the no-normie zone - 2017-08-29

Anonymous

17:30

@BalarkaSen Hey!

Anonymous

Let's continue :D

Anonymous

Diagonalization stuff today!

Balarka Sen

Ah right. I was typing up some calculations to send to Ted, sorry about forgetting.

Ok, diagonalization it is.

Anonymous

You may complete sending your calculation to Ted :)

Anonymous

We can continue a bit later if you wish

Balarka Sen

17:35

It'd take some time, don't worry about it. I have the whole night to do that.

LaTeXing is an annoying this to do.

Anonymous

Write it on paper and send a snapshot :P

Balarka Sen

lol. the most lazy thing to do.

Anonymous

I wish they come up with a software to texify pictures

Balarka Sen

So we mentioned yesterday that an nxn matrix has at most n eigenvalues, right?

Anonymous

Yes

Anonymous

17:36

Because it is a polynomial of order $n$

Balarka Sen

"it" being the characteristic polynomial.

Anonymous

Yes!

Anonymous

$p(t)$

Balarka Sen

So here's a nice way to think about this.

Suppose that $T : V \to V$ is a linear operator and $W$ is a $T$-invariant subspace of $V$; which is to say, $W \subset V$ is a vector subspace such that $TW \subset W$.

Anonymous

Alright so far

Balarka Sen

17:40

$A$ be the matrix of $T$ with respect to some arbitrary basis we don't care about.

Now, choose a basis $\{w_1, \cdots, w_k\}$ of $W$.

Anonymous

Okay

Balarka Sen

Construct a basis of $V$ out of this basis of $W$ by choosing "complementary basis vectors" $v_{1}, \cdots, v_{n - k}$ and appending it to that basis of $W$

So that our new basis of $V$ is $\{w_1, \cdots, w_k, v_{1}, \cdots, v_{n-k}\}$.

Anonymous

@BalarkaSen Okay..makes sense

Balarka Sen

Let us inspect the matrix of $T$ in this new basis

Anonymous

Okay?

Anonymous

17:45

We have chosen the basis of V now

Balarka Sen

I wonder if you can write down the matrix for me?

Anonymous

Umm...but I first need to know what operation $T$ does given a vector in $V$

Balarka Sen

Nope. :)

What is, by definition, the matrix of $T$ in the basis $\{w_1, \cdots, w_k, v_1, \cdots, v_{n-k}\}$?

Anonymous

$[T(w_1)T(w_2)....T(v_1)...T(v_{n-k})]$ ?

Anonymous

That? ^

Balarka Sen

17:48

Yes, good.

Here $T(-)$ is not an element of $V$, mind, but the vector representation of it in the basis of interest.

Anonymous

What is T(-) ?

Balarka Sen

T(whatever vector here) I mean

T(w_1), T(w_2), ...

Anonymous

How's that possible? Isn't it $T:V\rightarrow V$

Anonymous

I must be forgetting something

Balarka Sen

Yes, but we're using different notation.

Anonymous

17:51

I'm confused....

Balarka Sen

$T(w_1)$ is an element of $V$.

But it can be represented as an nx1 column vector in $\Bbb R^n$. (How?)

That is what is meant by $T(w_1)$ in $[T(w_1) \, T(w_2) \, \cdots ]$

Anonymous

@BalarkaSen By using the standard basis?

Balarka Sen

How so?

Anonymous

@BalarkaSen Wait, you just said T(-) is not an element of V

Balarka Sen

In this context.

I am just asking you how to write down matrix of a linear operator, man

I give you a linear operator $T : V \to V$, I give you a basis $B = \{b_1, b_2, \cdots, b_n\}$. How do you write down a matrix?

This is the down-to-earth question.

Anonymous

17:56

I already wrote it, didn't it? [T(b_1)T(b_2)...]

Balarka Sen

What does T(b_1) mean?

Anonymous

We are applying the linear operator T on the vector [1,0,0,0,....] in the basis you stated

Anonymous

i.e. $b_1,b_2,b_3,...,b_n$ basis

Balarka Sen

??? what does it mean to apply T on [1, 0, ...]? [1, 0, ...] is not an element of V

See, to say "[T(b1) T(b2) ... T(bn)] is a matrix", you need to say T(bi) is an nx1 column vector in R^n

Do you see that?

Anonymous

Oh...I was thinking of it as a column vector so I didn't mention that explicitly

Anonymous

18:00

But yes...I get it

Balarka Sen

Ok. So how is T(bi) a column vector in R^n?

It's an element of V, like you said.

How do you interpret it as a column vector?

Anonymous

A n*n matrix (corresponding to the operator T) multiplied by a n*1 column vector $b_i$ would surely give a n*1 column vector

Balarka Sen

bi is not an nx1 column vector.....

it's an element of V

an arbitrary vector space

Anonymous

Yes, but we can represent it as a column vector isn't it?

Balarka Sen

How?

That is the question I have been asking you for the last 15 minutes

How do you represent an arbitrary vector in V as an n x 1 column vector in R^n?

Anonymous

18:03

What's that word...isomorphic with $R^n$

Anonymous

Lemme type more explicitly

Anonymous

One sec

Balarka Sen

Yes, good. Can you tell me how to explicitly write a vector v in V as an nx1 column vector without referencing the isomorphism explicitly?

I just want to know what formula I am going to use to translate between V and R^n

Anonymous

We define a map $f:V\rightarrow \Bbb R^n$

Anonymous

With basis $\{v_1,\dots,v_n\}$

Anonymous

18:06

$$f(a_1v_1+\dots+a_nv_n)=(a_1,\dots,a_n).$$

Balarka Sen

Alright, very good.

Thanks.

Anonymous

Phew :P

Balarka Sen

@Blue Now let us get back to the earlier question. $T : V \to V$ is a linear operator, $W$ is a $T$-invariant subspace and I have the basis $\{w_1, \cdots, w_k\}$ of $W$, which we extend to a basis $\{w_1, \cdots, w_k, v_1, \cdots, v_{n - k}\}$ of $V$.

The matrix of $T$ in this new basis is, like you said, $B = [T(w_1) \, \cdots \, T(w_k) \, T(v_1) \, \cdots \, T(v_{n-k})]$.

Anonymous

Yes

Balarka Sen

Where $T(w_1)$ eg is, by abuse of notation, the nx1 column vector in $\Bbb R^n$ corresponding to $T(w_1) \in V$ using the isomorphism (with basis $\{w_1, \cdots, v_{n-k}\}$) you just wrote down.

$T(w_1)$ in this context, in the matrix $B$, is not really the element $T(w_1) \in V$, but the coordinate representation of it in $\Bbb R^n$.

It is indeed the unfortunate choice of notation.

We should have perhaps said $B = [f(T(w_1)) \, f(T(w_2)) \, \cdots]$

Where $f : V \to \Bbb R^n$ is the isomorphism coming from the basis $\{w_1, w_2, \cdots \}$

Anonymous

18:11

@BalarkaSen Agreed!

Balarka Sen

Alright, good issue we cleared up here.

Now, what is the coordinate representation of $T(w_1)$ in the basis $\{w_1, \cdots, w_k, v_1, \cdots, v_{n-k}\}$?

Remember that $W$ is a $T$-invariant subspace ($TW \subset W$)

You can say something special about the coordinate representation of $T(w_i)$'s in this basis.

Anonymous

$T(w_i)$ would be an element of $W$ ?

Balarka Sen

Correct. What does that mean about it's coordinate repn?

Anonymous

Then it can be represented in $\Bbb R^k$ ?

Anonymous

As $W$ has only basis $w_1,w_2,...,w_k$

Anonymous

18:15

I mean as an isomorphism obviously

Balarka Sen

Indeed. But we represent it in $\Bbb R^n$, here, writing $T(w_i) = a_1 w_1 + \cdots + a_k w_k + a_{k+1} v_1 + \cdots + a_n v_{n-k}$.

What does that say about the column vector $[a_1, \cdots, a_k, a_{k+1}, \cdots, a_n]^T$?

(I am writing the T above to denote transpose; I wrote a row vector but I consider it as a column vector)

Anonymous

$a_{k+1},...,a_n$ would be all $0$...because we don't need $v_1,...,v_{n-k}$

Balarka Sen

Cool, exactly right.

Anonymous

yay :)

Balarka Sen

That means the matrix $B$ can be written as a block matrix like this:

$$B = \begin{bmatrix} \mathbf{M} & \mathbf{N} \\ \mathbf{O} & \mathbf{L} \end{bmatrix}$$

Where $\mathbf{O}$ is the zero matrix of dimension $(n-k) \times k$

Do you agree?

Because for i = 1, ..., k, the i-th column in B has zeroes in the last n - k entries, because of what you said.

Anonymous

18:20

I'm confused....you are writing a matrix $O$ inside a matrix $B$ ?

Balarka Sen

@Blue Yeah, I'm saying you'll get a $(n-k) \times k$ dimensional submatrix consisting of zeroes in the matrix $B$, in the lower left corner.

1 min ago, by Balarka Sen

Because for i = 1, ..., k, the i-th column in B has zeroes in the last n - k entries, because of what you said.

That is the relevant statement. It's just a matter of seeing things the right way

Anonymous

Okay...and what are $M$, $N$, $L$ ?

Anonymous

I really need to take an example at this point it seems

Balarka Sen

Do you understand why there's a submatrix full of 0's in the lower left corner?

M, N, L are the rest of the stuff the matrix of B contains which we can't extract further information about

Anonymous

@BalarkaSen I understand the concept. But I need to see a solid example to actually see what's happening there

Balarka Sen

18:24

It'd be easier to explain if we were doing this face-to-face with pen and paper so I'd be able to draw you the matrix B in full

@Blue Ok, that is easy. Let's see.

Anonymous

Let's just take $\Bbb R^3$

Balarka Sen

Too hard.

How about something dumb like $A = \begin{bmatrix} 2 & 5 \\ 0 & 8 \end{bmatrix}$?

That has an invariant subspace $W$ as the x-axis in $\Bbb R^2$.

Because $A \cdot (x, 0) = (2x, 0)$

So $A \cdot W \subset W$

Anonymous

$\begin{bmatrix} 2 & 5 \\ 0 & 8 \end{bmatrix}\begin{bmatrix} x \\ 0 \end{bmatrix}=\begin{bmatrix} 2x \\ 0 \end{bmatrix}$

Anonymous

Give me just 2 minutes

Balarka Sen

Since $\{e_1\}$ is a basis of $W$ and the standard basis $\{e_1, e_2\}$ is precisely the appendment of that basis to the basis of $V = \Bbb R^2$, we just need to see what the matrix $A$ as is looks like.

And indeed: it has a (2-1)x1 dimensional submatrix consisting of 0's in the lower left corner :P

@Blue Sure

Anonymous

18:31

huh...got it now :)

Anonymous

Was just going through the discussion once more

Anonymous

Go on :D

Balarka Sen

In general you would not be so lucky, though. So you have to change basis to "{a basis of the invariant subspace W, more stuff}" to get that form for A

But upto similarity/conjugation (that's what basechange does to matrices), you'd get a submatrix of 0's in the lower left corner of A

@Blue So you see that knowing about invariant subspaces of linear operators tells us what their matrices would look like!

Anonymous

@BalarkaSen Yep!

Balarka Sen

Let's apply this to the realm of eigen"stuff"

Anonymous

18:35

BTW I think I have to leave now :P I have to complete writing my physics practical file for tomorrow morning class....

Anonymous

It's such crap :P

Balarka Sen

Ah, no matter, we can finish later.

Anonymous

Okay...thanks :) Hope we can move on to diagonalization tomorrow :D

Balarka Sen

Me too.

Anonymous

See you tomorrow then :) Bbye

Balarka Sen

18:36

Bye.

Transcript for

♦ $Mathematics$ the no-normie zone

Transcript for

♦ the no-normie zone

♦ $Mathematics$ the no-normie zone