Thread: Symmetry of two similar matrices

1. Symmetry of two similar matrices

Hi!

Assume $B = P^TAP$, where $P$ is the rotation matrix (below) with angle $\theta$ and position $r, s$. Show that $B$ is symmetric when $A$ is, and that we then also have: $B_{rs} = B_{sr} = sin\theta cos\theta (A_{rr} - A{ss}) + (cos^2\theta - sin^2 \theta) A_{rs}$.

The rotation matrix:
$
P =
\left[
\begin{array}{c c c c c c c}
1 & 0 & \cdots & \cdots & \cdots & 0 & 0 \\
0 & 0 & \cdots & & & \cdots & \cdots \\
\cdots & \cdots & cos \theta & \cdots & sin \theta & \cdots & \cdots \\
\cdots & \cdots & & & & \cdots & \cdots \\
\cdots & \cdots & -sin \theta & \cdots & cos \theta & \cdots & \cdots \\
\cdots & \cdots & & & & \cdots & \cdots \\
0 & 0 & \cdots & \cdots & \cdots & 0 & 1
\end{array}
\right]
$

where $P_{rr} = P_{ss} = cos \theta, P_{rs} = -P_{sr} = sin \theta, P_{ii} = 1$ (for $i ~= r, s$) and else $0$.

I don't know where to start!

Help is greatly appreciated!

2. Woho! I produced the
$B_{rs} = B_{sr} = sin\theta cos\theta (A_{rr} - A{ss}) + (cos^2\theta - sin^2 \theta) A_{rs}$
simply by multiplying $A$ from both sides with $P^{-1}$ and $P$.

So what's left is to show that B is symmetric when A is... Suggestions on that?

3. I think I got it!

If $A$ is symmetric, then $A = A^T$. If $B$ is to be symmetric, then $B = B^T$ must also be true. Hence I need to show that $B^T = B$ when $A$ is assumed symmetric.

Assume $A$ is symmetric.
We have: $B = P^TAP$. We then get:

$B^T = (P^TAP)^T = P^TA^TP^{TT} = P^TAP = B$

Hence, B is symmetric when A is symmetric.

Did I do it correctly?

4. Yep, that seems perfect