What's the rank of matrix
[0 0]
[0 1]
#linear-algebra
wintry steppe
native rampart
What's the rank of matrix
[0 0]
[0 1]
@wintry steppe 1
magic light
I think it's only 0 if everything is 0
may be wrong here
rank = number of independent columns
u have 1 independent column aka 0 1
half storm
$a \begin{bmatrix} 3 & -1 \ 4 & 0 \ \end{bmatrix} + b \begin{bmatrix} -2 & 5 \ 1 & 3 \ \end{bmatrix} + c\begin{bmatrix} 0 & 1 \ -1 & 0 \ \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \ \end{bmatrix}$
magic light
damn
wintry steppe
lmao
stoic pythonBOT
TheDon:
half storm
That should compile
half storm
there we go
wintry steppe
so that's a good answer then 🤔
magic light
wait
half storm
Alright, do you see how you get a system of equations out of that?
wintry steppe
if the rank of that matrix is 1
magic light
uh
I have no idea how to solve that unfortunately
and also it being equal to 0 isn't that going to be the kernel?
half storm
Thinking of a way to word this.
That is how you check of linear independence by definition
Right?
half storm
A set of vectors is linearly independent if and only if the scalars of the linear combination are all zero.
magic light
is that the idea here?
half storm
There's four of them right
yea
Then you solve the augmented matrix.
By using the row operations that you know about for solving matrices.
magic light
3a - 2b = 0
-a +5b + c = 0
4a + 1b - c = 0
3b = 0
essentially like the matrix
3 -2 0
-1 5 1
4 1 -1
0 3 0
?
magic light
ok I can already tell probably 2 of them are going to get cancled
magic light
nvm I'll just solve it first
half storm
If it comes out that a = b = c = 0. then they are linearly independent and you're done.
dapper lance
how can i find the solution set for x+2y+3z=4
magic light
well b = 0
3a - 0 = 0 => a =0
4a + b - c = 0 => c=0
magic light
but I can cancled out some of these
I got it from
3a - 2b = 0
-a +5b + c = 0
4a + 1b - c = 0
3b = 0 => b=0
then the first equation 3a = 0
then either one of the remaining 2nd/3rd
wintry steppe
can someone help me with this question
I did a but I came to the conclusion that both lines don't intersect
so how do I do b and find an equation that's perpendicular for both?
half storm
It doesn't matter how you do it. You can either do it the first way by just noticing that b has to equal zero because 3b = 0
or you can notice it from cancelling the equation
either is fine
@dapper lance the solution set sort of describes it self via that equation. Are for the vector equation of the plane that equation describes?
@dapper lance if so hop in a questions channel.
I'll help you there.
just @ me
@magic light but do you see how we can now conclude that the matrices are linearly independent?
@wintry steppe put htis in a questions channel and I'll help you out
wintry steppe
which one?
magic light
I see that but
half storm
You said the second one right?
magic light
I don't understand how to eliminate if they weren't
dim venture
(I've got a question in Zeta if someone can hop in)
magic light
like say we got a=1 or whatever
if I try to solve by elimination:
R2: 4R2 + R3, it becomes 0 21 3
then R2: R2 - 7R4 it becomes 0 0 3
R3: 4*R1 - 3R3 it becomes 0 -23 2
so we have
3 -2 0
0 0 3
0 -23 2
0 3 0
working a little we'll R3 to be 0 0 0
first eliminating it with R4 then R2
we would get
3 -2 0
0 3 0
0 0 3
so I did eliminate one of them
unless I did a mistake somewhere
half storm
I'm not sure what you're asking exactly but you take the matrix that you're given and get it in the row reduced echelon form.
magic light
we found out that the basis we found is independent
half storm
If you're asking, if they weren't linearly independent, how do I get a linearly independent basis out of that set?
magic light
because a=0=b=c
right
but what do I do if they're dependent ?
yes, sort of
I mean I should know which one to reduce
in regular vectors you just find the columns that don't reduce
actually now that I think about it perhaps I should've used columns instead?
that is
3 -1 4 0
2 5 1 3
0 1 -1 0
half storm
To get a basis out of that is a little bit more spicy and i'm going to use more linear algebraic concepts to explain it.
magic light
wait
before you do
so we've got T(1, 0, 0)... the regular basis
to find the image
but because these are matrices, we're doing it differently I suppose
I'm a bit confused on how to find the kernel either
I'm really losing focus on whats going on lol
half storm
Oh snap you're right
Wait no no no
This is right
That is what you do if you want to use a basis from the space your mapping from to to get a basis for the set you're mapping to.
There is a much easier way of doing this. Would you like me to explain it to you or would you like to keep going along with this?
magic light
easier way of solving this?
I don't mind it's just I'm not sure whats going on anymore
im so out of focus on this
I know the definition of linear dependency gives a solution a1v1 + a2v2 ... = 0 iff a1=a2...=0
but this isn't usually how I solve these things
usually what happens is that we get vectors, for example T(1, 0, 0) = (2, 3, 4) or whatever
then I take these and I put em in a matrix and eliminate them
then I take the columns of whats left
I think thats how I do it
not sure anymore t_t
ok
half storm
O.k. so you've gotta brodaen you're perspective about what it means for things to be linearly independent and linearly dependnent.
The idea of linear indepdence/ dependence is not restrict to just "vectors" in the way that you think of them i.e. them being points.
The idea of linear independence and dependence is much more general than that.
magic light
yeah I know
I didn't even really think about the vector stuff before you mentioned a[matrix] + b[matrix]... = 0 matrix
but I don't understand how practically I take matrices and show they're independent
half storm
When you show that a set is linearly independent or dependent it doesn't matter whether or not they are points or not.
You do it same the way you show that it's true when the vectors are points.
Literally the exact same way.
magic light
can I just treat it like a 4 variable vector?
half storm
the only difference is that we're dealing with matrices instead of points i.e. fact points are matrices but with only one row / one column.
magic light
like instead of
a b
c d
it'll be (a b c d)
half storm
Kind of yea.
magic light
so
half storm
It's a bit more subtle than that
magic light
T(1, 0, 0) =
3 -1
4 0
T(0, 1, 0 ) =
-2 5
1 3
T(0, 0, 1) =
0 1
-1 0
can I just do
representative matrix:
3 -1 4 0
-2 5 1 3
0 1 -1 0
?
half storm
No.
That doesn't work
But you can represent them as this
oh hold on
yea you're right
sorry
dim venture
(question-Zeta if someone can help)
magic light
Is that the Transformation matrix?
magic light
how do I know
so I get this confused
how do I know if to put the vectors in columns or rows
magic light
OK
so now that if I solve this and eliminate, I don't get any 0's
this means that they're independent
half storm
You mean that you don't get a row of zeroes?
magic light
yeah
half storm
Not necessarily.
magic light
but if they're dependent they would cancel out
half storm
I'm trying to parse what you're trying to tell me because it's not very precise. There's certain vocabulary that's used to describe these thigns which is why I'm getting a little bit confused.
If the rows are dependent then you get a row of zeroes yea.
But that doesn't necessarily mean that the columns are.
half storm
There's a much more straight forward way of oding this.
half storm
Cool
half storm
The image are the set of matrices that are equal to that.
Right? All matrices have to look like that
if you want a basis for the image then you need to rearrange the terms of the matrix.
Here's what I mean
$\begin{bmatrix}3x - 2y & z + 5y - x \ 4x + y - z & 3y \end{bmatrix} = x \begin{bmatrix} 3 & - 1 \ 4 & 0\ \end{bmatrix} + y \begin{bmatrix} - 2 & 5 \ 1 & 3 \end{bmatrix} + z \begin{bmatrix} 0 & 1 \ -1 & 0 \ \end{bmatrix} $
Do you see how I got there?
half storm
I didn't do that.
You get these matrices by doing that though; that is by coincidence.
It doesn't always happen to be like that
All i did is used what it meant for matrix to be in the image of T
and then rearranged the equation accordingly into a linear combination of matrices
stoic pythonBOT
TheDon:
half storm
It has to do with the TeX bot
magic light
anyways yeah I guess I see how you got to that
half storm
Alright, it can be shown that this will always give you a basis for the image.
half storm
Those matrices
half storm
The set of those matrices
magic light
Im(T) = ?
half storm
the span of those matrices
magic light
how do I write that
half storm
so you're write it as $$Im(T) = span \left { \begin{bmatrix} 3 & - 1 \ 4 & 0\ \end{bmatrix}, \begin{bmatrix} - 2 & 5 \ 1 & 3 \end{bmatrix}, \begin{bmatrix} 0 & 1 \ -1 & 0 \ \end{bmatrix} \right }$$
hidden ember
If we have two systems like Ax = c and Ax = d and they both are consistent, then does that mean Ax = c+d is also consistent?
stoic pythonBOT
TheDon:
magic light
but what about the kernel?
also
ok
so I don't know if these are linearly dependent though
right
nvm
I'm getting too tired and confused lmao
half storm
oh i understand lol
magic light
how do I find the kernel here, solve for 0?
half storm
@hidden ember yes.
O.k. so you have to use what it means for a vector to be an element of the kernel right
magic light
T(v) = 0
half storm
That means that $T\left( \begin{bmatrix} x \ y \ z \ \end{bmatrix} \right) = \begin{bmatrix} 0 & 0 \ 0 & 0 \ \end{bmatrix}$
stoic pythonBOT
TheDon:
half storm
Yup you've got it.
hidden ember
@half storm
half storm
@hidden ember please post in questions channel.
half storm
Alright so that means $$\begin{bmatrix}3x - 2y & z + 5y - x \ 4x + y - z & 3y \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \ \end{bmatrix}$
magic light
I need to find the KerT = something
stoic pythonBOT
TheDon:
Compile Error! Click the 
reaction for details. (You may edit your message)
half storm
You solve the corresponding system of equations right? You put it in an augmented matrix and solve for x y and z
magic light
I don't that is what it means though
half storm
so all vectors that have x = some formula, y = some formula, x = some formula are in the kernel.
magic light
I need to multiply this by a vector
magic light
ok
half storm
So then you get a system of equations where all of them are equal to zzero.
half storm
Yup
magic light
and so on
half storm
Yup
half storm
But once you solve for x y and z, that's it.
That's what the kernel has to be, the set of all vectors that look like x = some formula, y = some formula , z = some formula .
fossil rampart
I would really appreciate it if someone can give me like a tip/hint where to begin with a question
gusty hollow
@fossil rampart What question?
fossil rampart
Starts as the yellow part " Find a basis ..."
The main thing that's confusing for me is the W = span{v1 , ... } not sure what I should do with this information
half ice
W is a vector space, and they define it with the span of 4 vectors
The orthogonal complement is another vector space orthogonal to W @fossil rampart
half forge
can someone explain why this isnt Linear indepence
after solving by rrf i ended up with
fossil rampart
@half ice so to solve the problems do I need to find the basis for the Null space ?
half forge
@dusky epoch
half forge
wym?
stuck stratus
@half forge Because there's free variables. That means it's linearly dependent
wintry steppe
which one is true and false?
gray dust
what'd you try
gray dust
why do you think
wintry steppe
lead me in the right direction
for b its just 1/2 of 2b
whick is b
so makes sense to me
but im not sure
and for i its consistent
gray dust
show me all the steps for showing ii is true
gray dust
why b+b?
wintry steppe
vector addition
gray dust
show me ALL steps from M(u+v) to b+b
wintry steppe
bro im out of time soon
gray dust
test?
gray dust
this affects your grade?
wintry steppe
not much
gray dust
it's a timed quiz. asking for help on it is rule breaking
wintry steppe
ok
assignment
but timed
so false true?
its over now
i just want to know
and explanation
gray dust
you're asking for help on a timed assessment. there's a rule against it
marble lance
Send a picture of the screen that says it's done?
gray dust
for future reference never ask for help on any quiz/test
half forge
can we take the determinat of the matrix?
gray dust
not fully, you must look through notes to find what you need
look up defn of consistent system & properties of matrix multiplication
gray dust
no answer. look up this stuff then decide
gray dust
you looked for 5secs. take a good look through
the algebra you did for ii is ok but you didn't justify it
wintry steppe
@gray dust i looked at the rules and i think ii is true
solely based on the fact that half of 2b is b
and that ur allowed to move them around like that
and i think i is true too cus if its consistent thats one of the properties for it to be in the column
gray dust
you didn't justify M(u+v)=2b
gray dust
why Mu+Mv
gray dust
cus u multiply it in
wdym
gray dust
that doesn't apply
gray dust
c is a scalar
gray dust
show me properties of matrix multiplication
gray dust
you can't articulate your reasoning
wintry steppe
you can only do it with other matrix
so for that reason its false
the distributive property
correct?
gray dust
you can view a vector as a nx1 or 1xn matrix
wintry steppe
yes
a 1 column matrix
so then the distributive property would apply
??
im trying my best here
gray dust
yes
gray dust
i want to see all the steps
wintry steppe
ill articulate it in words
gray dust
start at M(u+v), end at 2b
wintry steppe
Mu = b Mv = b
M( U + V) = MU + MV = Mu + Mv = b + b = 2b
gray dust
watch the caps
wintry steppe
i cant just turn it back to a vector?
basically
Mu = MU
is that correct?
or can u not do that
gray dust
why capitalize?
gray dust
it's also bad bc you introduce U,V out of nowhere. be consistent with variable names
better
gray dust
justify each step on the side
wintry steppe
okay
and a vector is the same as a 1 column matrix
so we can use this
and we know when they are multiplied they are equal to vector b
which is also a matrix
so b + b
ah heres where i have issue
gray dust
in each step write the algebra then the justification
gray dust
you're going too fast
wintry steppe
its 2 steps
wintry steppe
for 2 reasons: we can treat vectors u and v as 1 column matricies, and second reason we can apply matrix distribution as they are all matricies so we can apply A(B +C) = AB + AC
so then
Mu + Mv is the result
that's my train of thought
gray dust
ok. Mu+Mv=b+b, why?
wintry steppe
because simply Mu = b from what was declared they are the same
and Mv = b
so i can just use b for Mu and b for Mv
gray dust
yes sub in
wintry steppe
essentially I am doing matrix addition on b and b
which is the same as multiplying b by the scalar 2
gray dust
b+b=2b. you don't need a reason for that. so M(u+v)=2b
wintry steppe
so its true my friend
gray dust
see the process we just went though, that's ideal
stuck stratus
does anyone know how to do the proof I sent above 😶
wary ravine
Hello guys can I get some help with this
I solved a similar one where I had only an x and Y but now I have like 3 variables what do i do
rightmove,
zoopla and
purplebricks. The housing.txt file contains the data of housing prices in the area:
:the first column is the size of the house (in square feet)
:the second column is the number of bedrooms
:the third column is the price of the house (in GBP)
You would like to use this data to help you determine a good price for your house using Linear Regression learning algorithm.```didn't get what I'm supposed to understand from that but I found a tutorial on youtube thanks though
wintry steppe
torpid horizon
hi
is there anyone here who can tutor me in linear algebra
i am willing to pay for a tutor
im really struggling in this class
half ice
We don't endorse cash transactions. You can feel free to ask any questions you want on the server though
Best source you can probably start with in linear algebra is 3b1b's linear algebra series on YouTube. It's not too long, watch through it
dim venture
For 3a, I already have a a mapping that satisfies the requirements and the question is asking me to show that it is linear and unique, right?
dim venture
could you help me understand part b
(I actually have the solution, but I don't understand it and want to give it another attempt)
half storm
Right so you need to show that the set of linear maps from V into W is isomorphic to to the set of m x n matrices whose entires are taken from the reals.
That's basically how you're going to end up showing that the rank of the linear map T is mn.
How do you show tthat map T is isomorphic, ideally you have to show that it's surjective and injective.
Actually, since the only thing that it wants you to show is that the matrix is surjective, you need to show only that. That it's a surjection. Though that map is indeed an isomoprhism - it is also one to one.
What you need to do is start with an arbitary matrix $A \in M_{m,n} ( \mathbb{R})$ and show that there exists a linear map say U $\in L( \mathbb{V}, \mathbb{W} ) $ s.t. $T(U) = A$.
stoic pythonBOT
TheDon:
dim venture
okay, just read this. I will give it a try
Do u mind checking if this is sufficient for 3a
we know U exists because each element in the arbitrary matrix A can be represented by a linear combination of the vectors from the basis of W (since L maps from V to W and the entries are from the reals).
now we must show that rank(T) = dim T(U) = m*n
pearl elm
Some tips for setting this one up?
void relic
write out what 12,25,26, and 37 are equal to in terms of the initial populations
void relic
0.4 times the chamber itself, and 1/(number of rooms other chamber is neighbor to) times other neighboring chambers
pearl elm
oh...
half forge
can someone explain too me why it doesnt span
wintry steppe
geometrically, the two vectors (-1, 3) and (4, -12) both lie on the same line through the origin, so they cant possibly span R^2
flat sedge
It spans the sub space
half forge
did she make a mistake?
wintry steppe
no
half forge
i thought it would spanned since i followed this similar example
flat sedge
The span is a subset of V but doesn’t necessarily span V
half forge
wintry steppe
well if you did that and you got that they span R^2, you made an error
flat sedge
You would need like an additional vector whose not scaled by that vector and independent to span R2
Usually you’d want to check for linear independence first
then if it is, you can gain some intuition for the span set.
If not and it’s linearly dependent, then you’re done
half forge
oh so then how do i know that it doesnt span
dusky epoch
dim(span(S)) = 1.
flat sedge
Well, for this example, look at ℝ². Its a plane. If we have those two vectors you have one is a vector (-1, 3) and the other is a scaled vector from (-1,3). They’re on the same line like ttera said
dusky epoch
span(S) = span{(-1,3)}
(4,-12) is just -4 * (-1,3), so it does not add anything new to the span
flat sedge
To span, you would need a vector that’s non scaled from (-1, 3) cuz then, you could take a whole bunch of linear combinations and generate that plane
half forge
hmm okay, im confused though
whats wrong with example 3 for spanning? is that another method they did?
wintry steppe
ah, ☐^3
half forge
can i follow the steps for my problem?
dusky epoch
the system looks consistent at a glance but that cannot be derived just from observing that there are as many unknowns as equations.
also is this like,,, chegg or sth
dusky epoch
i would recommend putting those vectors into a matrix and then taking its rref
i would also recommend not going on chegg
half forge
so the way they did it, it wasnt specific enough?
also, for spanning can we find the determinant?
wintry steppe
the only time taking the determinant would be conclusive is if you got something non-zero, in which case there would be one and only one solution
if the determinant were zero, you could have either zero or infinitely many solutions
assuming you take the entries of the vectors and set up a matrix / system of equations / whatever like they did in the previous picture
versed hearth
we define the dimension of a polyhedron $P \subseteq R^n$ as the dimension of the smallest dimensional affine subspace containing $P$:
$dim(P) := \min{k \in N: \exists A \in R^{nxn}, rank(A) = n-k \land Ax = Ay \forall x,y \in P }$
Why can we define an affine subspace like $Ax = Ay$? So we have one single $A$ such that we can transform each point onto each other?
stoic pythonBOT
lyinch:
errant mist
Was wondering if it is true that if two matrices A,B will commute (AB=BA) if they have the same eigenbasis?
earnest vessel
Scalar matrices commute with everything and certainly not every matrix has the canonical basis as an eigenbasis
So the answer is no
errant mist
What about if they had the same canonical basis as an eigenbasis for the two matrices? Would they then commute?
keen patrol
So the column rank is always the same as the row rank?
native rampart
So the column rank is always the same as the row rank?
@keen patrol yes
keen patrol
Is there a good explanation for WHY?
native rampart
You can show rank(A)=rank(A^t)(Defining rank as dim of range A)
And that's what we want
earnest vessel
If both matrices have the same eigenbasis, then you can write them in that basis and they would be diagonal, so they commute @errant mist
native rampart
Also,show rank(A)= columnrank(corresponding matrix)
errant mist
@earnest vessel thanks, I understood.
versed hearth
<@&286206848099549185> any comment on my question from earlier?
earnest vessel
You can always write a linear subspace as $\ker A = {x \in \mathbb{R}^n\ |\ Ax = 0}$ for some linear map $A$. Affine subspaces are the same but instead of $0$ you write an element $y$ of the affine space: $ {x \in \mathbb{R}^n\ |\ Ax = Ay}$. So in some sense $A$ gives you the direction and $y$ gives you a point to start drawing.
@versed hearth
native rampart
How is that related to $ \exists A|Ax=Ay$
stoic pythonBOT
DrunkenDrake:
stoic pythonBOT
edelopo:
versed hearth
but in my case $x,y \in P$ and in yours $x \in \mathbb{R}^n$ and $y \in P$ ?
stoic pythonBOT
lyinch:
earnest vessel
Because in your case what you want is that the affine space contains all of the points in the set P
You could write it as $P \subset {v \in \mathbb{R}^n\ |\ Av= Ay}$, where $y$ is some point in $P$
stoic pythonBOT
edelopo:
versed hearth
that means that $\forall v \exists y: Av = Ay$? If I understand my definition it says that :$\forall x, \forall y: Ax = Ay$
stoic pythonBOT
lyinch:
versed hearth
I think that's the confusing part for me
earnest vessel
The first one is a tautology, I don't understand what you mean
versed hearth
okay I need to think a bit about what you wrote
earnest vessel
Your condition is that $A$ satisfies that $Ax = Ay \forall x, y \in P$
stoic pythonBOT
edelopo:
versed hearth
yes exactly
earnest vessel
That is equivalent to saying that for all $x \in P$ you have $Ax = Ay$ for some fixed $y \in P$
stoic pythonBOT
edelopo:
earnest vessel
Because equality is transitive
And that is precisely the definition of $P$ being contained in the set of all $v \in R^n$ such that $Av = Ay$
stoic pythonBOT
edelopo:
earnest vessel
So the final step to understand this is that "the set of all $v \in R^n$ such that $Av = Ay$" is the affine space with direction $A$ passing through the point $y \in P$, and we are requiring that P be contained in this
stoic pythonBOT
edelopo:
dry pulsar
If a matrix 5x5 has 1,-3 and 15 as a eigenvalues, then a) A is diagonalizable b) B is not diagonalizable c) we cant say anything
gray dust
recall defn of diagonalizable
dry pulsar
If we can find a Matrix P invertible consisting of eigenvectors
gray dust
is it guaranteed by what's given?
dry pulsar
Mm i think no cause we dont know how many eigenvectors has associated every eigenvalue
gray dust
how many lin indep eigenvectors do you need to build P?
dry pulsar
5
gray dust
how many lin indep eigenvectors are you guaranteed?
dry pulsar
3
gray dust
diagonalizable guaranteed?
dry pulsar
Noo
gray dust
defective guaranteed? @dry pulsar
dry pulsar
Noo
gray dust
why no for both?
half forge
question, can you check spanning by checking the number of vectors?
dry pulsar
we dont know the multiplicity of eigenvalues
gray dust
so we may have 5 LI eigenvectors but we don't know for sure. what can you say?
dry pulsar
Nothing
gray dust
that's it @dry pulsar
past galleon
Just make sure theyre orthogonal
half forge
wym
past galleon
Dot product for each pair should be zero
keen patrol
So why do they call singular matrices singular? Just because they have no inverse?
soft burrow
not entirely sure but I'd guess it comes from "singularity"; a matrix $M$ is non-invertible iff $\ker M \neq {0}$, i.e. if $Mv=0$ for some non-zero $v$
stoic pythonBOT
bastian.uwu:
vagrant mica
Any recommendations for videos to watch as a review for finals?
strange dove
shadow mist
Im terrible and proofs, trying to learn, anyone have any tips to lead me in the right direction?
gray dust
definition of set equality
shadow mist
sorry?
gray dust
what
gray dust
did you ever prove 2 sets are equal
shadow mist
never before 😦
whole solstice
basically you want to prove that the set of linear combinations of u and v are equal to the other set
right @gray dust?
gray dust
using defn of set equality
whole solstice
what's the definition
i just think of set equality as "if they both contain the same elements they are equal"
gray dust
that's the intuition. there's a precise defn
warm kite
whole solstice
oh i thought you were calling me dumb 
this channel is in use sorry
@gray dust what's the precise definition
gray dust
@whole solstice did you ever prove 2 sets are equal?
keen patrol
I'm look at https://www.math.ucla.edu/~yanovsky/Teaching/Math151B/handouts/GramSchmidt.pdf on page 2 and don't understand how they got (0,1,1)-1/sqrt(2)(1/sqrt(2),1/sqrt(2),0)-1/sqrt(6)(1/sqrt(6),-1/sqrt(6),2/sqrt(6)) to be (1/sqrt(3), 1/sqrt(3), 1/sqrt(3))
I got (-2/3,2/3,2/3) instead. (Sorry for the formatting. Don't know latex yet)
I got it reduced to (0,1,1)-(1/2,1/2,0)-(1/6,-1/6,1/3) which is (-2/3,2/3,2/3)
keen patrol
Yeah I think it's a typo. They mixed up e3 and u3.
Spent the whole half hour trying to figure out why my answer was wrong and it just turned out to be a dang typo :/
limber sierra
this is not linear algebra
try #prealg-and-algebra , #precalculus , or one of the ten questions channels
surreal thistle
bruh i've thought of this for a while and idk the answers to either
i had a couple answers but their domain didnt include ARN
void relic
arn?
surreal thistle
all real number
keen patrol
A power of the identity matrix is just the identity matrix right? Like I^n=I ?
void relic
yea gooddeath
hmm what were your answers? let's see if we can extend them to all reals
for a) we can make g(x) have a small range so that f(g(x)) doesn't double up on the same outputs f(x) does
surreal thistle
void relic
nvm lol
surreal thistle
ur first example doesnt work
if u square g(x) u still dont get an injection
because g(x) has multiple x of the same y
surreal thistle
does anybody know about any R->R injective functions that have infinite domain
but is not surjective
gray dust
arctan @surreal thistle
surreal thistle
what channel should i ueese
gray dust
you see my answer?
gray dust
what is it
wintry steppe
test
rustic terrace
PleSe unmute me I did not do anything
Can u unmute me
Unmute me
I have a question
I’m trying to learn
quiet crane
can anyone explain to me how to do this? im pretty sure that these vector space axioms has something to do with it, but my book doesnt show me examples of what to do to prove them, but maybe if i figure out how to start A1. maybe ill know how to do the rest, so what my idea is that you pick one that looks similar to it and prove it?
i think that A2. and A5. works for the top and bottom one in that order
wait no that doesnt make sense for A2 the complex number is multiplied not added
maybe you start with A1 and end with A8 because it looks like its solving something i think?
i was thinking about doing something like this, showing how x and y can be backwards and equal to that equation but it doesnt look like it works ```
A1. x + y = y + x
x = (a + bi)
y = (c + di)
x + y = (a + c) + (b + d)i
= (c + a) + (d + b)i
= (c + a) + (di + bi)
= y + x```
keen patrol
So for the identity matrix like EVERY vector is an eigenvector right? With eigenvalue 1 because I * v = 1 * v
That's pretty nuts. They're ALL eigenvectors right?
soft burrow
of course they are, here 1 is the only eigenvalue and its eigenspace is the whole space
keen patrol
Neato
tepid violet
Hi, i use formula for zooming y = exp(-x * x) / radius Are there any other options to do that?
thick egret
@quiet crane Are you trying to prove that C is a vector space using the vector space axioms? Ur A1 proof is ok
mild hound
how can I find the vector using it's magnitude and direction
wintry steppe
what's the vector in? R^2?
wintry steppe
Any good video lectures for linear algebra ?
native rampart
You might like strang
fluid crescent
Me in early 8th grade seeing this with algebra test today 😐😐
old flame
Here is my proof attempt. Suppose $T \in L(V)$ and $U_1,...,U_m$ are subspaces of V invariant under T. Then for $u_1 \in U_1,...,u_m \in U_m$, $Tu_1 \in U_1,...,Tu_m \in U_m$. Define $U_1+U_2+ \cdot \cdot \cdot + U_m= {u_1+\cdot \cdot \cdot + u_m \mid u_1 \in U_1,...,u_m \in U_m }$. Since $U_1+\cdot\cdot\cdot+U_m$ is the smallest subspace containing $U_1,...,U_m$, pick $r \in U_1+\cdot\cdot\cdot+U_m$, so we could express $r$ as a linear combination, $r=a_1u_1+...+a_mu_m$, where $a_1,...,a_m \in F$. Then, $T(r)=a_1T(u_1)+...+a_mT(u_m)$. Since each $Tu_i \in U_i$, for $i={1,...,m}$, then $a_1Tu_1+...+a_mTu_m \in U_1+\cdot\cdot\cdot+U_m$. Hence the subspace $U_1+\cdot\cdot\cdot+U_m$ is invariant under T.
stoic pythonBOT
Otoro:
old flame
<@&286206848099549185>
native rampart
It's fine
old flame
thank you
fossil rampart
Good Day , I need help to find the orthogonal complement W prep of the vector space W
W= span{ v1 , v2 ,... ,v4} , vi element of R^6
What I think I should do :
Let V be the matrix defined by the set of vectors spanning W
V* < x1 , ... ,x6 > = 0 to find the rank of V
Then produce a basis by using Gram-Schmidt so that the resulting vectors are orthogonal to the vectors spanning W
Is my understanding correct , if not please help 😅
earnest vessel
Gram-Schmidt doesn't give you a basis out of nothing, you need a basis to start with. What you have to do to find W perp is to compute those vectors perpendicular to W (obviously). To do this you set the equations <v_i, u> = 0 for i=1,2,3,4 and solve for u. That is, you have a system of 4 equations in 6 variables and you need to find the space of solutions (a basis for it).
fossil rampart
I hope I'm following you correctly ? A
earnest vessel
Yeah, you have to find the ker of that matrix
fossil rampart
What I understood from what you were saying , I need a basis for W using the vectors spanning W , once I have that basis , Then I can use Gram Schmidt ?
native rampart
Yes
fossil rampart
Thank you all so much 🙌🙌
earnest vessel
You don't need Gram-Schmidt unless you want the result to be an orthogonal basis. You can just give the basis of W perp and have it be non orthogonal.
fossil rampart
Oh okay , if the question was give an orthogonal basis then definitely use Gram Schmidt
earnest vessel
Yeah
fossil rampart
Much Appreciated 🙌
proud otter
I'm trying to figure out when to stop iterating my program, QR Factorizing my square matrix and flipping them, and I saw on the wikipedia that Gershgorin circle theorem is used to determine whether or not my A_k value is a triangular matrix.
Why is it used?
wintry steppe
anyone know how to do number 2?
old flame
Hello, again heres my attempt. Suppose $S,T \in L(V)$ such that $ST=TS$. So $S,T$ are both invertible, this also means that both of them are injective and surjective. Because $T$ is injective, $T(0)=0=\lambda \cdot 0$, for $\lambda \in F$, so $null (T-\lambda I) = {0}$. Then, $S(0)= \lambda \cdot 0 \in null(T-\lambda I)$, because S is injective. Therefore, $null (T-\lambda I)$ is invariant under S, for every $\lambda \in F$.
stoic pythonBOT
Otoro:
native rampart
How do you know S and T arw invertible?
old flame
ah Im very sorry, I should use that T is a scalar multiple of the identity instead, as it is proven in another exercise
okay let me retry
native rampart
That is only if it commutes for all S
old flame
what do you mean ?
native rampart
You fix a T,now if You choose any S such that ST=TS,then T has to be a scalar multiple of Identity
Here,T and S are both fixed
old flame
oh so that isn't permitted as well then I guess
native rampart
Hint:||(Using a in place of lambda) Let v be such that. (T-a I)v=0
Then,(T-aI)(Sv)=TS(v)-aS(v)
=ST(v)-aS(v)=S(T-aI)v=
S((T-aI)v)=S(0)=0 ,So the space is invariant under S ||
old flame
may I ask (T-aI) need not to be an operator right ?
its just a map that works for the calculation ?
old flame
so as a sense, we defined another operator ?
native rampart
Yes
old flame
I see, so the usage of defining something applies here again....
wasnt even expecting this
native rampart
I mean the question requires you to define that
Because without T-aI, the null space of T-aI wouldn't make sense
(Also,I used a instead of lambda because I wanted to hide it)
old flame
yeah I was thinking about that, but I may have confined myself in thinking T has eigenvectors
so in this case, v themselves are eigenvectors already ?
old flame
sigh, I guess I need to be more observant on how to define things suitable for the question
thank you
lyric topaz
can someone help with 11b
wintry steppe
<@&268886789983436800> I believe this is not allowed?
lyric topaz
Oh am i not supposed to link
wintry steppe
thank you LINEAR_ALGEBRA_GUY
limber sierra
are you familiar with lagrange interpolation?
tranquil hazel
Nope
Someone at my school's math centre told me lagrange interpolation too, but I couldn't understand what they meant
rotund jetty
given a linear system of inequalities, what is an equality subsystem?
edgy folio
can somone show me the steps for this one?
half ice
It really is just a complete row reduction
Keep track of where you divide, and note you can never divide by 0
If you can't get to rref, there's not exactly one solution
OR if you know determinants, that will work too
half ice
That's not rref lol
Almost there though
In order to finish, you will need to divide by some expression of k. That is of course not possible if you're dividing by 0
brave scaffold
hey guys, if T and U are linear transformations in Rn such that T(U(x)) = x for x in Rn, does that mean that U(T(x)) = x?
will there be a case where they wont be?
native rampart
hey guys, if T and U are linear transformations in Rn such that T(U(x)) = x for x in Rn, does that mean that U(T(x)) = x?
@brave scaffold yes if UT=I,TU=I
half storm
hmmmmm
native rampart
half storm
Yea, that's it.
yea if it's from R^n into R^n
I see that now.
If they were mappings between vector spaces of different Hamel dimension, then I was gonna say "it depends"
brave scaffold
right cos my thinking was there might be matrices that have a left inverse but no right inverse
so if its in the same dimensions i guess it makes sense that that will be true
random shoal
can someone help me understand why in the very first step they would want to change r1c1 to a 0?
instead of leaving it as 1
like is it arbitrary or am I missing something?
dusky epoch
they got the first col to a state where there's a single 1 there and everything else is zero
it doesn't matter exactly where the 1 is cause it can always be relocated to the corner if one wishes
dusky epoch
two questions in one channel 
random shoal
Thanks @dusky epoch I understand that part I just find it a little weird that we were taught to go in a specific order. Maybe it was moreso intended to show that it doesn't matter...
dusky epoch
if you're doing it by hand it doesn't matter the exact sequence of row operations you take
random shoal
Thanks. I'll keep that in mind.
dusky epoch
which part? a or b?
there's a very elegant way to reframe part a which will make the task way less daunting
@inner oxide
inner oxide
a
dusky epoch
ok
instead of proving that b has either 0, 1 or infinitely many preimages
prove that if b has at least 2 distinct preimages then it actually has infinitely many
or to put it another way: suppose there exist $\bd{x}_1, \bd{x}_2 \in \bR^n$ such that $A\bd{x}_1 = \bd{b}$ and $A\bd{x}_2 = \bd{b}$ but $\bd{x}_1 \neq \bd{x}_2$. can you show that the equation $A\bd{x} = \bd{b}$ actually has infinitely many solutions?
stoic pythonBOT
Ann:
dusky epoch
👻
pallid rampart
well R means an arbitrary ring obviously
wintry steppe
ah yes $\bR$ is an arbitrary ring 
stoic pythonBOT
TTerra:
flat sedge
Lmao
dim marlin
i need help with
A7 = 7 - 5n
dusky epoch
i can say with like 95% certainty that you have the wrong channel, and with 100% certainty that you fucked up the notation so that it's impossible to understand what you meant.
dim marlin
its sequences
earnest vessel
Sequences of what?
dawn remnant
their question is now in #prealg-and-algebra
pure dragon
https://cdn.discordapp.com/attachments/692452672179273758/759372828989456444/IMG_20200926_190615.jpg
what is this called?
What do you guys call on this graph?
dawn remnant
if you mean what that function looks like - a piecewise linear one. https://en.wikipedia.org/wiki/Piecewise_linear_function
wintry steppe
Hey does anyone know a linear algebra book that isn't about computations and more catered towards intuition? Like how 3blue1brown explains in his videos
native rampart
So, You want an abstract LA book?
pure dragon
thanks @dawn remnant
native rampart
Also, You can just libgen
quaint marlin
native rampart
Is it just A?
stoic pythonBOT
Whoever:
dawn remnant
$A(A^{k-1} - \frac{10^k-1}{9}) = 0$
stoic pythonBOT
ConfusedReptile:
dawn remnant
if this is a linear algebra problem, the solution for A is actually not obvious from here. A matrix can result in an identity matrix when put to a power without itself being a diagonal matrix, I think.
blissful vault
a yz-plane is when x = 0 iirc, which explains why the place where x intersects is zero
but i don't understand how to get the other two coordinates
i used t=29/8 and just plugged it back into the line equation
blissful vault
nvm i found the answer
yz plane means x=0
so when you do 0=(equation for x)
you get 0 = 3t+5
then you can find t and the rest is trivial
robust pond
this is a dumb question i think but in a transform the dimensions of the standard matrix dont need to match the dimensions of its input right
it just needs to still be valid for multiplication
dawn remnant
What do you mean?
robust pond
like theres no reason that if the standard matrix of a transform is 3x3 that the input couldnt be a vector
it would just mean the codomain of the transform would be 1x3 vectors
dawn remnant
a 3x3 matrix can transform a 3d vector into a 3d vector when multiplying A v and a 3d covector (row vector) into a 3d covector when multiplying v A
if you write vectors as column ones (that's the norm I believe), you have linear operators be multiplications from the left by some matrix, A(v) = A v.
limber sierra
are you using the $\begin{bmatrix}a&b\c&d\end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$ formula?
stoic pythonBOT
Namington:
limber sierra
if so, you forgot to divide by the determinant
indeed, it should be easy to observe that $\begin{bmatrix}-1&0\0&1\end{bmatrix}$ is NOT an inverse for $E_3$, as when you multiply these together, you do not get an identity matrix.
stoic pythonBOT
Namington:
random shoal
I see. I wasn't finding a new determinant for every elementary matrix. I was just using the one I found from the A matrix given originally.
Thanks @limber sierra
toxic pendant
can you span Pn(R) with less vectors than degree n?
it sounds impossible
if you represented the vectors as a matrix and tried to reduce them you'd have more variables than vectors
oh I realized that I was misreading the question which is why I had that question
thank you for the assistance either way
gritty frigate
Guys, I know how to calculate a determinant and I know that it tells me if a matrix has an inverse
But.. how ?
What is the relation between Laplace theorem and the inverse of a matrix ?
limber sierra
Well, the real relation comes from det(AB) = det(A)det(B)
This tells us that A has an inverse iff Det(A) does
And the only number without an inverse is 0
The more "fundamental" question is, why is det(AB) = det(A)Det(B) true?
There are proofs of this of course, but honestly it does feel like black magic
gritty frigate
I see
You are right
If I think that det(AB) = det(A)det(B)
B = A-1 and det(I) = 1
det(A)det(A-1) = det(A.A-1) = 1
so:
det(A)det(A-1) = 1
det(A-1) is not zero, so det(A) cannot be zero
@limber sierra I cannot believe that humanity found this..
gritty frigate
it was not the best expression hahaha
blissful vault
so what i did was
i made two vectors
AB and AC
then i did a cross product to find the normal for the plane
then i used the equation a(x-x0) + b
(y-y0) + c(z-z0) = 0
not sure why it's giving the wrong asnwer
urban seal
Guys am I proving axiom 4 correctly?
urban seal
Oops I did not mean to write the 0+0i part
half ice
The general element a + bi has a negative, -a - bi. This is because
(a + bi) + (-a - bi) = 0
And the algebra that follows from there does it
urban seal
Oh so I don’t need to show the distribution of the negative sign
half ice
I think distributing a negative sign is not one of the axioms. It's a true fact, but something one needs to prove. Don't use that if you don't have to haha
It's equivalent to
(-1)x + x = 0
limber sierra
i mean if you phrase it as
"Write -(a+bi) for the (hypothetical) inverse of a+bi. We want to show -(a+bi) exists; to do this, we will show that it is equal to (-a) + (-b)i by showing that satisfies this property."
this is fine
this is kind of overcomplicating the exposition a bit
but its fine to do
urban seal
Dang yours sound formal, I was thinking about something like “Adding the negative form of x to positive x will cancel out the numbers. Thus, making the equation equal to 0.”
But I think this should help me out more than my textbook, so 7. And 8. Is where I absolutely need to show distribution right?
half ice
7 and 8 are distribution, yes
limber sierra
oh shit i thought you were working in a group, didnt realize this is #linear-algebra
i mean my thing still works
but for simplicity in that context i'd do as kaynex hinted at
and just say "the inverse of a+bi is (-a) + (-b)i"
and then prove it in the way you did
skipping line 1
urban seal
Thx
thorny kraken
erm... does anyone know how i could go about doing this? I know how to calculate the formula to start creating the matrix, but I'm not sure what T is supposed to be ._.
soft burrow
the transpose
thorny kraken
ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh. so it's not power? ahh im so stupid omg
soft burrow
for instance $\begin{pmatrix}1&2\3&4\end{pmatrix}^T = \begin{pmatrix}1&3\2&4\end{pmatrix}$
stoic pythonBOT
bastian.uwu:
thorny kraken
ah thank you so much
old flame
Here I tried to prove by contrapositive, so if $U \neq V$ and ${0}$, then $U$ is not invariant for some operator. Define $T \in L(V)$, $T(v)=v+1, \forall v \in V$. Take a basis of $(u_1,...,u_m)$ from $U$ and $r \in U$. Now write $r$ as a linear combination, $r=a_1u_1+...+a_mu_m$ for $a_1,...,a_m \in F$. Then $T(r)=a_1Tu_1+...+a_mTu_m=a_1(u_1+1)+...,a_m(u_m+1)=a_1u_1+...+a_mu_m+(a_1+...+a_m)$ Then the first part is in U, but the second is not. Hence $U$ is not invariant to such operator.
stoic pythonBOT
Otoro:
native rampart
What is 1?
old flame
Im sorry, but I don't get what you're asking
native rampart
1 is usually not a vector(Assuming you are talking about the multiplicative identity)
old flame
then maybe I should replace 1 with like a constant vector $c$ ?
stoic pythonBOT
Otoro:
old flame
I wanted to define an operator that shifts the elements
native rampart
Yes
old flame
but the approach is something similar right ?
native rampart
Yes
old flame
ah alright, at least I got this
Could you elaborate on why constants are not a linear map ?, just because $T(0) \neq 0$, just means it is not injective though
stoic pythonBOT
Otoro:
native rampart
Definition of a linear map T is that
T(cx+y)=cT(x)+T(y) for all x,y in vector space and c in the field
This means T(0) will always be 0
old flame
oh so the shifting has to be an input for the map yeah ?
native rampart
If T(0) is not 0, you end up with something called affine spaces
old flame
whats the difference ?
old flame
I havent learnt affine spaces yet, but from what you've said, Im guessing vector space with eigenvectors are affine spaces right ?
native rampart
All vector spaces are also affine spaces.
old flame
so vector spaces are a subset of affine spaces then ?
native rampart
Yes
old flame
ooo okay then thanks
native rampart
(Hint: ||Let's say U is a proper nontrivial subspace of V. Which means there is a vector w in V,but not in U. So, Consider the linear map T mapping everything in U to w. U is clearly not invariant under T|| )
old flame
I thought of something
Let $(u_1,...,u_m)$ be a basis in U. Define T as $T(u_i)=v_{i+1}$ where $v_i's$ are the basis vectors of V. Then pick $r \in U$, write that as a linear combination, so $T(r)=a_1v_2+...+a_mv_{m+1}$ which is a vector in V but not in U. Is this okay ?
stoic pythonBOT
Otoro:
old flame
A basis vector in V ?
native rampart
The basis vectors are not fixed
You could have a basis part of which is basis of U
old flame
What if I defined T(u_i)=v_i, where v_i are just vectors not in U ?
old flame
I was thinking somewhere along that, which is to map vectors from U to outside of U
Ahhhhhh okay thanks
May I ask what is a nontrivial subspace ?
native rampart
Any subspace other than the zero subspace and the vector space itself
old flame
So if I were to rewrite the proof, it would go as follows right ?
Let $(u_1,...,u_m)$ be a basis in U. Define T as $T(ui)=v_i$ where $v_i$ are vectors in $V \setminus U$. Then pick $r \in U$, write that as a linear combination, so $T(r)=a_1v_1+...+a_mv_m$ which is a vector in V but not in U.
native rampart
Yes
stoic pythonBOT
Otoro:
old flame
Thank you :)
One thing, so your hint is to map elements of U to a particular vector in V that's not in U
So instead that's simpler than what I did though
warm niche
When we are testing linear independence of vectors
Why do we test if the linear combination of the vectors only have a trivial solution to the 0 vector?
Why can't we pick any abritrary vector and see if it only has a unique soluion?
Is it just because of simplicity? Because I've seen definition of lienar independence specifically state the 0 vector
dusky epoch
if you pick an arbitrary vector it might end up not in the span of the set you're testing for LI
slim gyro
im guessing y'all get a lot of axler here
i'm not even sure where to start with this, any ideas?
void relic
you just need to show that there's a zero vector and that linear combinations are in the set right?
derivatives being linear makes it all work
slim gyro
yeah the zero vector was easy to show
what is meant by derivatives being linear?
uhh it isnt taken i dont think
just designated for linear algebra
ocean sequoia
it means they are linear i.e the function for calculating derivatives is a subspace
do you know what linearity is?
and by taken i meant that you are asking a question 🙂
dont want to interrupt
do you know what linearity is? @slim gyro
ocean sequoia
hey im trying to calculate a variance - covariance matrix and im getting negatives on the diagonal
i have to be doing something yea
I have positive eigenvalues should it should be positive definite yea?
tall surge
hi i have a question
how to calculate the eigenvalues of a matrix in the following form?
slim gyro
@ocean sequoia linearity is when the graph is like a straight line right?
so you're saying if you think of taking the derivative of a function as a function itself, its graph would be linear?
ocean sequoia
ah thats where you are confused
go re-read what linearity is
this is super important which is why its worth going back through on your own
tall surge
I have positive eigenvalues should it should be positive definite yea?
@ocean sequoia i mean as long as the eigenvalues are positive and it's symmetric it's positive definite
ocean sequoia
yea hm wtf