Are < b m a t r i x > and < b m a t r i x > orthogonal in ℳ2 × 2 endowed with the inner product

Step 1: Recall that two matrices M and N are orthogonal if their inner product is zero, i.e., (M

Are < b m a t r i x > and < b m a t r i x

Are $\begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix}$ and $\begin{bmatrix} 6 & 6 \\ -3 & -3 \end{bmatrix}$ orthogonal in $\mathcal{M}_{2 \times 2}$ endowed with the inner product \\ $<M, N> = (MN^T)_{1,1} + (MN^T)_{2,2}$?

Transcript

00:01 So let's call i -m -n the quantity integral from 0 to pi of cosine m -t, sine n -t d -t, where m and n are positive integers.

00:12 And of course, our goal is to show that i -m -n is zero.

00:15 So we do this integral by parts.

00:19 So we see this as the integral from 0 to pi of the derivative of sine mt divided by m.

00:26 And of course, this is just cosine m -t.

00:28 That multiplies sine n t d t and now by parts this is sine mt divided by m that multiplies sign and t evaluated from zero to pi minus the integral from zero to pi of sine mt divided by m that multiplies the derivative of their function so n cosine n d t d t now this all first bracket is zero and then we have minus the integral the integral to pye let's isolate i n divided by m and then we have sine mt cosine and t let's rewrite in the other side the other way so cosine n t sine m d t and we recognize that this integral is nothing else but minus n divided by m i n so observe that i mn is equal to something i nm so we're switching the indices but we can do the same integral so i mn by parts by seeing the other function as a derivative.

01:33 So this is the integral from 0 to pi of cosine mt.

01:37 And then sine nt, we see it as the derivative of minus cosine nt divided by n d t.

01:44 And now again by parts, this is cosine mt that multiplies minus cosine and t divided by n, evaluated from 0 to to pi, minus.

01:52 So the derivative of the first function, which is minus m sine mt, then multiplies minus cosine and t divided by n d t and again the whole bracket is zero and then we have minus the integral from zero to two minuses cancel each other so we have m divided by n cosine n t sine mt and again we recognize this expression to be just minus m divided by n i and m so we have two different expressions for i mn let's denote the first one they obtained up there in red by one and these other equality down here in green as 2 and we're going to use them right away so minus n divided by m i nm is by the first equality in red i mn which is by equality 2 in green equal to minus m divided by n i nm and this is true for every n and m greater equal than 1 in particular we we write i and m as by moving the n divided by m on the right -hand side, m squared that divides n squared, i and m.

03:10 And this expression in blue is a number which depends on m and n...

Question

Are $\begin{bmatrix} 2 & 0 \\ 5 & 4 \end{bmatrix}$ and $\begin{bmatrix} 6 & 6 \\ -3 & -3 \end{bmatrix}$ orthogonal in $\mathcal{M}_{2 \times 2}$ endowed with the inner product \\ $<M, N> = (MN^T)_{1,1} + (MN^T)_{2,2}$?

Please give Ace some feedback