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Here's my feeble attempt:

A = A* is the condition for a Hermitian matrix. So I try expanding both sides in terms of their SVD:

A = USV* and A* = (USV*)* = ... = VSU*

So equating I get:

VSU* = USV*

Now maybe through this I can conclude something about how V and U relate to each other but I don't know how...

Thanks for any help.

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  • $\begingroup$ It will help to conclude that $U=V$. What happens if you multiply $VSU^*$ by $USV^*$? Note that order will matter, so consider both. $\endgroup$
    – Alex R.
    Nov 7, 2012 at 6:47

3 Answers 3

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The process is fairly straightforward given the fact that any square matrix has at least one eigenvalue and eigenvector.

Suppose $Ax = \lambda x$. Then $\langle x , Ax \rangle = \langle Ax , x \rangle = \overline{\langle x , Ax \rangle} = \lambda \|x\|^2$, hence $\lambda \in \mathbb{R}$.

The point about being Hermitian is that if $x$ is an eigenvector of $A$, then both $\text{sp} \{x\}$ and the subspace $\{x\}^\bot$ are invariant under $A$. To see the latter, suppose $v \bot x$, then $\langle x , Av \rangle = \langle Ax , v \rangle = \lambda \langle x , v \rangle = 0$, hence $A v \bot x$. Now let $v_k$ be a basis for $\{x\}^\bot$, then in the basis $x, v_1,...,v_k$, $A$ must have a block diagonal form: $$ A \sim \begin{bmatrix} \lambda & \\ & \tilde{A}\end{bmatrix}$$ (By $\sim$ I mean that the two matrices are similar.) Now find an eigenvalue of $\tilde{A}$ (which must also be Hermitian) and repeat the process until $A$ is diagonalized.

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  • $\begingroup$ Nice, you made that seem easier than previous proofs I've read. $\endgroup$
    – littleO
    Nov 7, 2012 at 8:11
  • $\begingroup$ Thanks! Well, I skipped some details like writing out the similarity transformation that reduces $A$ to block diagonal form. However, I personally find the details more straightforward when I have the 'big picture'. $\endgroup$
    – copper.hat
    Nov 7, 2012 at 8:19
  • $\begingroup$ Yeah, you made a nice choice of which details to skip, in order to show how simple the proof really is. $\endgroup$
    – littleO
    Nov 7, 2012 at 8:31
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A more efficient route is to consider the Schur decomposition, $A=QTQ^*$, where $T$ is upper triangular and $Q$ is unitary. Note that the diagonal of $T$ consists of the eigenvalues of $A$. The equality $A^*=A$ is $QTQ^*=QT^*Q^*$, from where we deduce $T=T^*$. As $T$ is upper triangular and $T^*$ is lower triangular, we get that $T$ is diagonal. Moreover, as the diagonal of $T^*$ is the conjugate of that of $T$ and they are equal, we have that the eigenvalues of $A$ are real.

The eigenspaces of $T$ are orthogonal to each other (easy to see since $T$ is diagonal). But $Q$ takes the eigenspaces of $T$ to the eigenspaces of $A$, so these have to be pairwise orthogonal too.

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Use the eigenvalue decomposition of the hermitian matrix $A$.

$A=S \Lambda S^{-1}$

$ A^H = (S \Lambda S^{-1})^{H} $

$ A^H = ( S^{-1})^{H} \Lambda^H S^H$

Since all eigenvalues for a hermitian matrix are real, $\Lambda = \Lambda^H$, and since $A$ is hermitian we have

$S \Lambda S^{-1} = ( S^{-1})^{H} \Lambda S^H$

This shows $S^{-1}=S^H$ implying $SS^H=I$ as required.

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  • $\begingroup$ This assumes $S^{-1}$ exists. Not a general proof. $\endgroup$ Dec 11, 2013 at 9:40

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