linear-algebra

[limber sierra]

so let's let "k" be our stand-in variable for that scalar

clearly $6k = 4k + 2k$ has solution $k = 1$, and similar for $3k = 3k + 0k$

[stoic python]

Namington:

[limber sierra]

but as for $-4k = -4k - k$

[stoic python]

Namington:

[waxen jacinth]

k isn't 1

[limber sierra]

can you solve that equation?

for k

[waxen jacinth]

k=0

[limber sierra]

right, and that's the only solution

so $T\begin{pmatrix}x\y\z\end{pmatrix} = \begin{pmatrix}x\0\z\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

i.e. it maps the top entry to itself, the bottom entry to itself

but it multiplies the middle entry by 0

and when you have a 0 in your transformation's definition, it's not one-to-one

since for example, $T\begin{pmatrix}0\1\0\end{pmatrix} = T\begin{pmatrix}0\-3519539\0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

since they get mapped to the same thing

(0 0 0)

[waxen jacinth]

Mhm

It's a bit abstracted to be fair

[limber sierra]

yeah, but the point is

we've shown that this transformation cant be one-to-one

(sorry, using synonyms out of habit)

since it has to have a 0 entry

[waxen jacinth]

It's k, Im more used to saying injective than one-to-one

Those were the terms I used to use back in my home country

So dw about it

[limber sierra]

if you want to think about it as a matrix, it's like the transformation had an entire row that was just 0s

$T\begin{pmatrix}x\y\z\end{pmatrix} = \begin{pmatrix}1&0&0\0&0&0\0&0&1\end{pmatrix}\begin{pmatrix}x\y\z\end{pmatrix}$

[stoic python]

Namington:

[waxen jacinth]

Yeah that makes sense

[limber sierra]

and as you can see, since it has an entire row that's just 0s, it also has an entire column thats just 0s

(since its square)

and actually that hints at our solution for part 3

[waxen jacinth]

But how did u know u had to find a k scalar and not some summation or combination of the two for example?

Is it just a practice-born sense? xd

[limber sierra]

well, that's a valid question

i was only really considering one possibility to show that it wasnt always one-to-one

but it might still be sometimes one-to-one, maybe if we choose something else

that said, we have a bit of an advantage here:

T is linear

that means that it "distributes over addition" so to speak

so recalling my earlier equation

$T\begin{pmatrix}6\-4\3\end{pmatrix} = T\begin{pmatrix}4\-4\3\end{pmatrix} + T\begin{pmatrix}2\-1\0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

we can actually rewrite this as

$T\begin{pmatrix}6\-4\3\end{pmatrix} = T\left(\begin{pmatrix}4\-4\3\end{pmatrix} + \begin{pmatrix}2\-1\0\end{pmatrix}\right)$

uh

texit?

[stoic python]

Namington:

[limber sierra]

that is

[waxen jacinth]

One more question, the operation u take in summation, and the one that is defined by the exercise: T(c)=u+v are different things right?

[limber sierra]

$T\begin{pmatrix}6\-4\3\end{pmatrix} = T\begin{pmatrix}6\-5\3\end{pmatrix}$

texit is being slow 😦

[stoic python]

Namington:

[limber sierra]

uh you mean in part 1 and part 2

well, part 2 is a special case of part 1

but yeah

[waxen jacinth]

Okay

[limber sierra]

theyre distinct

[waxen jacinth]

Sry, the T(c)=u+v rly confuses me

I know its an entry

But I still confuse it as the transformation rule

[limber sierra]

anyway, we now have $T\begin{pmatrix}6\-4\3\end{pmatrix} = T\begin{pmatrix}6\-5\3\end{pmatrix}$

bleh

[stoic python]

Namington:

[limber sierra]

there's one more trick we can pull: try and put both of these on the same side of the equation

$T\begin{pmatrix}6\-4\3\end{pmatrix} - T\begin{pmatrix}6\-5\3\end{pmatrix} = \begin{pmatrix}0\0\0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

and again, we know T is linear, so we can do this:

$T\left(\begin{pmatrix}6\-4\3\end{pmatrix} - \begin{pmatrix}6\-5\3\end{pmatrix}\right) = \begin{pmatrix}0\0\0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

and evaluating that subtraction:

$T\begin{pmatrix}0\-1\0\end{pmatrix} = \begin{pmatrix}0\0\0\end{pmatrix}$

[stoic python]

Namington:

[waxen jacinth]

Mhm

And then we try to find T so that it satisfies it

Ofc we fail to do that

But this makes it heaps clearer

[limber sierra]

well, there is a way to satisfy it

but no matter what, it will involve "not using" that -1 at all

since if you "add" the -1 to any of the entries

well, you have no way to get rid of it

unless you multiply it by 0

(in which case, you might as well not have had it at all)

the point there being: the linear transformation T necessarily "ignores" the middle entry

[waxen jacinth]

Yeah, in order to keep it injective we have to avoid 0

[limber sierra]

no matter what we set the middle entry of the vector to be

it cant have any effect

on what we get out of T

[waxen jacinth]

Oki

[limber sierra]

so we know T must not be one-to-one

since changing the middle entry cant change our answer

so (0, 5, 0) and (0, 194398, 0) will be mapped to the same thing by T

[waxen jacinth]

Gotcha

[limber sierra]

so, our answer for part 2 is that it cannot be one-to-one

under any circumstances

[waxen jacinth]

Yep

[limber sierra]

now as for part 3, to tie this up

you may or may not have seen a theorem that deals with exactly this, but since you're being asked this question

i'm assuming you havent seen the theorem

there are a variety of ways to prove it, but here's one that may appeal to you, particularly if you like thinking in matrices:

since $T$ is a transformation from $\bR^3$ to $\bR^3$, we know it can be represented by a $3 \times 3$ matrix

[stoic python]

Namington:

[limber sierra]

and in terms of matrices, we know the matrix of an onto function has no 0 rows (after row reducing)

and similarly the matrix of a one-to-one function has no 0 columns after row reducing

so basically, what we actually want to determine is:

if a 3x3 row-reduced matrix has no 0 rows, how many 0 columns does it have?

but remember that we're dealing with a row-reduced matrix here

so it looks something like:

$\begin{pmatrix}a&?&?\0&b&?\0&0&c\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

since there has to be a pivot in each row

(a, b, c)

and those can't be 0

(since otherwise it'd be a 0 row)

but hey, would you look at that

we've "filled in" our columns "for free"

i.e. if a 3x3 matrix has no 0 rows (after being row reduced), it has no 0 columns

[i'm assuming you know what "row reduced" means]

[waxen jacinth]

Yeah ofc

[limber sierra]

in other words, we've just justified:

[waxen jacinth]

So wt ur saying is, lemme put this into pleb vocab

[limber sierra]

if a 3x3 matrix is onto, it is one-to-one

[waxen jacinth]

Alright so, just to recap in pleb

[limber sierra]

yeah sure

[waxen jacinth]

Because the matrix is always consistent

Since it's onto

Then there's no space for any non-pivot column

[limber sierra]

basically, but there's a caveat here

this only applies to square matrices

thankfully, we know the matrix is square, since its a transformation R^3 -> R^3

[waxen jacinth]

So, say this went from R^3 -> R^4

[limber sierra]

i.e. it should take in a vector with 3 entries, and spit out a vector with 3 entries

right, if we changed the domain/codomain to not agree with each other, this would no longer hold

since the matrix would not be square

and indeed we wouldn't be able to draw any conclusions about whether a function is one-to-one based on whether it's onto

[waxen jacinth]

Would we not have enough info or would it def not be injective?

Im guessing the former

AH oke

[limber sierra]

oh wait

hold on R^3 to R^4

that means we'd have a 4x3 matrix

which means it can't be onto; its straight up impossible

indeed, you can never have an onto function from R^n to R^m when m > n

and simialrly, you can never have a one-to-one function when m < n

sorry, i should clarify linear function

(this kind of goes out the window if you allow infinite dimensional vector spaces though, but i'd imagine you're not really dealing with those)

[waxen jacinth]

No not yet at least

[limber sierra]

anyway, to summarize, the answers are C, A, B respectively

unless ive made a big brain fart haha

kinda sleep deprived

[waxen jacinth]

No, ur correct

Immense thanks <3

Im screensharing w a friend rn to solve these, as we are in the same class and we were both equally confused, so u have his thanks too :p

[limber sierra]

note that i've been kind of bad and have been leaving off the word "linear" when i say function/transformation

but everything i say only applies to linear stuff, naturally

[waxen jacinth]

Yeah dw

I get that its only summation and scalars

[limber sierra]

for example, there is a surjection from R into R^2, but it's not linear

[waxen jacinth]

Otherwise we'd be bending stuff

I'd imagine

It would be bending right?

xd

I guess if u took x->x^3

Thatd be injective but non-linear

Actually itd be a bijection

If Im correct in my supposition

But non-linear since I'm multiplying by a variable

Havent taken these, so Im just supposing thats wt itd be like

Anyway, my brain is loaded as it is alrdy

:p

I have one more problem

If thats okay

This one is shorter

Looks like an IQ test question if Im gonna be honest

[limber sierra]

(a more interesting result is that there's a bijection between R and R^n for any n, but that's more a part of analysis, and is a bit complicated to explain)

anyway, sure

[waxen jacinth]

No clue wt I should do here

Lul

[limber sierra]

well, we know that 2 green + 1 red + 1 blue = $29

and similarly, 1 blue + 1 green + 2 red = $27

etc

[waxen jacinth]

Oh my god xd

I was thinking in terms of matrices

[limber sierra]

so this suggests that we should be setting up a system of linear equations

[waxen jacinth]

Jesus

[limber sierra]

ah haha

[waxen jacinth]

I was like

HOW CAN I DO THIS IF I DONT HAVE A SYSTEM

Cuz I was thinking, darn, why cant I have the prices of rows

But of the entire matrix

Im retarded

Just assign shapes to variables and solve the system, thx xDDD

Lul so retarded :v

Im embarrassed

Ive done Calculus 2

And I cant solve this.

[limber sierra]

No worries, it happens

[waxen jacinth]

Give me back my integrals and infinite sums

Im an algebra-illiterate

[limber sierra]

though just an FYI, we'd rather you not use "retarded" as a pejorative on this server

[waxen jacinth]

In my case, I think it applies :p

[limber sierra]

(and calling yourself negative things is never productive in any case)

[waxen jacinth]

Ur right on that

[limber sierra]

everyone makes "dumb mistakes/misinterpretations" that they should really know better than every once in a while

the key thing is being able to fix & learn from them

not being 100% perfect all the time

[waxen jacinth]

Tell that to mom n dad :p

[limber sierra]

(sadly, the emphasis on grades and tests in the education system encourages people to value perfection over improvement...)

(but alas, solving THAT problem is probably a bit more complicated than this server can handle)

[waxen jacinth]

True that

Then there's Finland

Anyway, getting sidetracked

[primal fable]

@waxen jacinth psst hey you
Try 3Blue1Brown's video playlist on Linear Algebra, it's super good and really concretes how to understand LinAlg because hardly anyone teaches it well imo and it's not your fault and you're chill ❤️

https://m.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

Ngl, disappointed this channel doesn't have it pinned.

[chilly solstice]

Can some one fix the problem from the 2nd to the 3rd inequality here:

<@&286206848099549185> (I post about this yesterday, so I hope it is OK for me to use the tag 🙂

[eager burrow]

Could it be that P is supposed to be a substochastic matrix instead? I don't quite know all the terms, but some light googling for the definitions makes that rephrasing somewhat likely...

[chilly solstice]

yes

this is also what I mean! @eager burrow

OH!! I have written it wrong!

😦

[eager burrow]

Yeeeee

[chilly solstice]

But P is sub-stochastic

I have come with another attempt. I can write it down it you have time.

[eager burrow]

I'm cookin' right now but I might give it a thonk afterwards

[chilly solstice]

Great. I do not have time to cook. xD

Assume that $A=(a_{ij}){i,j=1,\ldots,n}$ is a sub-intensity matrix and $\eta = \max_i (a{ii}) > 0$. I will like to show that $P=I+\eta^{-1}A$ is a sub-stochastic matrix. \

By assumption: $a_{ij},i\neq j$ and $a_{ii} \le \sum_{i\neq j= 1}^n a_{ij},i=1,\ldots,n$. It is obvious that $\eta^{-1}A$ is a sub-intensity matrix also, i. e. $\eta^{-1}a_{ij},i\neq j$ and $\eta^{-1}a_{ii} \le \sum_{i\neq j= 1}^n \eta^{-1}a_{ij},i=1,\ldots,n$. This implies that $0\le - \sum_{i\neq j= 1}^n \eta^{-1}a_{ij},i=1,\ldots,n$ or equivalent thatn $1 + \sum_{i\neq j= 1}^n \eta^{-1}a_{ij} \le 1$. So now I have that \eta^{-1}a_{ij},i\neq j and $1 + \sum_{i\neq j= 1}^n \eta^{-1}a_{ij} \le 1$. What is left now to show that $\eta^{-1}a_{ii}$ and that $1 + \sum_{ j= 1}^n \eta^{-1}a_{ij} \le 1$. If I can show this then I'm done. But I can not show this, so I need some help for that.

[eager burrow]

I'm still not 100% sure on the definition, but can't you do your inequality by just adding 1 on both sides?

If you want sub-stochastic-ness, you want your inequality to be (.....) <= 1, rathern than (....) <= 0 anyway

So then it seems like you'd end up with the right thing

So, you wouldn't get the inequality that you've written down, but instead you'd get the inequality plus a one on the right side.

[chilly solstice]

@eager burrow thanks, but I got it! 😄

i had forgotten that eta = max (a_ii)... in my mind I just thoght eta > 0 lolo

[old flame]

Here are my attempts. For question 10. Let $v$ be an eigenvector corresponding to the eigenvalue $\lambda$. Thus, $Tv=\lambda v$. Since T is invertible, $v=T^{-1}(\lambda v)=\lambda T^{-1}(v) \Longleftrightarrow \frac{1}{\lambda}v=T^{-1}(v), \lambda \neq 0$.

[stoic python]

Otoro:

[old flame]

For question 11, Let $\lambda \in F$ be an eigenvalue for S and T. Then let v be the corresponding eigenvector with $\lambda$. Then $Tv=\lambda v \Rightarrow S(Tv)=S(\lambda v)=\lambda Sv=\lambda^{2}v$. If $Sv=\lambda v \Rightarrow TSv=T(\lambda v)=\lambda Tv=\lambda^{2}v$. Thus they both have the same eigenvalue.

[stoic python]

Otoro:

[old flame]

For question 12, Suppose $T \in L(V)$ such that $\forall v \in V, Tv=\lambda v$, then $(T-\lambda I)v=0$. Since $v \neq 0$, this implies that $T-\lambda I=0 \Rightarrow T=\lambda \cdot I$. (Scalar multiple of the identity)

[native rampart]

Here are my attempts. For question 10. Let $v$ be an eigenvector corresponding to the eigenvalue $\lambda$. Thus, $Tv=\lambda v$. Since T is invertible, $v=T^{-1}(\lambda v)=\lambda T^{-1}(v) \Longleftrightarrow \frac{1}{\lambda}v=T^{-1}(v), \lambda \neq 0$.
@old flame correct

[stoic python]

Otoro:

[old flame]

@native rampart ty !

[native rampart]

For question 11, Let $\lambda \in F$ be an eigenvalue for S and T. Then let v be the corresponding eigenvector with $\lambda$. Then $Tv=\lambda v \Rightarrow S(Tv)=S(\lambda v)=\lambda Sv=\lambda^{2}v$. If $Sv=\lambda v \Rightarrow TSv=T(\lambda v)=\lambda Tv=\lambda^{2}v$. Thus they both have the same eigenvalue.
@old flame T and S need not have the same eigenvalues. Take an eigen value of T as say,a and eigen value of S as say,b.

Do something very similar to your method

[old flame]

ah alright, good point

So if $\lambda_1$ for S and $\lambda_2$ for T, then at the end would have $\lambda_1\lambda_2$ instead

[stoic python]

Otoro:

[native rampart]

Wait,there could be an eigenvector of ST that is not an eigenvector of S or T

Don't use this method. I don't think you can prove it via this method)

[old flame]

oh...

alright, let me try again with Q11

what about q12 ?

[native rampart]

Lambda could be different for different v

[old flame]

so for q12 should it be something like, $Tv_i=\lambda_i v_i, \forall v_i \in V$. Then $(T-\lambda_i I)v=0$. Since $v \neq 0$, $T-\lambda_i I = 0, \forall i$. Thus, $T=\lambda_i I$. Then $\lambda_1=\lambda_2, \lambda_2=\lambda_3,...$. Therefore, $\lambda_1=...=\lambda_n=...$ Hence, $T= \lambda_1 I$. Which is a scalar multiple of the identity

[stoic python]

Otoro:

DrunkenDrake:

[old flame]

why ?

[native rampart]

Take $T-\lambda_1$I = 0 . This would mean Tv=$\lambda_1$ v for all v,but then there may be a v such that Tv=$\lambda_2$ v such that the lambdas are not equal

[stoic python]

DrunkenDrake:

[native rampart]

Show there cannot be 2 distinct eigenvalues

[old flame]

oh, if that happens, then what does that imply though

[native rampart]

That would imply all vectors have same eigenvalue and T is scaled identity

[old flame]

oh

ok

got some ideas for q12

Suppose $\exists i,j \in {1,...,dim V}$, such that $\lambda_i \neq \lambda_j$. Then $Tv_i=\lambda_iv_i$ and $Tv_j=\lambda_jv_j$. Subtracting both equations, $T(v_i-v_j)=\lambda_iv_i-\lambda_jv_j$. Since $v_i-v_j \in V$, but it is not an eigenvalue, since there is no $\lambda \in F$, such that $T(v_i-v_j)=\lambda(v_i-v_j)$. Therefore, contradiction and thus all the eigenvalues are the same $\Rightarrow (T-\lambda I)v=0,\Rightarrow T=\lambda I$, since $v \neq 0$.

[stoic python]

Otoro:

[native rampart]

Why is there no such lambda?

Your proof will be complete if you show that

[old flame]

@native rampart because you cant factor out some scalar for the maths

I could argue that for the case of $v_i \neq v_j$, then there is no $\lambda$

[stoic python]

Otoro:

[native rampart]

Do so

Why can't you factor out?

[old flame]

oh if I factor one, for example, $\lambda_i(v_i-\frac{\lambda_j}{\lambda_i}v_j)$. Then $v_i-v_j$ is not an eigenvector

is this the argument ? same for factoring out $\lambda_j$, so WLOG

[stoic python]

Otoro:

Otoro:

[old flame]

which does not satisfy the condition of all v in V being a eigenvector of T

[native rampart]

The eigenvalue of that term could be some other number,say k

[old flame]

so I replace \lambda_i with k ?

what you do mean

[native rampart]

Take $T(v_i-v_j)=k(v_i-v_j) and use T(v_i-v_j)= \lambda_i v_i - \lambda_j v_j $ to show what you need

[stoic python]

DrunkenDrake:

[native rampart]

v_i and v_j will be linearly independent (supposing different eigenvalues) so,(k-lambda i) =(k- lambda j)=0

[old flame]

so I guess something like this then

from the definition of T, $\exists k \in F$, such that $k(v_i-v_j)=\lambda_iv_i-\lambda_jv_j$. Since $\lambda_i \neq \lambda_j$, then comparing coefficients, $k=\lambda_i$ and $k=\lambda_j$. Contradiction

[stoic python]

Otoro:

[native rampart]

(Don't forget to mention this is because we assumed v_i and v_j to have different eigenvalues)

Otherwise,we might not have been able to compare coefficients

[old flame]

ohhhhh okay thanks

but isnt that assumption, $\lambda_i \neq \lambda_j$ ?

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

I thought I mentioned it

[native rampart]

Well,nvm it's fine to use that theorem implicitly

[old flame]

ohhhh ok no problem, which theorem though ?

[native rampart]

Eigenvectors of different eigenvalues linearly independent

[old flame]

ohhhhh I see then

[native rampart]

Q11 uses a weird property

[old flame]

I have never learnt that

[native rampart]

(replace matrices with operators)

[old flame]

so Im guessing your hint is that ?

[native rampart]

Yea, it's weird

Yes

[old flame]

are you saying I can use q8 immediately as a result ?

[native rampart]

I mean, you still have to show it

[old flame]

btw how do you know q8 is related though ?

ah so its an extra proof ?

[native rampart]

Let's say (TS-xI) is non invertible,this theorem tells us (ST-xI) is also non invertible

i.e.,x is an eigenvalue of both TS and ST

[old flame]

noted, I will get back to you on this later thanks

btw, I just attempted q13, here it is. Let $(v_1,...,v_{dim V -1})$ be a basis for $U \subset V$, where $dim U = dim V -1$. Let $r \in U$, then writing it as linear combination, $r = a_1v_1+...+a_{dim V -1}v_{dim V -1}$, then $T(r)=a_1Tv_1+...+a_{dim V-1}Tv_{dim V -1}=a_1\lambda_1v_1+...+a_{dim V -1}\lambda_{dim V-1}v_{dim V-1}$. Since T is an operator for $T_{U}, \forall U$ with $dim V-1$, then this implies that $\lambda_1=...=\lambda_{dim V-1}$, which therefore concludes that T is a scalar multiple of the identity

[stoic python]

Otoro:

[old flame]

detail : each $\lambda_i$ is itself an eigenvalue, but since $T$ is an operator for all U, the set of basis is invariant under T as well, thus they all have to be equal

[stoic python]

Otoro:

[native rampart]

Do you know why v1,v2... Should be eigen vectors of T?

(If you show that,you could use that to show all vectors in V are eigenvectors and use q12)

[old flame]

You mean q12 could be helpful ?

[native rampart]

Yes

[old flame]

Ah alright I will look into it and get back to you later ty

[stoic python]

Anton S.:
Compile Error! Click the reaction for details. (You may edit your message)

[copper pulsar]

How can I solve $(X^T\cdot B^T)^{-1}=(C-AX)^T$ for $X$, where $X$ is a matrix?

[stoic python]

Anton S.:

[dawn remnant]

$(AB)^T = B^T A^T$, so $(C-AX)^T = C^T - X^T A^T$, $(X^T\cdot B^T)^{-1} = (B^T)^{-1}\cdot (X^T)^{-1} = (B^{-1})^T \cdot (X^{-1})^T$

[stoic python]

ConfusedReptile:

[dawn remnant]

So then you can transpose both sides and get:
$$
X^{-1} \cdot B^{-1} = C - A X
$$

[stoic python]

ConfusedReptile:

[copper pulsar]

I see

thanks

then it becomes difficult though

[pallid rampart]

Is there an easy method to see if a matrix is normal?

[native rampart]

Maybe with char equations?

[pallid rampart]

How so

[native rampart]

If T has eigen value c,T* will have eigen value c*

[pallid rampart]

Well that test only tells you when a matrix is not normal

But I guess yes that's a good method to keep in mind

[native rampart]

Are there operators with this property,which aren't normal?

[pallid rampart]

No

Wait

what's c*?

[native rampart]

Conjugate of c

$\bar c$

[stoic python]

DrunkenDrake:

[pallid rampart]

Ah

Well

Actually I think every operator has that property

It's that every eigenvector of T is an eigenvector of T* with eigenvalue c*

[native rampart]

I think you are mistaking T* for transpose of (T*)

[pallid rampart]

Isn't T* the conjugate transpose of T over finite dimensional space?

[native rampart]

Yes

[pallid rampart]

And if T is normal then every eigenvector of T is an eigenvector of T* with eigenvalue c*

[native rampart]

Yes

[pallid rampart]

And for every operator T, c is an eigenvalue of T iff c* is an eigenvalue of T*

I proved this assuming finite dimensional space I think

[native rampart]

So,Why are normal operators special?

[pallid rampart]

Because of spectral theorem?

[native rampart]

ok

[pallid rampart]

But T**=T

So you can just reverse it

$\text{null}(T^-\overline{\lambda}I)=(\text{range}(T-\lambda I))^\perp$, so if $\text{null}(T-\lambda I)$ is nontrivial then $\dim\text{range}(T-\lambda I)<\dim V$ so \dim(\text{null}(T^-\overline{\lambda}I))>0$

[native rampart]

I know it's T T* =T* T

[stoic python]

Whoever:

[pallid rampart]

TT*=T*T is normal

[tall surge]

how to find the determinant of a 3x3 matrix where every component is a block matrix

there is an easy formula for 2x2

but are there any generalizations?

[gray dust]

generalizing 2x2 to nxn? no, just look up laplace expansion or leibniz formula

you can also consider row/col ops and their predictable effects on det

[tall surge]

wait laplace's law doesn't hold for block matrices

you can't have like

$\begin{pmatrix}A & B&C \ D&E&F \ G&H&I \end{pmatrix}

as

[stoic python]

Jay007:
Compile Error! Click the reaction for details. (You may edit your message)

[tall surge]

that determinant

is not

$A \begin{pmatrix}E & F\ H&I \end{pmatrix}- ...$

[stoic python]

Jay007:

[tall surge]

again these are block matrices

i can't use a laplace expansion on them because i have no info about the individual components of the block matrices

i simply know what they are

with a 2x2 I was able to use the determinant formula
$\det\begin{pmatrix}A& B\ C&D \end{pmatrix}= \det(D)*\det(AB^{-1} C)$

[stoic python]

Jay007:

[tall surge]

but it's not applicable now

[gray dust]

how you applied laplace makes 0 sense, and that's not a formula for a 2x2 of blocks

laplace & leibniz still hold, you write em out in terms of individual entries, not the blocks. but if you got no info on the blocks then they're useless

[tall surge]

how you applied laplace makes 0 sense, and that's not a formula for a 2x2 of blocks
@gray dust this was my point. that expansion makes no sense and that's what I thought you were suggesting. on the note of the 2x2 of blocks, iirc it only holds if the matrix is invertible
laplace & leibniz still hold, you write em out in terms of individual entries, not the blocks. but if you got no info on the blocks then they're useless
yep as i thought. thanks for helping anyways

[gray dust]

for that 2x2, its det is det(AD-CB) given A,C commute, A invertible. and no prob

[tall surge]

sorry if D*** is invertible, then the formula holds

[neat sable]

can anyone give solution for this?

anyone?

Try row reducing the augmented matrix
@wintry steppe answer is b right?

😕

hmm fair enough

ok I'll try

[ocean sequoia]

@neat sable did you figure out your answer

[neat sable]

sadly no

i did something else and

[ocean sequoia]

ok what did you try

[neat sable]

took x as (x1 x2 x3)

[ocean sequoia]

thats a good first step

[neat sable]

then multiplied them and made algebraic expressions

[ocean sequoia]

well what do you notice about row 1 and row 3 in the matrix

[neat sable]

yeah they are same

with rhs having same values

[ocean sequoia]

so how many unknowns do we have

how many equations do we have

[neat sable]

umm

like we got 2 different equations with 1 and 3 being the same

[ocean sequoia]

yep

[neat sable]

rows

[ocean sequoia]

so we have 2 equations and 3 unknowns

[neat sable]

so I am getting unique values

yes

[ocean sequoia]

what happens if you have more unknowns than equations

[neat sable]

yeah thats a problem

but dont we have same x1

so we kinda cancel them with no problem?

and left with 2 unknowns

[ocean sequoia]

ah no

[neat sable]

xD

[ocean sequoia]

you can subtract row 3 from row 1

and row 3 becomes all zeros

[neat sable]

yeah

[ocean sequoia]

and you are left with 3 unknowns and 2 equations

[neat sable]

ok

[ocean sequoia]

which means do you know

like as a general rule of thumb

if I asked you to solve x + y = 10

how many solutions

[neat sable]

infinite

[ocean sequoia]

got it

[neat sable]

ohhh

okk

got it

[ocean sequoia]

does that make sense?

[neat sable]

yeah yeah

[ocean sequoia]

[neat sable]

thanks man

thank you so much

[ocean sequoia]

np

[neat sable]

you're a great mentor

[ocean sequoia]

hey btw generally good ediquette here is to ask for help not solutions

I know you werent trying to be rude more people will want to help you learn

instead of jsut giving answers

glad i could help

[neat sable]

ya sorry I am new here😓

[ocean sequoia]

nothing to apologize for

[neat sable]

😇

[ocean sequoia]

hang on let me double check this solution to make sure

[neat sable]

ok

[ocean sequoia]

cause i just eyeballed

[neat sable]

Actually I also want to know when this could be 1 unique solution and no solution

lemme try first

no solution when we got 2 similar equations but getting to different answers

[ocean sequoia]

ok

[neat sable]

only 1 solution when we get same number of unknowns and number of equations

am I right?

[ocean sequoia]

yes

[wintry steppe]

not necessarily

[neat sable]

where can this fail?

[ocean sequoia]

divide by 0

[wintry steppe]

for example

[ocean sequoia]

cases

[wintry steppe]

x+y=1
x+y=2

0 solutions

[ocean sequoia]

or that yea

[wintry steppe]

x+y=1
x+y=1

infinite solutions

[ocean sequoia]

sorry I was trying to stay general carla is obviously right

[neat sable]

ok

so what topic should I study to get a grip on this?

[ocean sequoia]

algebra

[neat sable]

just for these types of questions

[ocean sequoia]

thats just HS algebra and practice probably khan academy would do well

hang on did you try to solve the matrix

[neat sable]

actually I have an exam day after tomorrow and similar question to that has been since 2018

My algebra is okish level.I can understand basic to intermediate level.

[ocean sequoia]

practice your algebra

I have to go study my own stuff

btw i was right the answer is c

[neat sable]

like slimvesus told me to use row reducing method

[ocean sequoia]

ok

we can do that really fast

and yea hes right dude is smart

[neat sable]

thank you

[ocean sequoia]

lets go through it

what do we do first

[neat sable]

haha

[ocean sequoia]

which row reduction should we do

[neat sable]

row reducing the augmented matrix

[ocean sequoia]

yea what do we do first

do you know how to do it?

[neat sable]

no

I am asking that method would be enough to solve right?

[ocean sequoia]

yea

it will

go ahead and watch khan academy for that

[neat sable]

ok man

[ocean sequoia]

or google gaussian elimnation for solving systems of equations

[neat sable]

going right away

[ocean sequoia]

good luck!

[neat sable]

Thanks for the help

[ocean sequoia]

youll get it im sure

[neat sable]

haha Lets see

[ocean sequoia]

[steady cargo]

Can anyone tell me if leaving this as RREF is okay since I cant eliminate any more numbers or to be RREF do I have to move numbers as far right as possible

[gray dust]

a leading entry must be the only nonzero entry in its column

[steady cargo]

What do you mean?

Do you mean for row echelon or reduced? I tried to reduce it making it only a 1 put ended up with this

[gray dust]

rref

[steady cargo]

excl adding the constants

[gray dust]

THIS is rref

[steady cargo]

Ohh

Its already in RREF?

I wasn't sure if I could go any further with it since I know rref usually looks like

[gray dust]

there are rref criteria to have handy

[steady cargo]

See I get this The leading entry in each row must be the only non-zero number in its column.

But I can't arrive to that if you get me

It doesn't look possible, I'm not sure how any one row can only have a leading one and have only zero entries in rest

I've done rref but this just can't be solved because no 1 by itself in row 4

Well, at least I think so

It just looks impossible to me

And question states to put it into rref

[gray dust]

there are no leading entries to speak of in 0 rows

a leading entry must be the only nonzero entry in its column
only applies to rows that have leading entries

[steady cargo]

Ah okay thank you I think I get you

[nova crypt]

Quick question: For two matrices A and B, when is it true that dim(Ker(AB))=dim(Ker(BA))?

[wintry steppe]

when they commute

[pallid rampart]

When one matrix is invertible

[severe cedar]

I need to show these two determinants are equal/equivalent. How would you go about doing it? Is it possible to transpose or apply elementary matrix multiplication to switch rows/columns?

[robust pond]

just use commutative property?

[severe cedar]

hmm, how?

[dusky epoch]

wym commutative property

[robust pond]

just show generally where an inverse exists you can switch the elemnts around how they are here

[dusky epoch]

you can switch rows and cols

well

switch rows by pulling the last row up

then switch cols by pulling the last col to the left

[severe cedar]

that's what i tried to do but my teacher told me the moves were illegal

[dusky epoch]

why

[severe cedar]

bc the dimensions of the block does not fit

[dusky epoch]

i mean you can do it as a sequence of n-1 row switches

[robust pond]

am i dumb

[dusky epoch]

yes niku

[robust pond]

oh okay

[dusky epoch]

@severe cedar do n-1 adjacent row switches, pulling the [x, 1] row up one by one until it's at the top

then do n-1 adjacent col switches, pulling the now [1; -y'] column to the left until it's at the very left

the total switch count (2n-2) is even so the sign of the det does not change

or will your teacher reject that too

[robust pond]

i dont understand why what i posted is wrong

[dusky epoch]

x and y are row vectors

[severe cedar]

this is what i tried to flex, but it was rejected given that (1) the dimensions of my elementary matrices does not fit the dimensions of the blocks in the block matrix (2) the dimensions of the matrix after the second equal sign does not fit

[robust pond]

ah

[dusky epoch]

you need a different matrix sfl

uhh

n by n identity padded with a zero row and col on the left and bottom, with a 1 in the corner

[severe cedar]

after the first equal sign i mean

[dusky epoch]

$\bmqty{\bd{0}^T & I_n \ 1 & \bd{0}}$

[stoic python]

Ann:

[dusky epoch]

this is the matrix you should have had in your first equality

on the left

[severe cedar]

yeah that's what i thought

[dusky epoch]

this one will have a determinant of (-1)^(n-1)

[severe cedar]

but how about the second one

if i use the same matrix there they won't be completely equal so to speak

then i will have I_n instead of 1 in the bottom right corner

[dusky epoch]

bottom left

[severe cedar]

hmm yeah

that makes sense

however my teacher rejects the block matrix after the first equal sign

$\begin{pmatrix}
-y^{T} & I_n \
1 & x \
\end{pmatrix}.$

[stoic python]

sfl:

[dusky epoch]

wym

x and y are row vectors, so y' is a col vector

no?

what does your teacher say is wrong with this matrix

[severe cedar]

oh no, sorry

i forgot to say

they're column vectors

so y^T is row

[dusky epoch]

...

then [1, x; y', eye(n)] doesn't make sense either LOL

[severe cedar]

no i guess not

only if y^T has only one element i guess

[dusky epoch]

your teacher rejects one but not the other??

[severe cedar]

$$\begin{pmatrix}
I_n & -y^{T} \
x & 1 \
\end{pmatrix}$$

[stoic python]

sfl:

[severe cedar]

this one is well defined right

[dusky epoch]

no

you said x and y were col vectors

this doesn't match up at all

[severe cedar]

my teacher at least left a "Good!" comment on it

[dusky epoch]

what

[severe cedar]

yeah

what i am basically trying to do here is to show that det(I_n +xy^T)=det(I_m +y^Tx) = 1+y^Tx, by using block matrices and the fact that the determinant for a block matrix {A B; C D} = det(A)det(D-CA^(-1)B)

[dusky epoch]

uhh

wait. what

hm

[severe cedar]

http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/proof003.html

3.1-3.3

so obviously i can use the relation det(A)*det(D-CA^(-1)B) = det(D)*det(A-BD^(-1)C) as shown in the proof i linked

the problem is i have not shown the property det(D)*det(A-BD-1C)

and i don't think you're supposed to use that fact here

but some other technique

[warped garden]

Hello everyone, if a set of vectors are linearly independent, say {u1, u2, u3, u4}, is it true that {u1, u2} is linearly independent as well?

[native rampart]

Yes

[warped garden]

thanks!

[dawn remnant]

definition of linear independence of ${u_1,\dots,u_n}$ is that:
$$
\forall \alpha_1,\dots,\alpha_n (\alpha_1 u_i + \alpha_2 u_2 + \dots + \alpha_n u_n = 0 \implies \alpha_1, \dots, \alpha_n = 0)
$$

[stoic python]

ConfusedReptile:

[dawn remnant]

in other words, it's impossible to represent a zero-vector as a linear combination of these vectors except by having all the coefficients be zero.

From this follows that it's also impossible to do that for any subset of ${u_1,\dots,u_n}$ - otherwise you could use this to prove that the larger set is also not linearly independent.

[stoic python]

ConfusedReptile:

[warped garden]

Thanks for the detailed explanation! I somehow thought there would be a case whereby a linear combination of {u1, u2} would have a zero-vector without coefficients being zero (non-trivial) once it's taken out of {u1, u2, u3, u4}.

[dawn remnant]

suppose {u1,u2} isn't linearly independent. Then:
$$
\exists \alpha_1\exists \alpha_2 (\alpha_1 u_1 + \alpha_2 u_2 = 0)
$$
(and the alphas aren't both zero)
But then:
$$
\alpha_1 u_1 + \alpha_2 u_2 + 0 u_3 + \dots + 0 u_n= 0
$$
so the larger set isn't linearly independent either

[stoic python]

ConfusedReptile:

[warped garden]

this makes great sense!

[sleek helm]

Okay so I am probably going to regret asking this because I feel very dumb

But my pset says to take a generic jordan block matrix J

and write it as J=I+N where N is upper triangular with trivial diagonal

is this even possible? Don't we at least need like

\lambda I + N

oh wait

i can prove that is true

i wonder if it was intended that we prove that as well. its weird how its written

whatever hahaha

wait no i cant prove that

they have to be roots of unity but not 1

this is weird

im working over C 😦

[pulsar bay]

hey there

can anyone help me with this one ?

i believe the a) is y= 4 but that's as far as i've gotten

i can figure out c) myslef i just don't know how to tackle b)

[pulsar bay]

<@&286206848099549185>

[rough bobcat]

Can someone please help me

[winged belfry]

What method do you guys use to solve 4x4 determinants

[slim knot]

By hand or by computers/calculators?

[winged belfry]

By hand

Well

I guess I could just use my calculator

Idk might need to show work

So safe to know by hand

[slim knot]

I think the determinant of minors expression is easier than the sign of permutation definition but that's all I can really say. Never had to do 4x4 determinants except for good-looking ones.

[half ice]

Honestly just Laplace expansion

It's a bit of work, but not impossible to do

If there's a row of almost all 0 then Laplace is an obvious choice

[chilly solstice]

Correct ? (Problem + solution)

Remember the definition of exp for matrices:

I'm not sure about the argument from going to 2nd to 3rd line

[pallid rampart]

Diagonalizable does not mean A has n distinct eigenvalues

It means there is a basis of eigenvectors

For counterexample: consider the identity matrix. The only eigenvalue is 1 but it is clearly diagonalizable since it is diagonal

[chilly solstice]

Diagonalizable does not mean A has n distinct eigenvalues
@pallid rampart But that is what is said in the problem:

So I just played along

So the formulation of the problem is wrong? lol

[wintry steppe]

i'm pretty sure this still works for diagonisable matrices that don't have n distinct eigenvalues

[chilly solstice]

Oh, interesting 🤔

[pallid rampart]

Yeah it still works

[chilly solstice]

But the solution is actually correct no matter the formulation of the problem, rihgt?

[pallid rampart]

Yeah

Also should've said x is a real number

Because I originally thought x is a vector and then exp(Dx) would've made no sense

[chilly solstice]

yes, but that fact is stated 3 pages previusoly lmao

[pallid rampart]

Oh oof

[chilly solstice]

your are quiet observant

Problem:

[pallid rampart]

Do basically the same thing with the proof you wrote above

[pallid rampart]

Nonononono

[chilly solstice]

okokoko

[pallid rampart]

You can't do that

You can't distribute the exponent

[chilly solstice]

Oh, okay. mhh...

[pallid rampart]

You can only distribute the exponent when multiplication is commutative

Which it very much isn't for matrix multiplication

[chilly solstice]

oh, yesyes

[pallid rampart]

But still you can find something very nice about it

$(VDV\inv)^2=VDV\inv VDV\inv=VDDV\inv=VD^2V\inv$

[stoic python]

Whoever:

[pallid rampart]

$(VDV\inv)^3=VDV\inv VDV\inv VDV\inv=VDDDV\inv=VD^3V\inv$

[stoic python]

Whoever:

[pallid rampart]

Can you see a pattern

[chilly solstice]

yes

cool! 😄

[pallid rampart]

This is the very reason why diagonalization is important

Because raising the matrix to the nth power is very easy

And raising a diagonal matrix to the nth power is also very easy

[chilly solstice]

i'm probably going to use that when implementing this in python

[pallid rampart]

nice

[chilly solstice]

How do I put the sum between V and V^-1?

How have I just put V and V^-1 arround the sum?

@pallid rampart sorry for the tag.

[pallid rampart]

Well you are allowed to just do that

[chilly solstice]

says who?

[pallid rampart]

$\sum_{n=0}^\infty\frac{x^n}{n!}VD^nV\inv=V\br{\sum_{n=0}^\infty\frac{x^n}{n!}D^n}V\inv$

[stoic python]

Whoever:

[chilly solstice]

yes, but what means?

[pallid rampart]

Well

This is the distributive property

Namely $A(B+C)=AB+AC$

[stoic python]

Whoever:

[pallid rampart]

So $\sum AB_n=A\sum B_n$

[stoic python]

Whoever:

[pallid rampart]

And x^n/n! is a scalar so it commutes with everything

[chilly solstice]

oh, yes! this was it!

[pallid rampart]

Lmao alright

I guess if you want to be like more rigorous, you can write

$\sum_{n=0}^\infty\frac{x^n}{n!}VD^nV\inv=\lim_{k\to\infty}\sum_{n=0}^k\frac{x^n}{n!}VD^nV\inv=V\br{\lim_{k\to\infty}\sum_{n=0}^k\frac{x^n}{n!}D^n}V\inv=V\br{\sum_{n=0}^\infty\frac{x^n}{n!}D^n}V\inv$

[stoic python]

Whoever:

[chilly solstice]

oh,Okay,

[pallid rampart]

That's if you don't believe you can pull out a "constant" from an infinite sum

[chilly solstice]

only if it converges

[pallid rampart]

Yeah

And you know the series for matrix exponentiation always converge

[chilly solstice]

🔥

Problem 2.3:

I have used 3 days on this.

My epsilon is weak.

drunkeDrake what are you doing lmao !!

[native rampart]

But everything changed when the fire nation attacked

[wide haven]

Don't be silly @chilly solstice ε is e, and e is basically 3 and 3 as basically pi, ergo that whole thingy there in part 3 is obviously less than pi....you can just tell by looking at it!!1!

(I'm so sorry, I'm of no help)

[chilly solstice]

I do not believe that there is a problem that no one on this can't solve

[pallid rampart]

Here's a problem: Find a problem that no one can solve

[chilly solstice]

powerful talk

[wide haven]

Here's a problem: Find a problem that no one can solve
@pallid rampart create an equation that solves roots of quintic (5th degree) polynomials

no googling

[pallid rampart]

Let $f(a_0,a_1,a_2,a_3,a_4)$ be the function that gives the roots of the polynomial $a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+x^5$. Then $f$ is the answer

[stoic python]

Whoever:

[wide haven]

Okay that made me laugh

[pallid rampart]

I mean

[wide haven]

One such as the quadratic formula, I mean

[pallid rampart]

The statement is that you can't find an equation in terms of the radicals

[golden drum]

That function exist, anyways

Here's a problem: Find a problem that no one can solve
@pallid rampart

Continuum Hypothesis

[pallid rampart]

Lol

Because technically you can just create more and more of such functions

[golden drum]

Prove Continuum Hypothesis, I mean

In ZFC

[pallid rampart]

First we created the solution to $x+a=0$, then to $ax=b$, then to $x^2=a$, then to $x^n=a$, then to $x^2=-1$

[stoic python]

Whoever:

[wide haven]

@pallid rampart I was just messing with you. A quintic root equation is not only one that none of us can solve, but one that no one can solve. It's been mathematically proven that no such equation/formula can't exist

[fallen karma]

Can I post a proof to see if it makes sense

[pallid rampart]

Lmao I don't think you get what I said

[wide haven]

can't*

[pallid rampart]

Yes I do know that lmao

[wide haven]

Wait

Okay lol my bad

LOL I just saw

[golden drum]

It's like, there exist a function that choose every prime

Find it, hahaha

f : N -> P

[pallid rampart]

I know a function that outputs a prime for every natural number

f(n)=3

[golden drum]

Me too

HAHHAAHA

Suppose that f is inyective

[wide haven]

I believe this is more indicative of this general convo direction lol

I'm with it though

[warped garden]

hey, need some help with this, is the basis the span?

[wintry steppe]

a basis is a spanning set that is linearly independent

[fallen karma]

Yes@wintry steppe

[half forge]

what textbook are you using zynoZII?

[warped garden]

does this mean dim(W) is 4?

@half forge linear algebra: concepts and techniques on euclidean spaces

[wintry steppe]

are those vectors linearly independent?

if they are then dim(W) is 4. else no.

[half forge]

how did you know the dim was 4

[fallen karma]

Dim W should be three I think

W is spanned by the first second and fourth

[half forge]

@warped garden whats the name of the author?

[warped garden]

ah shit. i realise why.

there's a redundant vector...

[severe cedar]

The third column is a combination of the first and second

So you can exclude it

Dim(w) = 3

[wintry steppe]

you don't need to verify any of that, you can get the answer directly from the RREF.

[warped garden]

it was a silly mistake, thanks for your help guys!

🙂

just realise there's a non-pivot, silly me

[dusky epoch]

i think you're overcomplicating it ngl there's a way easier proof

[fallen karma]

Just bear with me

It's not the most parsimonious but is it convincing?

[dusky epoch]

idk why you needed to use induction tbh

this all feels unnecessary

[fallen karma]

I use induction bc I've only demonstrated the case for nonzero constant polynomials that the degrees of the images of T and T inverse are the same as the degree of the polynomial

So thought I would build it from there

[dusky epoch]

are you not allowed to use the result that dim T(V) = dim V for T an injective map and V a findim space?

[fallen karma]

But the space of all polynomials is infinite

[dusky epoch]

i never said you had to consider T(the entire space)

write $V_n := \mathrm{span}{1,x,\dots,x^n}$. (note that $\dim V_n = n+1$.)

now assume there exists a polynomial $p^$ with $\deg(Tp^) < \deg(p^) =: n^$. clearly then we have $T(V_{n^}) \subseteq V_{n^-1}$ (since ${1, x, \dots, x^{n^-1}, p^}$ is a basis for $V_{n^*}$) but this contradicts $T$ being injective

you haven't answered me on this tho

are you not allowed to use the result that dim T(V) = dim V for T an injective map and V a findim space?

cause maybe your prof doesn't want you to use that

[stoic python]

Ann:

[dusky epoch]

fixed a typo

[fallen karma]

Believe it or not I do this for fun lol I'm not in school

[dusky epoch]

oh ok

[fallen karma]

So I come here for feedback which I appreciate btw

[dusky epoch]

i mean, yeah, you made a massive detour with the induction thing

's all i can say

[fallen karma]

I dont see the contradiction in your proof

[stark acorn]

Im confused on how to do this;

[pale mortar]

why does the span of the columns of a matrix change when you reduce it (go to echelon form)

[limber sierra]

row operations preserve spans of rows, but there's no guarantee they preserve spans of columns

consider, for example, $\begin{pmatrix}1\1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

clearly the span of this column is the span of $\begin{pmatrix}1\1\end{pmatrix}$, i.e. all matrices of the form $\begin{pmatrix}a\a\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

but upon row reduction, we get $\begin{pmatrix}1\0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

and the span of this column is certainly different

it's the matrices of the form $\begin{pmatrix}a\0\end{pmatrix}$, not those of the form $\begin{pmatrix}a\a\end{pmatrix}$

[stoic python]

Namington:

[calm hamlet]

@stark acorn there is maybe an easier method, but here is mine :
If it works for any vectors, it works for a=3.x and b=2sqrt(2).x +2sqrt(2).y with (x,y) the canonic base

[stark acorn]

What does canonic mean

[calm hamlet]

Oh it's maybe only a thing in my language
The standard base of R²

[stark acorn]

Can you write it out? Sorry im confused about what the .x notation is

@calm hamlet

[calm hamlet]

$\overrightarrow{a} =3\overrightarrow {x} $ and $\overrightarrow{b} = 2\sqrt{2}\overrightarrow {x} + 2\sqrt{2}\overrightarrow {y} $

[stoic python]

Svet L'octogone:

[calm hamlet]

@stark acorn

[stark acorn]

Im stil confused

How do you find out what X is?

[calm hamlet]

When you draw your vectors, you draw them in a base right?

[stark acorn]

yi

[calm hamlet]

$\vec{x} \ \text{and}\ \vec{y}$ are just vectors of the standard base of R² that you use probably all the time

[stoic python]

Svet L'octogone:

[stark acorn]

ah I see

would the cross product of 2a crossed with a

be 0?

[raw sand]

yeah

[stark acorn]

How would this formula change if b was negative:

[raw sand]

just use the negative values?

[stark acorn]

would it be - [ |a| |b| sin(theta) ]

[raw sand]

what do you mean by b is negative

you dont need to change the formula

[stark acorn]

I have A x -B

[raw sand]

why dont you just distribute the negative into the vector

and use the new vector as b

?

[stark acorn]

What do you mean?

can you elaborate

we dont know what vector b

is

[raw sand]

can i see what you are trying to do?

[stark acorn]

just its magnitude

Still tryng to do this one

i multiplied eveerything out

and was left with (a x -b) + (b x 2a)

[chilly solstice]

guess who solved the super hard question from before ?
https://discordapp.com/channels/268882317391429632/540211747613704221/762378570772512768
chasing in on three days of labour @pallid rampart

[stark acorn]

@raw sand

[raw sand]

yeah just do what you were doing

sorry i was really confused by your question

[stark acorn]

im unsure

how to proceed after i distribuyte

Im goiung to post into questions

[chilly solstice]

How do one check if A is diagonizable? It should be simple, so it can be implemented in a Python.

[raw sand]

check for each eigenvalue that the dimension of the eigenspace is equal to the multiplicity of the eigenvalue perhaps

[chilly solstice]

Thanks!

[half forge]

can someone help me with latex

i wanna move that next to the let statement

[wintry steppe]

what is your latex code

it should work with

Let $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$

[half forge]

\begin{equation*}
A =\begin{pmatrix}
1 & 1 \\
1 & 1\\

\end{pmatrix}
\end{equation*}
\end{statement}```

[wintry steppe]

don't put it in an equation environment if you want the matrix to be inline

[half forge]

oh okay

how do i do this in latex

[wintry steppe]

M(\mathbb{R})

[half forge]

thank you

[pallid rampart]

You're welcome

[warm kite]

Can anyone tell me what I did wrong?

Ignore the equation on the top right

[vernal pebble]

I can't think of another way to solve this besides creating a dummy matrix A = [a b c; d e f; g h i] and setting up a system of equations to manually find a,b,c,d,e,f,g,h,i

is there an easier/more time efficient way of solving this?

[wintry steppe]

🤔

usually when i solve problems like this i solve by hand

like finding the elements' matrix or something like that

[vernal pebble]

yeah my initial thought as well

but doing it by hand takes way too long. i'm thinking there's probably a trick I'm not seeing here

[wintry steppe]

@warm kite i'll try to write it

moment

[warm kite]

@wintry steppe thank you so much I have no idea what’s wrong

[vernal pebble]

@warm kite i can solve it using physics lol

but that probably won't help too much lmao

[warm kite]

@vernal pebble haha I know there is an equation, I wrote it there but my teacher wan#yes it solved the trig way

And idk what I’m doing wrong

[wintry steppe]

@warm kite is there no time given?

[warm kite]

Nope

That’s all the q says

I know the speed is (c1 + c2)^1/2

@wintry steppe +

?

[wintry steppe]

@warm kite

I found it i guess

first I did: Voy = Vox (velocity speed of x is equal to velocity speed of y, since tg(45°) = 1)

so, by that the entire velocity is:
V² = (Voy)² + (Vox)² ("vectorialy" it means V = Voy + Vox, since V is initial velocity)

We reach that Vox = 10/t by the definition of velocity (size divided by time)

and our Vox is towards x, and it ends at 10m from 0m

and we find another value by another definition (the definition of acceleration, that give us: V = Vo + at, in this case 0 = Vox - g(t/2), because its displacement is just one half, and due to that the time is one half too, by symmetry)

[warm kite]

Hmmmmmmm I still don’t get it lol, the answer is sqrt (980)

[wintry steppe]

the remainder part i think you can get, it's thorough algebraic

wait what?

[warm kite]

Yeah

Idk why

This is how we solved a previous similar problem @wintry steppe

[fallen karma]

@vernal pebble doesn't take too long to do what you said, for instance let your first row be 0 0 1, second row be -1 2 0 and so on

[wintry steppe]

wait i did a mistake there

@warm kite

almost the same

(980)^(1/2) is near 30

that happens

[warm kite]

Ohh sorry I meant sqrt 98

[wintry steppe]

so yeah:

7(2)^(1/2) = (49.2)^(1/2) = (98)^(1/2)@warm kite

[warm kite]

Wait where e V0y -gt/2 come from

[wintry steppe]

@warm kite from acceleration

a = dV/dt

adt = dV

[warm kite]

Ohhh yes yes

[wintry steppe]

V - Vo = a.t
V = Vo + at

[warm kite]

@wintry steppe I think I figured it out

Thank you so much

!!

[wintry steppe]

🙏

[halcyon pollen]

need some help anyone here i can understand the geometry of linear equations

if (2x - y = 0) and ( -x + 2y = 3) how does he plot that in the graph?

please some one help me this is the reason i kinda failed in math i dont wanna do that again

[fallen karma]

They're equations for lines do you remember how to do those

[wintry steppe]

@halcyon pollen wdym?

[halcyon pollen]

how do they plot it man?

[limber sierra]

let's take some example points

for example, if x = 0, then where should y be on each line?

2x - y = 0, so we replace x with 0

2*0 - y = 0 - y = -y = 0

so -y = 0, i.e. y = 0

as for the other line

-x + 2y = 3, replace the x with 0

-0 + 2y = 2y = 3

2y = 3, so dividing both sides by 2

y = 3/2

so one point of the first line is at (0, 0), and one point on the second line is at (0, 3/2)

we can find another point by using a different x value, say x = 1

then 2x - y = 0 becomes 2 - y = 0

so y = 2

meanwhile, -x + 2y = 3 becomes -1 + 2y = 3, so 2y = 4; i.e. y = 2 again

so our points are (1, 2) and (1, 2)

so, our first line goes through (0,0) and (1,2)

our second line goes through (0, 3/2) and (1/2)

both of these are linear (straight lines), so we can just "connect the dots"

and voila.

my choices of testing where x = 0 and x = 1 were arbitrary

i couldve chosen other points

[halcyon pollen]

woah thanks man

that is really great of you

@limber sierra got an doubt tho can you help me?

they solve that linear equations they gave

so the first line gives (0,0) and (0,3/2)

[old flame]

@native rampart Is there any hints on how to start with (I-AB) is invertible, not sure how to continue with this first step