# Matrices (Hermitian and Unitary)

Hey! I have some short proofs I’m quite stuck on below. I know the definitions but get stuck on how to use them to prove what’s required. If possible, can you please explain how to apply the definition to get these proofs so that I can try again? Thanks!

Q1. Show that if there exists a unitary matrix P such that $P*AP = D$ where $P*$ is the conjugate transpose of $P$ and $D$ is a diagonal matrix then $A$ is a normal matrix. Show that the columns of $P$ form an orthornormal basis of $Cn$ if and only if $P$ is unitary.

A1. If $P$ is unitary, $PP*=P*P=I$. A is normal if $AA*=A*A$. I'm not sure where to go from there.

Q2. Show that eigenvectors of a Hermitian matrix corresponding to distinct eigenvalues are orthogonal.

A2. Hermitian if $A=A*$. Eigenvalues of $A$ are given by the equation: $Av=(\lambda)v$. Again, what do I do with this?

Q3. Show that the eigenvalues of a Hermitian matrix are real, and that the eigenvectors corresponding to different eigenvalues are orthogonal.

A2. Similar to the above.

Q4. What is meant to say that a matrix is unitarily diagonalisable? Prove that if $P$ is a unitary matrix then all of the eigenvalues of $P$ have modulus equal to one. Further, prove that the column vectors of $P$ form an orthornormal set (with respect to the Euclidean inner product).

A3. Is it unitarily diagonalisable when it can be expressed as a matrix with orthornormal vectors? I’m not sure about what the rest of the question even means! :(

Q'1. Show that if there exists a unitary matrix P such that PAP = D where P is the conjugate transpose of P and D is a diagonal matrix then A is a normal matrix.

A'1. We have $P^*AP=D$. By applying $()^*$ to both sides, we get $P^*A^*P=D^*$.

Now, $(P^*AP)(P^*A^*P)=DD^*$. On the other hand, $(P^*A^*P)(P^*AP)=D^*D$. But $DD^*=D^*D$, since $D$ is diagonal. Hence $(P^*AP)(P^*A^*P) = (P^*A^*P)(P^*AP)$. Therefore $(P^*AA^*P) = (P^*A^*AP)$. Lastly, premultiply both sides by $P$ and postmultiply both sides by $P^*$ to get $AA^*=A^*A$.

Q3. Show that eigenvectors of a Hermitian matrix are real and that eigenvectors corresponding to distinct eigenvalues are orthogonal.

A'3. First, let $\lambda \in \mathbb{C}$ be an eigenvalue with associated eigenvector $v \neq 0$. Look at the inner product $\langle v,v \rangle$. We have:

$\lambda \langle v,v \rangle = \langle Av,v \rangle = \langle v,A^*v \rangle = \langle v,Av \rangle = \langle v, \lambda v \rangle = \overline{\lambda} \langle v,v \rangle$.

Since $\langle v,v \rangle \neq 0$, we must have $\lambda = \overline{\lambda}$.

This shows that the eigenvalues of a Hermitian matrix are real.

Now let $\lambda$ and $\mu$ be distinct eigenvalues, with associated eigenvectors $v$ and $w$ respectively.

Look at the inner product $\langle v,w \rangle$. We have:

$\lambda \langle v,w \rangle = \langle \lambda v,w \rangle = \langle Av,w \rangle = \langle v,A^*w \rangle = \langle v,Aw \rangle = \langle v,\mu w \rangle = \mu \langle v,w \rangle$

But $\lambda \neq \mu \implies \langle v,w \rangle =0$.

Q'4. Prove that if P is a unitary matrix then all of the eigenvalues of P have modulus equal to one.

A'4. Let $\lambda$ be an eigenvalue of $P$ with associated eigenvector $v \neq 0$. Then:

$\langle v,v \rangle = \langle PP^*v,v \rangle = \langle Pv,Pv \rangle = \langle \lambda v, \lambda v \rangle = \lambda \overline{\lambda} \langle v,v \rangle$.

Hence $(1-\lambda \overline{\lambda})\langle v,v \rangle=0$. Since $\langle v,v \rangle \neq 0$, $\lambda \overline{\lambda} = 1 \implies |\lambda|^2=1 \implies |\lambda|=1$.

• Much appreciated!!!! :) I will try these again and get back to you if im still stuck. – user4645 May 20 '11 at 11:21
• The only question I have about the above is how you went from <Av,w> to <v,A*w>? I understand all the other steps! :) – user4645 May 23 '11 at 10:17
• $<Av,w>=<v,A^*w>$ by definition of the inner product. Much like $<az,wy>=\overline{w}<az,y>$. – MathIsHard Oct 6 '17 at 16:56