Question

In lecture 12, Prof Strang says that A'CA is symmetric ( where A' is transpose of A)- but isn't this only possible if C is symmetric?

Answer 1

Probably you just try it out by yourself, in the easiest case, a 2x2 Matrix. Then you will see, it really doesn't depent on C.
For if you get stuck somewhere here it is:
\[A ^{T}CA=\left[\begin{matrix}a11 & a21 \\ a12 & a22\end{matrix}\right]\left[\begin{matrix}c11 & c12 \\ c21 & c22\end{matrix}\right]\left[\begin{matrix}a11 & a12 \\ a21 & a22\end{matrix}\right]\]
Let's call the first colomn in A a1 and the secound a2, as well as in C you got c1 and c2.
So know you have:\[\left(\begin{matrix}a1 \\ a2\end{matrix}\right) (c1 c2) (a1 a2)=\left[\begin{matrix}a1c1 & a1c2 \\ a2c1 & a2c2\end{matrix}\right](a1 a2) = \left[\begin{matrix}a1a1c1+a1a1c2 & a2a1c1+a2a1c2 \\ a2a1c1+a1a2c2 & a2a2c1+a2a2c1\end{matrix}\right]\]
As you can see, we didn't use any special property of C to solve this problem, but still the new Matrix is a symmetrical one.
So to answer your question: No, C doesnt't have to be symetrical, but \[A ^{T}CA\] always will be symmetrical.

Answer 2

Oh, can you read the full answer? Here is the missing and most important part (what at least I can't read) \[A^{T}CA =\left[\begin{matrix}a1a1c1+a1a1c2 & a2a1c1+a2a1c2 \\ a1a2c1+a1a2c2 & a2a2c1+a2a2c2\end{matrix}\right]\]