Linear algebra happy fun time with Dan and Bernardo

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Bernard Meurer

03:12

OKAY

@DanielSank Let's rock it

DanielSank

Bernard Meurer

So, let $V$ be a real euclidean space of dimension $\mathbb R^n$

let $B=(b_1,b_2,\ldots,b_n)$ be a basis of $V$

DanielSank

Bernard Meurer

let $x,y\in V$ s.t. $x_B = (\alpha_1, \alpha_2, \ldots, \alpha_n)$ and $y_B = (\beta_1, \beta_2, \ldots, \beta_n)$

DanielSank

ARG!

$$x_B = (x_{B1}, x_{B2},\ldots)$$

Why the hell would we use $\alpha$ for elements of $x$ in basis $B$?

>:(

Bernard Meurer

03:16

So it doesn't overload the notation with what I'm about to right I think

DanielSank

@BernardMeurer fine, whatever.

Bernard Meurer

$\langle x, y\rangle = \langle\alpha_1 b_1 + \alpha_2 b_2 + \ldots + \alpha_n b_n, \beta_1 b_1+ \beta_2 b_2+ \ldots+ \beta_n b_n\rangle$

DanielSank

That looks wrong.

Oh oh oh

Yeah that equation doesn't make sense.

The thing on the right is an inner product, but there are four commas in it.

I think you turned some plus signs into commas.

Bernard Meurer

Dammit, my bad

yeah

One sec, I'm making the gran-finale

@dmckee Thanks for the fix!

(still writing my gran-finale)

dmckee

@BernardMeurer I edited that line at Bernard's request.

DanielSank

03:26

@dmckee Thank you, oh grand moderator with fancy powers and probably a fancy hat too.

dmckee

@DanielSank Very soon we all get fancy hats for a few weeks.

DanielSank

$$x = \sum x^b_i b_i \qquad y = \sum y^b_j b_j$$

So $$\langle x, y \rangle = \sum \sum x_i^b x_j^b \langle b_i, b_j \rangle$$

Which is the same as $$x^t G y$$ where $G_{ij} = \langle b_i, b_j \rangle$.

Bernard Meurer

which subsequently yield us $$\langle x, y\rangle = \begin{bmatrix}\beta_1 & \beta_2\ & \ldots & \beta_n \end{bmatrix}\begin{bmatrix}\langle b_1,b_1 \rangle & \langle b_2, b_1\rangle & \ldots & \langle b_n, b_1\rangle \\ \langle b_1,b_2 \rangle & \langle b_2,b_2 \rangle & \ldots & \langle b_n,b_2 \rangle \\ \vdots \\ \langle b_1,b_n \rangle & \langle b_2,b_n \rangle & \ldots & \langle b_n,b_n \rangle\end{bmatrix}\begin{bmatrix}\alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n
\end{bmatrix}$$

DanielSank

lololol

Yes!

Bernard Meurer

Where the middle matrix is denoted $G$ and is called the Gram or Gramian matrix

And we have the properties:$$G = G^T \\ \forall x\neq 0 \in V; \quad x_B^T Gx_B > 0$$

And from this we can define that a real, symmetric, matrix will be called a positive-definite matrix when all of it's eigenvalues are positive.

DanielSank

03:37

@BernardMeurer Yes.

Uh huh.

Bernard Meurer

And with this we can define a non-usual inner product as:

Let $A$ be a real symmetric matrix of order $n$. The following statements are equivalent. $$\text{(i) The equation }\langle x, y\rangle = y^T Ax \text{ defines an inner product in $\mathbb R^n$} \\ \text{(ii) $A$ is a definite-positive matrix.}$$

I broke it

There

DanielSank

Yah, that's right.

Bernard Meurer

Ta-daa we now have a nice definition of any inner product

OH GOD

I gotta be up in 4 hours

@DanielSank We will resume this eventually, but now we have a nice definition of inner product here :)

DanielSank

@BernardMeurer Yep. Nice.

Good job!