fails closure under addition?
mint sentinel
i think so, because if U = {(a,1, 0) | a in R^3}, (1,2,0) cannot be in U
dusky epoch
R^3 is a vector space and its dimension is indeed 3.
mint sentinel
alright, so U not being a vector space at all implies that U isnt a vector space in R^3
dusky epoch
"U not being a vector space at all implies that U isn't a vector space"
yes, congratulations, you've just said a tautology.
dusky epoch
in mathematics, only a very low degree of imprecision is tolerated. and at your level, that degree of imprecision is ZERO.
mint sentinel
i was confused between whether the vector "generated" from a subspace is in R^3 or not
dusky epoch
"generated"?
dusky epoch
what
dusky epoch
but you aren't interested in that.
unkempt robin
Is there a particular case where the following is true given two n-vectors:
stoic pythonBOT
skippi:
unkempt robin
I initially thought if the vectors were orthogonal that this would be true, but this is the difference of squared norms 🤔
dusky epoch
the condition for this is not b orthogonal to a
but what if you write this as $|b|^2 = |a|^2 + |b-a|^2$
stoic pythonBOT
Ann:
unkempt robin
Ugh, rn I can only visualize this with two aligned vectors
I'm going to think about it on my drive to school
dusky epoch
no this is an equality that doesn't involve norms at all
it's always true
bc you're just
adding and subtracting a
do you... not understand that
unkempt robin
I definitely understand $b = (b - a) + a$
stoic pythonBOT
skippi:
rocky hill
how could I go about showing that for any given matrix A, the spectral radius of the diagonal matrix from A's diagonal entries is larger than the spectral radius of a lower triangular matrix from A?
is that even true?
Where L and D are both obtained from A
scarlet hamlet
^^ anyone?
north sierra
so this is a square matrix
and is it not one to one because T(u) = T(v) for only some pair of distinct vectors u and v?
half ice
Yes, not one-to-one. Also not bijective
north sierra
if T(u) = T(v) for all pair of distinct vectors u and v then itd be injective?
slow scroll
not at all! injectivity says T(u) = T(v) implies u=v
half ice
Every output is unique to its input
north sierra
so how do we know here that it's not injective?
slow scroll
another way of looking at it is that T(u) - T(v) = T(u-v) = 0 so u-v != 0 is an element of the kernel. Since the kernel is non trivial, T can't be an injection
half ice
If T is injective, you can take it off both sides of the equal sign.
T(u) = T(v)
u = v
But u isn't v, so T isn't injective
Yes, u and v map to the same place under T, so T isn't injective
half ice
Two different ways to think about it
half ice
And finally, if T is linear and not injective, then more than one thing maps to 0
half ice
Injective ⇔ Nullspace is trivial
By kxrider's proof
Well, they proved one way, but both hold
prisma jewel
Hello I was wondering if any one knew how to approach this question?
Suppose that x is an element of G with order 16. What is the order of x^3? What is the order of x^(−4)? What is the order of x^6?
sonic osprey
This really should be #groups-rings-fields, but its busy so use a questions channel
This is not linear algebra
prisma jewel
My bad. I saw Algebra and just went for it.