Geometrical Meaning of Linear Algebra operations

Question

Need some intuition, I'm new to the world of Linear Algebra... (I know how to do most of the linear algebra operations but don't know the complete geometric meaning of them).

So, I have a $3\times3$ matrix A, which I'm multiplying on the left with the rotation matrix in the z direction, and on the right with the inverse rotation matrix in the z direction.

That is, $$ T = \begin{bmatrix} \cos(\theta) & \sin(\theta) & 0 \\ -\sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

And, I want to compute:

$$ B = T\times A\times T'$$.

So, when I multiply $T\times A$, I get all the columns of $A$ rotated clockwise by $\theta$ about the $z$-axis.

Now, what do I get when I multiply this new matrix, $T\times A$ with $T'$?

What does it mean geometrically?

(I know that $A$ and $B$ have the same trace, and although I can do it through computation, I need to feel it geometrically to understand it thoroughly.)

Answer 1

Assuming that by $T'$ you mean the transpose or inverse of $T$ (those are the same in this specific case), you're thinking about the multiplication in the wrong order. Let's apply $B$ to a vector and see what we get:

$T'$ rotates the vector by $-\theta$. However, it is more common to turn this around and say that it rotates the basis by $\theta$ (i.e. it rotates the coordinate system without moving any vectors). Then we apply $A$ in this new coordinate system, and then $T$ rotates the coordinate system back.

All in all, $B$ is what you get if you look at what the linear transformation $A$ does to any points in your space, but impose a different coordinate system.

For instance, let $$A = \begin{bmatrix}2&0&0\\0&1&0\\0&0&1\end{bmatrix}$$Then in our current coordinate system, $A$ stretches space along the $x$-axis.

Now, put a new coordinate system on our space. One such that the coordinate axes are rotated $-\theta$ about the $z$-axis compared to the old one. Then the same linear transformation doesn't stretch space along the $x$-axis any more, but rather along some diagonal line in the $xy$-plane. This same linear transformation in the new coordinate system is represented by the matrix $B$.

Answer 2

Away from linear algebra...

Suppose you have bot that can read messages in English, and write a response. You want to create a bot that can read and write messages in Russian, but you don't want to write an entirely new bot. Instead you write a translation program.

Algebraically you could say $(f\circ g\circ f^{-1})(x)$

$f^{-1}$ translates message $x$ from Russian to English. $g$ takes a message in in English and creates a response in English. $f^{-1}$ translates the response back to Russian.

Or more abstractly, $f^{-1}$ translates object $x$ into some form such that $g$ can now act on it. $f$ translates it back to its original form.

On to linear algebra. Remember that matrices are linear transformations, which makes them a special class of function.

$B = PAP^{-1}$

$P$ represents a change of basis from the basis of $A$ to the basis of $B$

$P^{-1}$ takes in the object in terms of the basis of $B$ and rotates it so that it is suitable to work on in terms of the basis of $A.$

$A$ operates on it. $P$ takes the output and puts it back into $B$'s basis.

$A$ and $B$ are similar matrices.

For example, suppose $A$ compresses horizontally and dilates vertically.

$P^{-1}$ takes an object that is in a rotated frame. Squares it up for $A$ to act on it. $A$ does its action. $P$ rotates the transformed object back to the rotated frame

$PAP^{-1}$ would make A compress and dilate to the same degree, but it would be compressing and dilating in directions that have been rotated.