User:J D Bowen/Math725 HW11

1a) Let $M \ $ have singular value decomposition $M=UDV^H \ $.  Observe that $M \ $ is injective if and only if $M \ $ has a left inverse.   The left inverse of $M \ $ would be given by $V D^+ U^H \ $ if it exists at all, but if and only if this product is well defined. Observe $VD^+U^HUDV^H = VD^+ DV^H =I \ $ if and only if $V \ $ has right inverse $D^+DV^H \ $ and left inverse $D^H (D^+)^HV^H \ $, but only square matrices have inverses on both sides.

Then $M^{-1} = VD^+ U^H \ $ and since $I=I^H=(MM^{-1})^H = (M^{-1})^H M^H \ $, we must have $(M^H)^{-1}=(M^{-1})^H \ $. Since both $M, M^H \ $ are invertible, their product is as well.

2a) Suppose $MN, \ NM \ $ are both well-defined matrix products. Then $M, \ N \ $ are square matrices.  Call their size $n\times n \ $ and define $MN = ((mn)_{ij}), \ NM=(nm)_{ij} \ $.

We have $(mn)_{ij} \sum_{k=1}^n m_{ik}n_{kj}, \ (nm)_{ij} \sum_{k=1}^n n_{ik}m_{kj} \ $.

Then

$\text{Tr}(MN) = \sum_{l=1}^n (mn)_{ll} = \sum_{l=1}^n \sum_{k=1}^n m_{lk}n_{kl} = \sum_{k,l=1}^n m_{lk}n_{kl} \ $

and

$\text{Tr}(NM) = \sum_{l=1}^n (nm)_{ll} = \sum_{l=1}^n \sum_{k=1}^n n_{lk}m_{kl} = \sum_{k,l=1}^n n_{lk}m_{kl} \ $.

Since these two expressions are equal, $\text{Tr}(MN) =\text{Tr}(NM) \ $.

2b) Let $(V,B) \ $ be the vector space $V \ $ with basis $B \ $, and let $(V,B' ) \ $ be similarly defined. Let $\eta:V\to V \ $ be defined $\eta(B)=B' \ $.  Clearly, $\eta^{-1} \ $ exists since $\eta \ $ is an isomorphism.

Define $N=\mathfrak{M}^B_B(\eta) \ $, and observe that $\mathfrak{M}_{B'}^{B'}(T) = \mathfrak{M}_B^B (\eta T \eta^{-1}) = N\mathfrak{M}_B^B (T) N^{-1} \ $.

Then $\text{Tr}(\mathfrak{M}_{B'}^{B'}) = \text{Tr}(N\mathfrak{M}_B^B (T) N^{-1}) = \text{Tr}(\mathfrak{M}_B^B (T) N^{-1}N) = \text{Tr}(\mathfrak{M}_B^B (T)) \ $.

2c) Let $M \ $ be an $m\times n \ $ matrix. Then $M^H \ $ is an $n\times m \ $ matrix. Then $M^H M \ $ is an $n\times n \ $ matrix.

If we define $M^HM = (a_{ij}) \ $, we have

$a_{ij} = \sum_{k=1}^m \overline{m_{ki}}m_{kj} \ $.

Observe that $\text{Tr}(M^H M) = \sum_{l=1}^n a_{ll} = \sum_{l=1}^n \sum_{k=1}^m \overline{m_{kl}}m_{kl} = \sum_{k,l} |m_{kl}|^2 = ||M||^2_F \ $.

2d) Suppose a matrix $M \ $ has singular value decomposition $M=UDV^H \ $. Then

$||M||^2_F = \text{Tr}(M^H M) = \text{Tr}( (UDV^H)^H UDV^H ) = \text{Tr}(VD^H U^H U D V^H) = \text{Tr}(VD^HDV^H) = \text{Tr}(D^HDV^HV) = \text{Tr}(D^HD) = ||D||^2_F \ $.

Since $D \ $ is composed exclusively of the singular values of $M \ $, this final expression is just the sum of the squares of the singular values of $M \ $.

3a) Let $M \ $ be an $r\times c \ $ matrix of rank $n \ $ with singular value decomposition $M=UDV^T \ $. Then $U \ $ is an $r\times r \ $ matrix, $D \ $ is an $r\times c \ $ matrix, and $V^T \ $ is a $c\times c \ $ matrix.

Not counting duplicate numbers, the total number of numbers will be $\frac{r^2+c^2}{2}+ n \ $, while $M \ $ will have $rc \ $ numbers.