Question

Question B

7. (a) Let -1 0 0 (i) Find a unitary matrix U such that M-UDU where D is a diagonal matrix. 10 marks] (i) Compute the Frobenius norm of M, i.e., where (A, B) = trace(B·A). [4 marks] 3 marks] (iii) What is NM-illp? (b) Let H be an n × n complex matrix (6) What does it mean to say that H is positive semidefinite. (il) Show that H is positive semidefinite and Hermitian if and only if there exists an n × n positive semidefinite Hermitian matrix B such that H B (ili) Show that if H is positive definite and Hermitian, then there exists a positive definite Hermitian matrix Q such that QHQ = 1, where I is the identity miatrix. 3 marks] 5 marks]

Answer 1

b) (i) is said to be positive semidefinite if for all $x\in\mathbb C^n$ one has $x^\ast Hx\geq 0$ where $\ast$ denotes conjugate transpose.

(ii) Suppose is positive semidefinite and Hermitian. Then (being Hermitian is equivalent to this condition, namely) there is a unitary matrix such that $H=UDU^\ast$ for some diagonal matrix whose diagonal entries are real numbers. Note that

$0\leq x^\ast Hx=x^\ast(UDU^\ast)x=(U^\ast x)^\ast D(U^\ast x)$

for all $x\in\mathbb C^n$. In particular, letting $x=U_i$ the -th column of , we get $U^\ast x=U^\ast U_i=e_i$ , where $e_1,\cdots,e_n$ denote the standard basis vectors, and thus,

$0\leq (U^\ast x)^\ast D(U^\ast x)=e_i^\ast De_i=D_{ii}$

Thus, the diagonal matrix has non-negative diagonal entries only. Let $B=U\Delta U^\ast$ where $\Delta$ is diagonal, and $\Delta_{ii}=+\sqrt{D_{ii}}$ , that is, the diagonal entries of $\Delta$ are non-negative square roots of the corresponding diagonal entries of . Then, $B=U\Delta U^\ast$ is Hermitian and also, for all $x=U_i$ we have

$x^\ast Bx=(U^\ast x)^\ast \Delta(U^\ast x)=e_i^\ast\Delta e_i=\Delta_{ii}\geq 0$

But the columns of $U_i$ form an orthonormal basis of $C^n$. Hence, an arbitrary $x\in\mathbb C^n$ can be written as $x=a_1U_1+\cdots+a_nU_n$ ; then,

$x^\ast Bx=\sum_{i=1}^n |a_i|^2(e_i^\ast De_i)=\sum_{i=1}^n|a_i|^2D_{ii}\geq 0$

which shows that is positive semidefinite.

(iii) As in part ii) above, we have $H=UDU^\ast$ where is diagonal with strictly positive diagonal entries. Let

$Q=U\Delta'U^\ast$

where $\Delta'$ is a diagonal matrix whose diagonal entries are

$\Delta'_{ii}={\frac 1{\sqrt{D_{ii}}}}\hspace{2cm}(\ast)$

Then $Q=U\Delta'U^\ast$ is Hermitian positive definite (as in part ii) above). Moreover, we have

$\begin{align*}QHQ&=(U\Delta' U^\ast)H(U\Delta' U^\ast)\\ &=(U\Delta' U^\ast)(UD U^\ast)(U\Delta' U^\ast)\\ &=U\Delta' (U^\ast U)D(U^\ast U)\Delta'U^\ast\\ &=U\Delta' D\Delta'U^\ast\\ &=U(\Delta')^2DU^\ast\end{align*}$

But, by definition $(\ast)$ we have $\begin{align*}(\Delta')^2D=I\end{align*}$ ; therefore,

$\begin{align*}QHQ&=U(\Delta')^2DU^\ast\\ &=UIU^\ast\\ &=I\end{align*}$