Let Q be orthogonal. Show that Q^tAQ has the same characteristic polynomial. as A.

Question

Do I have to use the diagonalization of A?

No. $I = Q^tQ$, so just use the fact that determinant is multiplicative. — Prahlad Vaidyanathan, Mar 03 '15 at 05:09

score 4 · Accepted Answer · answered Mar 03 '15 at 05:25

4

The characteristic equation is

$$ |Q^TAQ-\lambda I|=0 \iff |Q^T AQ-\lambda Q^TQ |=0 \iff |Q^T|| AQ-\lambda Q |=0 \\ \iff |Q^T|| A-\lambda I ||Q|=0 \iff | A-\lambda I |=0. $$

since $|Q|=\pm {1}, |Q^T|=\pm {1} $.

answered Mar 03 '15 at 05:25

science

2,900

@AVP: You are welcome. – science Mar 03 '15 at 19:56

score 4 · Answer 2 · answered Mar 03 '15 at 05:29

You can also show that the eigenvalues are the same for $A$ and $Q'AQ$.

In one direction, suppose that $v\neq 0$ is such that $Av=\lambda v$. Then $\lambda$ is an eigenvalue of $Q'AQ$ with eigenvector $Q'v$: $$ [Q'AQ](Q'v)=Q'A[QQ']v=Q'Av=\lambda(Q'v). $$ Conversely, suppose that $v\neq 0$ is such that $[Q'AQ]v=\lambda v$. Then $\lambda$ is an eigenvalue of $A$ with eigenvector $Qv$: $$ A(Qv)=IAQv=QQ'AQv=Q(Q'AQv)=\lambda(Qv). $$

I like this approach also. Thanks a lot. – AVP Mar 03 '15 at 15:14 — AVP, Mar 03 '15 at 15:14

Let Q be orthogonal. Show that Q^tAQ has the same characteristic polynomial. as A.

2 Answers2