# Thread: Symmetry of two similar matrices

1. ## Symmetry of two similar matrices

Hi!

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

The rotation matrix:
$\displaystyle P = \left[ \begin{array}{c c c c c c c} 1 & 0 & \cdots & \cdots & \cdots & 0 & 0 \\ 0 & 0 & \cdots & & & \cdots & \cdots \\ \cdots & \cdots & cos \theta & \cdots & sin \theta & \cdots & \cdots \\ \cdots & \cdots & & & & \cdots & \cdots \\ \cdots & \cdots & -sin \theta & \cdots & cos \theta & \cdots & \cdots \\ \cdots & \cdots & & & & \cdots & \cdots \\ 0 & 0 & \cdots & \cdots & \cdots & 0 & 1 \end{array} \right]$
where $\displaystyle P_{rr} = P_{ss} = cos \theta, P_{rs} = -P_{sr} = sin \theta, P_{ii} = 1$ (for $\displaystyle i ~= r, s$) and else $\displaystyle 0$.

I don't know where to start!

Help is greatly appreciated!

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

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

3. I think I got it!

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

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

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

Hence, B is symmetric when A is symmetric.

Did I do it correctly?

4. Yep, that seems perfect