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.