@empty ibex what are you stuck on?
#linear-algebra
sonic osprey
sonic osprey
If v1 did equal pv2 what would happen?
Think about Av1
And remember that the eigenvalues have to be distinct
empty ibex
would it be that their eigen values would have to be the same in that case and since that isnt possible as said in the question, v1 cannot therefore equal pv2?
sonic osprey
yes
empty ibex
oh we could also prove it with diagonalization of A. The invertible matrix consisting of the eigen vectors would give a determinant of zero and would no longer be invertible, hence v1 cant equal pv2
livid falcon
Can someone help me in this
honest token
Should I take actual values?
no, it seems like they're talking about general non-zero vectors u and v
by "fix" i think they mean they're talking about the same vectors u and v in all parts
waxen jacinth
by "fix" i think they mean they're talking about the same vectors u and v in all...
Sorry, I still dont get it xd
Do you mean to say that they u and v share the same dimension?
In all cases?
honest token
Sorry, I still dont get it xd
well, they can be any (non - zero) vectors in R^n basically
waxen jacinth
Alright, I suppose.
The problem still throws me off though lol
I cant picture what the problem means by If u=OP, we can think of the above as the line through P, parallel to v.
lavish jewel
u = OP is a vector that points to a point p, with tail at the origin
waxen jacinth
This then?
glad acorn
Part c I have no idea how to show
waxen jacinth
Take these and multiply them by the factors and compute I suppose?
waxen jacinth
Take these and multiply them by the factors and compute I suppose?
@glad acorn
lavish jewel
waxen jacinth
Okay, thanks
lavish jewel
u + t v is a fan of vectors that have tail at the origin and head at some point on the line
so, many vectors like the purple one
waxen jacinth
u + t v is a fan of vectors that have tail at the origin and head at some point ...
Yes, I see it now, thanks
glad acorn
How do I show this system is consistent? Does two equivalent rows are enough to imply?
hollow void
Anyone have shortnotes of this
surreal otter
I am not sure if this is the right place by any means, but I was reading a book and found this; I understand that this is finite rotation, but then the author goes on to the infinitesimal version, and I am sure I am missing some sort of derivation for this. Does anyone know where he got the second matrix from? I've found some material online about Lie groups and generators for rotations, but it was not particularly helpful in this case.
nocturne jewel
I am not sure if this is the right place by any means, but I was reading a book ...
the one labelled (3.1.4)?
surreal otter
that one indeed!
nocturne jewel
just small angle approximations
$\sin(x)\approx x \ \cos(x)\approx 1-\frac{x^2}{2}$
stoic pythonBOT
Mosh
surreal otter
Oh snap, I didn't know this was a thing
if you don't mind me asking, where is this derived from?
nocturne jewel
taylor series about x=0 (aka maclaurin series)
surreal otter
OH I remember now
the sin(x) approximation is just inherent while the cos(x) approximation requires a MacLaurin series, is that it?
nocturne jewel
they're both derives from taylor/maclaurin series
$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}=x+O(x^3)\approx x$
stoic pythonBOT
Mosh
surreal otter
most interesting; supplementary question, but is that O the same as for asymptotic notation?
nocturne jewel
yes, it's Big O
for small value approximations, it just means everything else in the series is at least order x^3
surreal otter
correct me if I'm wrong but isn't that then an upper bound for the entire series?
surreal otter
Oh I know but I was asking out of curiosity and completeness ๐
wintry steppe
love me some asymptotics
surreal otter
I learnt about them not too long ago on my own and they really help envision some things, don't they
surreal otter
I think you are, I will check myself to see
I just had no clue the author used a MacLaurin series since he did not mention it
Otherwise, thank you @nocturne jewel ! I'm very grateful for your help, I've been stuck on this for several days and couldn't figure it out
primal schooner
can someone help me with my homework?
hollow void
what is conjugate of matrix, how to find it out
nocturne jewel
what is conjugate of matrix, how to find it out
You mean congujate transpose?
Or just the congujate?
wintry steppe
it's the conjugate, transposed
hollow void
If matrix is lower, upper triangular
Or diagonal
Det = product of it's diagonal
Is it right?
lavish jewel
mhm
hollow void
Eigen value = multiplication of diagonal
Sum of diagonal = sum of eigen value?
dusky epoch
the trace is the sum of the eigenvalues even if your matrix isn't triangular 
hollow void
the trace is the sum of the eigenvalues even if your matrix isn't triangular <:c...
Yes. But i didn't said that
lapis light
I am following steps from: https://www.mathsisfun.com/algebra/eigenvalue.html
However, I'm trying to use the eigenvalue 2 rather than the -1 eigenvalue they used on the page
but unsure how to solve these
Also I'm confused whether the results would be the same
lavish jewel
Yes. But i didn't said that
you did
yes
dude
you said the sum of the diag is the sum of the eigenvalues
the sum of the diagonal elements of a matrix is called the "trace"
and this is true for all matrices
not just triangular and diagonal
this is what ann told you
lavish jewel
yes
lavish jewel
the trace is the sum of the eigenvalues even if your matrix isn't triangular <:c...
see
this is what ann said
hollow void
lapis light
What would I need to do to figure out the answer here
Am I able to get help with this one pls
hollow void
I've figure out new method to find eigen values. Can you check it
sonic osprey
what is this new method?
sonic osprey
i cannot make sense of what you're writing at all
hollow void
Eq 1 | A | = product of eigen value
Eq2 trace = Sum of eigen value
Right?
sonic osprey
sure
hollow void
Ad - cb = L1 + L2. (L = EIGEN VALUE)
a +d = L1 ร L2 right?
sonic osprey
sure
hollow void
Rest is simplification
sonic osprey
although you have those in the opposite order, the first one should be L1 x L2 and the second one should be L1 + L2
Sure I get your point
This is well known for one
lavish jewel
this also only works for 2x2
sonic osprey
And secondly, this only works in 2d
hollow void
So this was already known.
sonic osprey
this was known like 400 years ago
hollow void
hollow void
Dunning kruger still hunting me
sonic osprey
people have been spending their whole lives doing math for thousands of years, its hard to come up with someone that no one has thought about
sonic osprey
No it's fine, being curious and looking for new things is a good attitude to have
hollow void
What about zero or not real
lavish jewel
that can also be
hollow void
Uhhhh i hate my teachers. If these things figured out 500 years ago.
Why still no one my teachers able to know it so well.
lavish jewel
they probably do if ur in uni
lavish jewel
then they knew
hollow void
I've so much frustration.
But i loves maths. But i don't know how to learn this complex maths.
lavish jewel
i'm guessing you didn't study any STEM
this is pretty elementary
for some people, high school level
hollow void
Yes
hollow void
I directly did polytechnic. Instead of stem. Coz studying it was 75k โน and studying in polytechnic was 750โน per month
750 = apx 10$ per sem
lavish jewel
use substitutions, you have 2 equations in 2 variables
hollow void
Y If A is a square matrix then|A^T|=|A|?
lavish jewel
sure
lavish jewel
i like to think of the eigenvalues as the amount a transformation shrinks or stretches in a specific dimension, and the det as the overall change in "volume"
hollow void
determinant of matrix are always real right?
stoic pythonBOT
TTerra
wintry steppe
the determinant of a matrix with real entries is real
the determinant of a matrix with complex entries may be real or complex
pine lion
the set of all real-valued functions defined on the natural numbers
Well that means that it fits the three criterias for a subspace right?
wintry steppe
i'm asking whether you've checked if that's the case or not
that's how you'd start
pine lion
Yeah I don't know how to start that
Like how do I show that the zero vector exists in here for example
wintry steppe
does it?
the zero vector would be the function that sends everything in N to zero, yeah?
so you want to check whether or not that's in your set U
wintry steppe
pretty much
wintry steppe
yeah
pine lion
Hmm
I always have a problem understanding what the set is trying to say that's my issue
But thank you, this does help me understand more
Just need to do more of these and ask more questions
wintry steppe
pine lion
Oh wait. Showing that the zero vector isnt enough to say something isnt a subspace right?
Like showing it is in the space shows its nonempty which is one of the criteria
wintry steppe
it is enough
to be a subspace, every one of the subspace axioms has to hold
if even one fails, it's not gonna be a subspace
pine lion
Oh oh okay
wintry steppe
also yeah if your set is empty it's not gonna be a subspace lol
but that's usually a silly thing to check
limber sierra
empty set discrimination
wintry steppe
cause checking whether or not it contains zero already pretty much does that
limber sierra
so sad
pine lion
Okay cause the way my teacher defined subspaces is that 1) the subset is nonempty (or in other words, the zero vector is in the subspace)
wintry steppe
replace that with just "contains 0" lmao
hollow void
the determinant of a matrix with real entries is real
ohhk
The sum of the products of the elements of any row (or column) of a
determinant with cofactors of the corresponding elements of any other
row (or column) is zero...... what does it means actually
pine lion
So I would interpret that as "its enough to show that a space is nonempty by showing that the zero vector is in there" but that doesn't immediately tell me that a space is empty just because the zero vector isnt...
wintry steppe
idk what a cofactor is so i couldnt tell you
your interpretation is correct
but if you don't contain the zero vector, it doesn't matter if you're empty or not since you're not gonna be a subspace
wintry steppe
it's weird phrasing and you really should just think of it as "contains 0"
instead of some weird empty/nonempty stuff
pine lion
Sounds good :)
So for this one, could I say it's not a subspace because I can find a scalar that makes the output not in the natural numbers?
west saddle
so for row space and column space, the coordinates are not supposed to be fractions, they are supposed to be whole numbers correct?
lavish jewel
wdym?
west saddle
so my matrix is (1,0,0),(-1,-2,0),(2,0,-2),(3,-3,-13)
lavish jewel
mhm, 4x3 matrix
west saddle
the answer in the back of the book for rowspace doesn't use fractions
lavish jewel
what's the field?
west saddle
I'm sorry i dont understand, i haven't heard that term
lavish jewel
it doesn't matter
west saddle
so for example a rowspace vector is (0,2,0,13) instead of (0,1,0,13/2)
ok so both is right
they would just prefer whole numbers
pine lion
I'm having a hard time understanding what this even means
So F(D) I guess is just every real valued function defined on {a,b,c}
But how do I find a linearly independent set of vectors that span this?
winter harbor
take f_1,f_2,f_3 \in F(D) such that f_1(x) = 1, if x=a and 0 otherwise; then f_2(x) = 1, if x=b and 0 otherwise and f_3(x) = 1, if x=c and 0 otherwise
Then, $\forall f \in F(D)$, we have that $f = f(a) f_{1}+ f(b) f_{2} + f(c) f_{3}$.
stoic pythonBOT
MisterSystem
winter harbor
Try to check this for yourself
Also
These functions are linearly independent
And I will leave you to check this also
pine lion
Cool! thanks a lot for the help!
winter harbor
Np
tranquil thistle
Anyone have any advice or a really good video for finding RREF?
I know people usually suggest to go by pivot and remove all non zero values above and below them
tired fossil
Anyone have any advice or a really good video for finding RREF?
Yes
Basically try to keep non fractions throughout your reduction
And keep a pivot 1 in each column if possible.
Then once you get pivot 1s throughout your entire matrix, then begin to eliminate the non zeros above the pivot 1s
So basically find REF first
elder kelp
Maybe im just stopid and this is easy but can anyone help me find the value of h in S=2 ฯ r^2+2 ฯ rh
wild fulcrum
what's this? cylinder surface area?
elder kelp
yea
wild fulcrum
(S-2pi*r^2)/(2pi*r) ? lol
nocturne jewel
Maybe im just stopid and this is easy but can anyone help me find the value of h...
Not Linear Algebra
elder kelp
oh its not? we are doing linear algebra and this was in our hw so I thought it was
Sorry
nocturne jewel
that's just algebraic manipulation so... not linear algebra
glad acorn
Does this count as a reduced row echelon form? If b1b2b3 are unknown constant
lavish jewel
yes
glad acorn
what are "the indices of the pivot column"?
lavish jewel
the columns that contain pivot elements
i think this image explains it fairly well
hollow void
Has anyone seen kota factory
strange delta
can someone explain to me why U + W makes (x,x,y,z)
lavish jewel
the actual letters they used might throw you off
strange delta
i don't get how z randomly appears
is the addition basically the intersection of both subspaces
dusky epoch
no
lavish jewel
no
strange delta
o
dusky epoch
intersection is intersection.
lavish jewel
the addition is related to all the linear combinations you can now get
the smallest subspace that contains the other 2
dusky epoch
you can think of U + W as {(x,x,y,y) + (x',x',x',y') : x, y, x', y' in F^4}
lavish jewel
yeah, use different letters to avoid getting mixed up
strange delta
yeah thats what i thought
lavish jewel
say a = x + x', b = y + x', c = y + y'
strange delta
ahh yes
lavish jewel
they kinda said "z = y + y lol" where the 2 y's are different
strange delta
i'm trying to self study using this book, doesn't seem to be helping
well using linear alg done right
lavish jewel
look up the definition and examples in different places. the accepted answer here seems handy https://math.stackexchange.com/questions/890215/trouble-understanding-sum-of-subspaces
strange delta
for anyone who has read linear algebra done right, does l(v) just mean the map from v to itself
dusky epoch
no, L(V) refers to the set of all linear maps from V to itself.
limber sierra
linear maps
dusky epoch
yes thank you
strange delta
ok thanks thats clear
strange delta
ok i'm a bit confused with this mapping
shouldn't T(w,z) map back to itself?
so wouldn't it be
(w,z) the output of the map
L(V): v---->v
lavish jewel
from the vector space V to itself
not every element of V to itself
that's just the identity map
lavish jewel
in V
hollow void
Any shortcut for finding rank of matrix?
lavish jewel
nope
hollow void
How do you convert echolan form.
If rank of matrix is "m"
You will never be able to zero last row
lavish jewel
same as always
you only need to ensure each row has more leading zeroes than the last, and all 0s rows (if any) go at the bottom
the procedure is the same regardless of rank
hollow void
But if rank of matrix A (mรn) is it's m.
You can't zero last row.
Coz rank is non zero row even after so many gauss elimination
hollow void
So is it ok. If last row is non zero?
lavish jewel
that's what i said twice ๐
hollow void
Hmm
lavish jewel
the presence of those 0s rows is related to the rank
hollow void
Yes
deft apex
why can we represent these constants 'a' as row vectors?
lavish jewel
notice that f_a(x) there has the same form as the dot product, for one
ionic belfry
hello
lavish jewel
then, using standard matrix-vector products, note that a^T x is the product of a 1xn matrix with an nx1 matrix, which is a 1x1 scalar given by the sum of the elementwise products
so a^T x is the same as the dot product, which is the same as the linear functional f_a
lavish jewel
you can try something like
if A and B are row equivalent, you can transform A into B with some set of elementary row operations which, when put all together, yield a matrix P
so B = PA
at the same time, you can also use elementary row operations to transform A into some RREF form, let's call it R
so that R = TA
and the transformations P and T are invertible
deft apex
then, using standard matrix-vector products, note that a^T x is the product of a...
thanks!
how i say that a function f is in a infinite dimensional vectorspace? just f in R^n?
hard drum
- R^n is finite (n) dimensional (over R)
- a function cannot be 'in' R^n
so i'm unsure what you mean
deft apex
so i'm unsure what you mean
i mean like how do i say that a vector for example is in a infinite dimensional vectorspace
with symbols
hard drum
i think the most sensible thing to say is probably just $v \in V$, where $V$ is an infinite dimensional vector space
deft apex
ah ok
stoic pythonBOT
potato
west orchid
deft apex
"how do i say something"
"just say it"
lmao
jolly tapir
winter harbor
You want to find $v_{1} = (x,y) \in \mathbb{R}^{2}$ and $v_{2} = (z,w) \in \mathbb{R}^{2}$ such that
\begin{align*}
(x,y)+(z,w) = (x+z,y+w) = (3,4)
\
\exists \lambda \in \mathbb{R}^{\times}, (x,y) = \lambda (2,3)
\
\langle (z,w), (2,3) \rangle = 2z+3w = 0
\end{align*}
So in fact, we want to solve the following system of linear equations:
\
\
\begin{cases}
2 \lambda + z = 3 \
3 \lambda + w = 4 \
2z + 3w = 0
\end{cases}
stoic pythonBOT
MisterSystem
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
winter harbor
@jolly tapir
bold plume
can someone help me with this? It's asking to find the value of k for which the matrix is rank 2 and i feel like im doing it right but dunno where im goin wrong
winter harbor
this matrix has rank 2 iff the column space has dimension 2
Which means that span{(7,1,-3), (3,1,4),(1,-1,k)} has dimension two
And since (7,1,-3) and (3,1,4) are linearly independent, what we need to do is
try to find k such that (1,-1,k) can be expressed as a linear combination of (7,1,-3) and (3,1,4)
bold plume
I'm confused.. isn't rank based on the # of rows with a leading 1?
That's why I was thinking to clear the last row
winter harbor
I'm confused.. isn't rank based on the # of rows with a leading 1?
The rank is defined as the dimension of the column space
But sure, you can try to apply elementary row operations and so on and this doesn't change the rank
I think they might not have discussed column space in your class yet
bold plume
Maybe, in that case, is there something in my work that looks wrong based on what my current knowledge is?
winter harbor
Yeah so, the idea is that you want to row reduce that matrix, such that the last row is full of zeros.
Which seems that is what you are trying to do
And that is correct
bold plume
Yea, I dunno where I went wrong pretty much
winter harbor
Maybe just some mistake on your calculations
The idea is correct tho
Just try to find where you made a mistake
bold plume
I did go thru it a few times and just right now, but there doesn't seem to be an error.. I also checked to make sure I got the right numbers for the prob and that too is fine
I hope it isn't too much to ask but could u go thru the problem too?
unborn salmon
Sorry guys, dumb question but what exactly is meant by mapping a vector to another in terms of linear transformatiins?
stable kindle
if f(v) = w, f maps v to w
unborn salmon
Is it literally just turning one vector into the target vector?
thanks, whats that mean though?
winter harbor
I hope it isn't too much to ask but could u go thru the problem too?
Sure
Hang on a second
Consider the matrix
\
\
\begin{bmatrix}
7 & 3 & 1 \
1 & 1 & - 1 \
-3 & 4 & k
\end{bmatrix}
\
\
Apply the following elementary row operation
$$
3 \textbf{r}{2} + \textbf{r}{3} \rightarrow \textbf{r}_{3}
$$
Then, we have the matrix:
\
\
\begin{bmatrix}
7 & 3 & 1 \
1 & 1 & - 1 \
0 & 7 & -3 + k
\end{bmatrix}
\
\
Apply to it now the following row operation
$$
- 7 \textbf{r}{2} + \textbf{r}{1} \rightarrow \textbf{r}{2}
$$
Then, we get the matrix:
\
\
\begin{bmatrix}
7 & 3 & 1 \
0 & - 4 & 8 \
0 & 7 & -3 + k
\end{bmatrix}
\
\
And last, apply to it the elementary row operation
$$
\dfrac{7}{4} \textbf{r}{2} + \textbf{r}{3} \rightarrow \textbf{r}{3}
$$
At last, we get as a result:
\
\
\begin{bmatrix}
7 & 3 & 1 \
0 & - 4 & 8 \
0 & 0 & 11 + k
\end{bmatrix}
\
\
Since the rank of a matrix is preserved under elementary row operations, for the original matrix to have rank $2$ we need the last row of the latter matrix to be zero. This is accomplished if and only if $k = -11$.
stoic pythonBOT
MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
winter harbor
@bold plume
To check k = - 11 is indeed the answer
We could also solve it the way I had mentioned before
Which would be by solving the system of linear equations (1,-1,k) = a * (7,1,-3) + b * (3,1,4)
And as we can see
,w (1,-1,k) = a * (7,1,-3) + b * (3,1,4)
stoic pythonBOT
winter harbor
k = -11 is indeed the solution to this system of equations
This is another way to solve the same problem and yields the same answer, as expected.
bold plume
Ah i see! i think i also understand how u got it with the other method u mentioned now as well, thanks so much and ill double check my own work too!
winter harbor
Np
white gate
Very picky Paula Perkins
white gate
Yep
midnight blade
<@532171623542030336> we was correct
yay
desert palm
Wait what's the explanation
midnight blade
you wont htink its black anymore
the question didnt ask how the pattern follows
to the left or to the right
or switching them
hmm
midnight blade
Wait what's the explanation
explain it yourself im not you
midnight blade
Please tell me <@532171623542030336> I haven't ate in hours
๐ฟ
say
its black because yes because black looks cool because after white comes black because yes because the pattern going to the left not to the right because outside always the right side because the right side is always the outside since no one cares about the left side being the outside because we all since we read the patterns as we read the texts from the left to the right...
and more
too lazy to think
lol
desert palm
Thank you
midnight blade
no
that is just to let you think how to explain
man
dont put my text to it
or you will get a 0
wintry steppe
this question is impossible right? Take for example X consisting of n copies of just a single column vector in C^n. Then if A takes that column vector to anywhere else there is no chance of finding a C since the range of XC is limited by the range of X which is 1-dimensional
midnight blade
Aii bruh glooks for helping me I'ma eat <@532171623542030336>
e
wintry steppe
My linear algebra is a bit rusty can someone explain how they went from -6 and -4 to 1 and 3 in the 3rd step
sonic osprey
this question is impossible right? Take for example X consisting of n copies of ...
I don't see how that line of reasoning makes it impossible
nimble mantle
Hi guys, I have this linear algebra system, the thing is that in its equation matrix there's an incognito value "R". By simulation I know that value is 75500, just because a adjusted that value still I1, I2, I3, and Ib (the variables of the matrix system, like X Y W and Z) corresponds to currents that again a I verified by simulation. The thing is that obviously I want to find R by solving the matrix system, and doing that, treating to R as a incognito variable, I found I1 = an equation that depends on R, I2 = another equation that depends on R , and the same for I3 and Ib, so I don't know what to do to find that R = 75500
these are the solution equations that equals to I1, I2, I3 and Ib, but I don't know what to do now to find R
that was calculated by using X = A^-1 * B as you should suppose
unborn salmon
If T1(x,y) = (x-2y, x+y) and T2 = (x,y) = (x-y, 3x+y), find the composite transformation T2 o T1.
Is this answer correct ot at least down the right path?
wooden pike
Yeah then just simplify
unborn salmon
awesome, thanks!
sonic plank
strange delta
can you direct sum two subspaces that have different dimensions?
quasi vale
@strange delta Yes, for example take R^3. We can take the direct sum of any line going through the origin in R^3(subspace of R^3 of dimension 1) and a plane which has the line as a normal vector containing the origin(another subspace of R^3 of dimension 2) which gives us R^3.
strange delta
hmm ok
im a bit confused on the example b
how did they know dim w = 1
is it because u and w are invariant under t
so it has to be 1
marble lance
If V is the direct sum of U and W, then dim V = dim U + dim W
So dim W = dim F^2 - dim U = 2 - 1 = 1
@strange delta
strange delta
ohh ok
btw
is dim u = 1 because
(x,0) the y coordinate is just 0
which is basically dimension 1
unborn salmon
Can someone please explain how to take a set of vectors that form a basis and change it to the standard basis in R^3? Struggling to follow my textbook and i think an example will help greatly
lavish jewel
an easy way is to notice that if you have a set of vectors that spans R^3, you can put these vectors as rows of a 3x3 matrix and do gauss jordan on them
the operations you do throughout gauss jordan can be put into a matrix
if we call the matrix whose rows are the original vectors M, and the gauss jordan operations a matrix T, then TM = I, where I is the 3x3 identity matrix
an example would be to take the vectors [1,1,0], [0,1,0], [0,0,1]
you can see that the operation you need is to subtract the second vector from the 1st
unborn salmon
i see, will this give me the transition matrix though?
lavish jewel
so T would be a matrix
1 -1 0
0 1 0
0 0 1
unborn salmon
Oh
lavish jewel
this is closely related to how you invert a matrix
you form an augmented matrix of [M | I] and then gauss jordan the rows of M
unborn salmon
Then the right matrix is the ttansition?
lavish jewel
then the I is transformed into T
unborn salmon
Awesome, ill give this a go
lavish jewel
the tl;dr is that inverse matrices can be interpreted as a change of basis into the standard one
unborn salmon
So matrix T starts out as our target vector?
lavish jewel
target matrix
do note that this T is the "transformations that need to be done to the vectors to convert them into the canonical basis"
if you want an actual change of basis, the matrix M should have the vectors as columns instead
and this will transform the coordinates in terms of those vectors into coordinates in terms of the canonical basis
not quite the same thing
lavish jewel
seems ok
dense stone
To be more precise, it should be u in U, then T(u) in range T by def, so in U.
strange delta
seems ok
i feel like this kind of stuff i can never tell whether it's right or wrong
dense stone
U invariant under T means T(U) is included in U
lavish jewel
do pay attention to what melo wrote regarding taking an element of U
dense stone
You want to prove the inclusion $T(U) \subseteq U$
stoic pythonBOT
Mรฉlo
dense stone
So it means an element of range T is in U, and your proof was fine
Since you took an element in range T
But it is quite a particular case
lavish jewel
you could've started with an element u in U, then T(u) is in range T by def, and range T is a subset of U, so both u and T(u) are in U
you started with an element in range T, which is also fine here since range T is a subset of U
just be aware of that, i guess
dense stone
In general, you have to apply T to see that U is indeed invariant, so here you sort of used a trick, which is that the exercise is a tautology
Range T subset of U already means U is invariant
And there is nothing to prove, in fact
The only thing left was to write it
median swift
Hey! L is a linear transformation.
L(u1)=(u2) and L(u3)=(u4)
Could someone explain why:
A(u1-u3)=(u2-u4) is true?
Let it be known that all the vectors U1, 2,3,4 are known.
median swift
Yeah, so the question they ask is is there any linear transformation L so that L(u1)=u2 and L(u3)=u4
and if yes, write such a matrix L
wild fulcrum
what are u1 u2 ...
median swift
vectors
lavish jewel
can you show the original problem? this should be a consequence of linearity, as you were already told. to say anything more specific, we'd need to see the vectors
wild fulcrum
ofc they are vectors lol, like are you given specific vectors to work with or anything
median swift
U1=(1,1,1) U2=(1,0,1) U3=(1,0,-1) U4=(1,1,2).
Is there any linear transformation L so that L(u1)=(u2) and L(u3)=(u4). If yes, how many such L's are there? Mention such a matrix L. If no, explain
That's the entire question. I'm just trying to understand how L(u1-u3)=(u2-u4) though.
lavish jewel
That's the entire question. I'm just trying to understand how L(u1-u3)=(u2-u4) t...
this is completely unrelated to the rest of the question, and the answer is just "linearity"
L is a linear operator
median swift
Forgot to add a part of the question, added it now
lavish jewel
still unrelated
median swift
I'm only asking because it was listed as a possible solution to the problem.
lavish jewel
as soon as you said "L is a linear transformation", and L(u1)=(u2) and L(u3)=(u4), L(u1-u3)=(u2-u4) was already true without any further proof needed
median swift
Yeah, sorry, mate. It's been ages since I did maths. Sorry, if I sound dumb now.
Just to be clear if something's defined with additivity also means f(x-y)=f(x)-f(y)?
Because I looked at the proof earlier and it just threw me off even more
lavish jewel
sure, it's the same as defining w = -y, then doing f(x + w) = f(x) + f(w) and substituting back
if that helps you
median swift
Ah, thanks. Makes sense now.
Glad I found the discord. Thanks for the help, wish you all a good weekend.
marble lance
Looks fine, but just to double check you are getting it
How does STu = 0 imply that null S is invariant under T?
@strange delta
Like what is it that you had to show to show that null S is invariant under T
strange delta
uhh
marble lance
What does invariant under T mean?
strange delta
i showed that putting something from u into ST gives me back something thats in the null s
marble lance
That's not what you have to show
A set X is invariant under T if for every x in X, Tx is in X
So you have to take an element u of null(S), and show that Tu is in null(S)
strange delta
so i basically showed every element u from null(s) is in ST/TS
marble lance
ST and TS are not sets
dusky epoch
^
dusky epoch
it sounds like you don't know what it means for a subspace to be invariant under an operator.
marble lance
A set X is invariant under T if for every x in X, Tx is in X
Please read this. Is this the definition as you understand it?
strange delta
yes
marble lance
Okay, so you have to take an element of null(S), say u, then show that Tu is in null(S)
That is what you have to show in order for null(S) to be invariant under T
strange delta
okay
marble lance
So what do you need for Tu to be in null(S)?
strange delta
for all Tu to be in null(s), where u is in null(s)
marble lance
Yes, but what do you need to show for Tu to be in null(S)?
Like by definition of null(S)
strange delta
every Tu range to be 0
marble lance
Not sure what you mean by Tu range
dusky epoch
no redd
strange delta
i mean the output
dusky epoch
what does it mean for a vector to be in null(S)?
marble lance
The output under what?
marble lance
what does it mean for a vector to be in null(S)?
Yes, y is in null(S) if and only if what?
strange delta
if a vector is in the null(s) doesn't it mean it's the zero vector?
marble lance
No
Do you know what the definition of null(S) is?
Are you saying y is in null(S) if y is 0? Or if what is 0?
strange delta
uhh lemme see
marble lance
You really need to be sure you know all the definitions involved in a problem before attempting it.
strange delta
say for some y that is in v, Ty = 0
dusky epoch
but we are talking about null S, not null T
marble lance
^
strange delta
for some u that is in null(s) Tu = 0
strange delta
yes
marble lance
So Tu is in null(S) if ...?
strange delta
u is in null(S)
marble lance
No
A vector y is in null(S) if its image under S , Sy, is zero
Now we look at the vector Tu
The vector Tu is in null(S) if what?
Use the same definition
The vector is now just Tu instead of y
strange delta
So a Tu is in null(S) if it's image under S, STU is zero
gray dust
what's y
dire thunder
it is better but what is y now?
marble lance
Yeah, y was just what we used in the definition
In this problem we have Tu, not y
No, Su is wrong
u is in null(S) if Su is zero. But we don't want u to be null(S) we want Tu to be in null(S)
dire thunder
So a Tu is in null(S) if it's image under S
i shall now highlight
So a Tu is in null(S) if it's image under S
marble lance
its* no apostrophe
strange delta
So Tu is in null(S) if it's image under S, STU is zero
strange delta
yay lol
marble lance
So it's a little more confusing because it is already the image under T, but remember Tu is just another vector in V, so we just need to use the same definition on this vector
Okay, so to recap
strange delta
yeah it's a bit harder because i am self teching myself this stuff
marble lance
null(S) is invariant under T, if for every u in null(S), Tu is in null(S). To show that Tu is in null(S), we need to show that STu = 0
So, all you need to do is show that STu = 0 for every u in null(S)
strange delta
right ok
marble lance
Okay, so first, do you understand it now? Like really understand why that is what you have to do?
And, can you show that STu = 0?
strange delta
hmm
marble lance
We are really just applying the definitions
So you need to become more comfortable with the definitions
strange delta
for some u in null(T), then Tu = 0
marble lance
We are working with null(S)
strange delta
for some Tu in null(S), STu = 0
marble lance
No, you cannot assume Tu is in null(S)
You need to show it is
So you need to show STu = 0 in a different way
You assume u is null(S). So you can use that to prove that STu = 0
marble lance
Yes
L(V) is the set of linear maps from V into V. So if you say S and T are in L(V), that means they are linear maps from V into V
strange delta
for some u in null(T), then Tu = 0. since the maps are linear, for all null(T), STu = 0
marble lance
Hold up
marble lance
We have to work with u in null(S)
strange delta
linear
marble lance
@strange delta ?
strange delta
hi
marble lance
Are you there?
marble lance
๐
Remember the property they give you
ST = TS
Ping me if you need help or to show what you did
wintry steppe
hi guys
one questions
Hi guys sorry can someone tell me if what I did is right or not?
wanton spoke
yes
marble lance
Can you give more details about why STu = 0?
And you don't need the is an element of null S
The important thing is that it is 0
@strange delta
strange delta
um all i'm thinking is that
because Su = 0, Tu must be the null of S
and hence STu = 0
marble lance
No
It's the other way around
You need to show that STu = 0
And that's why Tu is in the null of S
You have that u is null(S), that means Su = 0, and you have that ST = TS. You have to use these two things to show that STu = 0
marble lance
You need to show that for all u in null(S), STu = 0
marble lance
For all u in null(S), Tu is in null(S) [this is the definition of invariance] is what you have to show. And to show that Tu is in null(S), you need to show that STu = 0
marble lance
You need to use: ST = TS, Su = 0, and you need to remember that S and T are linear, because you will need that too.
So use that to show that STu = 0
marble lance
It's not enough
Because you don't understand why it's true
You can only leave out details if you know what they are
winged prairie
yes
why did they take 2 vectors from each subspace for this proof ? Wouldn't it be suffiecient to take 1 from each subspace ?
I underlined the part i don't understand
strange delta
i know another null is due to linearity, 0 would be a null
wanton spoke
It is showing that M+N is a vector space
marble lance
You have to show M+N is closed under addition, so you have to take two elements of M+N and show their sum is in M+N, but each element of M+N is of the form x+y, so you need two x's and two y's
marble lance
this is all i have so far
You have the order wrong. You need to show that STu = 0, and then conclude that Tu is in null(S)
gray dust
could anybody help me with some linear algebra?
ask, don't ask to ask
short magnet
south wadi
I have a question from my solids 2 course
can someone help me understand transformations in 3D and have a better understanding where the formula is coming from
winter harbor
Idk the notation he's using, but I suppose he's talking about the Cauchy stress tensor, right?
I am nto well versed in physics tbh
But I could try helping you understand tensors in general
And how the coordinates of a tensor change with respect to a change of basis
I suppose that's what your question is about
royal bluff
winter harbor
Does $P_{3} = {p(x) \in \mathbb{R}[x] , \vert , \text{deg} , p(x) \leq 2}$?
stoic pythonBOT
MisterSystem
winter harbor
I.e, is P_3 the set of polynomials with real coeffiecients and degree less than or equal to 2?
If yes
Then what you can try to do
Is use the isomorphism from this vector space of polynomials
To R^3
And view T as a linear transformation from R^3 to R^3
i.e, a 3 x 3 matrix with real coefficients
that makes things way easier to calculate
In this case, the map $T$ is represented by the matrix:
\
\
\begin{bmatrix}
3 & 1 & 0 \
2 & - 3 & 1 \
-1 & 0 & 0
\end{bmatrix}
stoic pythonBOT
MisterSystem
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
winter harbor
So all you need to do is find the inverse of this matrix
winter harbor
3 2 -1
1 -3 0
0 1 0
That's the transpose
Yeah so
All you have to do is
since we are viewing this map from R^3 -> R^3
Apply this map to the standard basis of R^3
i.e (1,0,0), (0,1,0) and (0,0,1)
the column of the matrix for T (in the standard basis)
royal bluff
yea
winter harbor
yeah so
the first column of this matrix will be T((1,0,0))
and as we can see
that is 3x^2 + 2x - 1
which is being identified with (3,2,-1)
royal bluff
oh shit now i see
winter harbor
so the first column of the matrix for T in the standard basis is going to be (3,2,-1)
and so on
and so we get that matrix for T in the standard basis
And all we need to do is find the inverse of that matrix
which like, should be easy enough
winter harbor
,w inverse of {{3,1,0}, {2,-3,1}, {-1,0,0}}
stoic pythonBOT
winter harbor
This seems to be the case
Nice
Now all you have to do is
Given this matrix
find the linear transfomation that is associated to it
and then you get the inverse linear transformation of T
stoic pythonBOT
MisterSystem
royal bluff
I wouldn't have even thought to use the isomorphism
winter harbor
damn thanks
Np
winter harbor
I wouldn't have even thought to use the isomorphism
That's why finite dimensional vector spaces are a bless ๐
royal bluff
That's why finite dimensional vector spaces are a bless <:stare:826920038623346...
Facts ๐คฃ
plain saffronBOT
dire thunder
@formal dawn anyway multiply everything by (x+1)(x-3)x and see what happens
(and technically this is not linear algebra)
formal dawn
Ok
visual siren
<@!753965095364591656> anyway multiply everything by (x+1)(x-3)x and see what ha...
Wouldnโt he multiply by (x+1)(n-3)n?
dire thunder
there is no n
marble lance
You are saying V is an elemeny of U1 + ... + Um
That makes no sense
You need to show that T(u1 + ... + um) is in U1+...+Um
That's the definition of invariant
No one said you had to show V = U1+...+Um
marble lance
"because U1+...+Um = V", but that's not necessarily true
Tuj is in Uj but that's because they say Uj is invariant under T
@strange delta
strange delta
<@591696293915787283>
yeah my bad, will i get it right if i change that part
marble lance
Well, it looks okay, but whether you actually understand why each thing you wrote follows from the previous thing, that only you can decide
strange delta
ok
do you think i should keep on using axlers book
i'm using it to self study so i don't really get much help
zinc copper
Can the special linear group be defined independently of the notion of a determinant, so that we can define the determinant as the canonical map $\operatorname{det}: GL_n(F)\to GL_n(F)/SL_n(F)\cong F^\times$?
stoic pythonBOT
๐ittle โarwhal โ
lusty bison
Just a quick question, can the zero matrix be considered? an elementary matrix
I would think no because it would require you to multiply each row by zero and wouldnโt that count as more than one row operation?
lavish jewel
elementary matrix as in, representing elementary row operations?
those matrices are all full rank, the 0 matrix is rank 0, so no
it turns the target matrix into something that isn't row equivalent
lusty bison
those matrices are all full rank, the 0 matrix is rank 0, so no
Okay thank you!
copper pasture
hi, so not sure where to post this.
I'm implementing an algorithm that gives me a orthonormal basis of the space orthogonal to some vector v. I use householder reflections for that.
So basically I give that algorithm a n-dim vector and get a "n x (n-1)" matrix back. I'd like to check it's correctness. I could just compare to some computed example but I guess there are some properties I could use to test the correctness no?
Can anyone of you think of something simple to test the correctness?
lavish jewel
for one, you can check orthonormality by taking that matrix, call it M, and computing M^T M to get an n-1 x n-1 identity
copper pasture
I guess I could just add v to the ONB and test for Q^TQ=Id?
lavish jewel
yep, that too
copper pasture
ah, a orthogonal matrix doesn't have to be quadratic eh. ๐
copper pasture
yeah I just confused myself when I thought about how to test it. nevermind
lavish jewel
the n x n-1 one won't satisfy M^T M = M M^T = I
only M^T M = I
once you put in the other vector, then yes
copper pasture
okay thanks, so I'll just check for || M^T M - ID || << eps
ah weird markdown formatting
norm(M^TM - Id) << eps. thanks ๐
lavish jewel
take the frobenius norm
if you just do norm and it defaults to 2-norm, it'll only check the first eigenvalue
copper pasture
yeah, I remember that we used the forbenius norm a bit when testing such things.
thanks for the pointer
have a nice day
lavish jewel
u too
dusk vine
tall nymph
Anyone wanna help, im so confused..
dusk vine
I just dunno how to do this lol, I tried plugging in (3s+2t) for x and (-s+5t) for y but I do not think that is right
dusk vine
Anyone wanna help, im so confused..
This isn't really linear algebra, you might want to ask in precalc or somethin
tall nymph
oh ok
quasi vale
@dusk vine Can you find the vector equation for the line?
dusk vine
Ah, so I need to parametrize then transform?
quasi vale
Yes once you find the parametric equation or vector equation for the line, you then multiply it with A
and then you get your transformed line
dusk vine
Awesome thanks
west saddle
Ok, i'm struggling with something very basic.
if I have S = {(1,-2,3),(2,1,4),(7,6,3)}
It's written out like
1 2 7
-2 1 6
3 4 3
is the {} what dictates going downwards
where as if I have , (1,-2,3),(2,1,4),(7,6,3)
It would be
1 -2 3
2 1 4
7 6 3
I know it is the transpose, but i'm running into trouble because it messes with spanning sets
nocturne jewel
if you have a set then no direction can be derived
west saddle
but when im looking for a set i would label c1(1,-2,3),c2(2,1,4) and thats like the first matrix
nocturne jewel
??
hard drum
Definitely not trial and error
Equate the corresponding elements of each matrices to one another to get a system of equations you can solve
For example c+d = b
As the matrices are equal to one another
nocturne jewel
you have 4 entries, so 1 equation per entry
which was stated already by potato
yes, you put the co-efficient of the variable(s) like normal.
unborn salmon
Hi
If I have a transformation T(x,y) and I have to find the matrix T with respect to bases B_1 and B_2, I think I know how to solve this - I basically just apply the T(x,y) transformation to bases B_1, then I can take the resulting set of vectors and augment them with B_2 and the resulting matrix is matrix T right?
Now after that, I have to use the matrix T to find T(w) where w is some new vector
Is that simply matrix T multiplied by vector w?
Hope that made sense
winter harbor
If I have a transformation T(x,y) and I have to find the matrix T with respect t...
What I think you meant is the following
This argument can be carried over for finite dimensional vector spaces in general, or even infinite dimensional vector spaces for that matter, but for simplicity (since you have asked for the dimension 2 case) I will be going over for dimension 2
unborn salmon
Thanks!
winter harbor
Let $T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ be a linear map and $B_{1} = {e_{1}, e_{2}}, B_{2} = {e'{1}, e'{2}}$ two ordered basis for $\mathbb{R}^{2}$.
\
\
Notice then that $\forall j \in {1,2}$, we have that $\exists \alpha_{1,j}, \alpha_{2,j} \in \mathbb{R}$ such that:
\
\
$T(e_{j}) = \alpha_{1,j} e'{1} + \alpha{2,j} e'{2}$
\
\
This is the case since $B{2}$ is a basis for $\mathbb{R}^{2}$
\
\
We then define the matrix $[T^{B_{1}}{B{2}}]$
, which is called the matrix of $T$ with respect to the ordered basis $B_{1}$ and $B_{2}$ as the $2 \times 2$ matrix whose entries are
$$
[T^{B_{1}}{B{2}}]{ij} = (\alpha{i,j}){i,j \leq 2}
$$
Notice that this completely species the transformation $T$, since $\forall v \in \mathbb{R}^{2}$, I have that $\exists ! v{1}, v_{2} \in \mathbb{R}$ with
$$
v = v_{1} e_{1} + v_{2} e_{2}
$$
So in particular
$$
T(v) = T(v_{1} e_{1} + v_{2} e_{2}) = v_{1} T(e_{1}) + v_{2} T(e_{2})
$$
So $T$ is completely specified by its values on the basis $B_{1}$ and $B_{2}$
@unborn salmon
You can argue like this in general
stoic pythonBOT
MisterSystem
winter harbor
I would recommend you to try to mimick this argument in general
Say, let
$$
T : V \rightarrow W
$$
Be a linear map between finite dimensional vector spaces $V$ and $W$, where $B_{1}$ is an ordered basis for $V$ and $B_{2}$ is an ordered basis for $W$, find the matrix for $T$ with respect to these given ordered basis.
stoic pythonBOT
MisterSystem
random axle
is this room free?
winter harbor
Yeah
random axle
So. I have three planes. For the values of k that we are considering, the planes do not meet at a unique point. Therefore, all the planes must intersect at a line or not at all. The question wants me to figure out whether they intersect at a line or not. To do this, I found the cross product of the normal vectors of the first two planes and the cross product of the last two planes. I got two vectors that was in the same direction. Why is this not enough to conclude that they intersect at a line?
winter harbor
So. I have three planes. For the values of k that we are considering, the planes...
I think you are misinterpreting the question a little bit.
There are in fact two possibilities, i.e either these planes intersect at a line
Or
They are all parallel
Which considers the case where they do not intersect at all
or are the same plane
And the question does not want you to find whether or not which one of these happen
because you don't have enough information to do this in the first place
but rather to find out
for which values of k
we have the first configuration (intersection at a line)
and for which values of k we have the second configuration (they are parallel)
And to do this
You can basically try to study the matrix associated to this system of equations
they intersect at a unique point iff the associated matrix has full rank
now think about the rank of the associated matrix
and how it corresponds to each one of these geometric configurations
and try to find via row reduction for which values of k we have a certain rank and etc
random axle
they intersect at a unique point iff the associated matrix has full rank
What do you mean by full rank? Sorry, I am studying a basic high school linear algebra/3d vectors course
But for the values of k found, k = 1 and k = 2. These planes are not parallel. But they don't intersect either
winter harbor
What do you mean by full rank? Sorry, I am studying a basic high school linear a...
I suppose you don't know what the rank of matrix is then
Do you know what row reduction is ?
random axle
Do you know what row reduction is ?
Could you explain what it is? I think I may have learned it but under a different name
winter harbor
Could you explain what it is? I think I may have learned it but under a differen...
Gaussian elimination?
random axle
is it just like doing row operations on determinants to make the determinant simpler to compute but not changing the value of the detemrinant/
winter harbor
Yeah so, row reduction is a method of applying elementary row operations to a matrix in order to solve systems of linear equations
you can do this to compute determinants too
in this case
we want to try to ''solve'' that system of equations
by putting it into reduced row form, which we can obtain by gaussian elimination/row reduction
That's one way to solve the problem
Because studying the intersection of planes has to do with studying the solution set of that system of linear equations
calm yoke
Can you define a inner product without defining a set subspace?
half ice
Odd wording. You need a vector space for an inner product, that's really it
calm yoke
Yeah, cost me almost 20% of my linear algebra test because I said it wasnt defined properly
half ice
What is a "set subspace"?
naive sonnet
What is a "set subspace"?
Seconded
calm yoke
What is a "set subspace"?
Vector space in R^n
half ice
Well, you'll need some kind of vector space
half ice
You can always use the dot product as an inner product for R^n
But it isn't the only inner product you can use
calm yoke
So, you can define an inner product without defining a vector space that belongs to R^3?
winter harbor
The wording seems a bit weird
But formally, an inner product on a real vector space $V$ is a symmetric, positive-definite bilinear map, i.e:
$$
g : V \times V \rightarrow \mathbb{R}
$$
Such that $g$ satisfies:
\begin{enumerate}
\item (Bilinearity)
\
\
$\forall u,v,w \in V$ and $\lambda \in \mathbb{R}$ we have
$$
g(u,v+\lambda w) = g(u,v) + \lambda g(u,w)
$$
and
$$
g(u+\lambda w, v) = g(u,v) + \lambda g(w,v)
$$
\item (Symmetry)
\
\
$\forall u,v \in V$ we have that
$$
g(u,v) = g(v,u)
$$
\item (Positive definiteness)
\
\
$\forall v \in V$ we have
$$
g(v,v) > 0 \iff v \neq 0
$$
\end{enumerate}
stoic pythonBOT
MisterSystem
winter harbor
This is what an inner product is
With this, we can answer your question
No, we don't necessarily for a vector space to be a subspace of R^3 in order to define what an inner product is
We can define an inner product even on infinite dimensional vector spaces
And it doesn't matter
And as you may know
We have a canonical inner product on $\mathbb{R}^{n}$ given by $\forall u,v \in \mathbb{R}^{n}$, we define it to be:
$$
\langle u,v \rangle = u^{T} v
$$
lavish jewel
that's so cursed, you normally work with row vectors?
winter harbor
that's so cursed, you normally work with row vectors?
why is this cursed lmao
lavish jewel
i've never ever seen that before haha
winter harbor
that's just the inner product
lavish jewel
for row vectors, yeah
stoic pythonBOT
MisterSystem
lavish jewel
i usually think of row vecs as 1 x N instead of N. doesn't really make a diff, i just don't see it often
winter harbor
Yeah, I actually meant to write it like this
for some reason I wrote it the other way around
thanks for noticing
lavish jewel
๐
winter harbor
**MisterSystem**
Yeah, but basically what I wanted to explain to him is that an inner product is defined like this
And we can have an inner product on a general vector space
and it doesn't need to be a subspace of R^3
calm yoke
Okay, thanks
hard drum
not sure what you mean by that exactly
limber sierra
most abused = in history
nocturne jewel
Multiplication of Eigen, my favorite
lavish jewel
multiplication of eigen sounds like another ange nickname
sonic beacon
how would i define what it means to add and scalar multiply linear maps? I want to show that the set of all linear maps between vector spaces form a vector space
winter harbor
how would i define what it means to add and scalar multiply linear maps? I want ...
You do it pointwise
For instance
Suppose that I have a linear map $T \in \text{Hom}(V,W)$ between two vector spaces $V$ and $W$.
\
\
If I have a scalar $\lambda \in \mathbb{R}$, then notice that
\begin{align*}
\lambda T&: V \rightarrow W \
v \mapsto& \lambda T(v)
\end{align*}
Is also a linear map between $V$ and $W$.
stoic pythonBOT
MisterSystem
winter harbor
What this map does for a vector v is just apply T(v) to it and then multiply by a scalar
sonic beacon
ah, i see
winter harbor
So we are doing scalar multiplication point-wise
How would you define sum of linear maps then?
sonic beacon
i want to say T: V->W, so addition would be defined as T(v) + T(w) = T(v+w) ?
winter harbor
No
We want to take two (potentially distinct) linear maps T, T' : V -> W
And define another linear map
T + T' : V -> W
How would we do it?
Well
Take a vector v in V
sonic beacon
what you said makes sense
winter harbor
What is the most natural thing we could ask for (T+T')(v) to be?
sonic beacon
im thinking
because let X be the set of all maps between vector space V, W then X = { T, T',..} and T: V-> W, T': V->W etc
and now you want to define what T+T' is
winter harbor
Yup, that has to be another linear map
Between V and W
So we must define
What it does to a vector v in V
marble lance
Just noting that this definition is not unique to linear maps. This is the same as the definition of the sum of two functions you would have seen in calculus or wherever.
winter harbor
How would we define T+T' applied to v?
Notice that T(v) is in W
And T'(v) is in W too
And in W we have a sum
winter harbor
How do we sum functions, say in calculus?
If I have another function like f (it could be x^2)
sonic beacon
im trying to look it up havent had calculus in forever