1. ## Matrices, Square matrix

(a) Prove that the transpose of the sum of two matrices A and B is the sum of their transposes.

(b) Verify this is so when
A = [1 2 3] and B = [4 6 8]
[4 5 6] [5 7 9]

(c) Prove that if A is a square matrix then S = 1/2 (A -A^t) is skew-symmetric.

(d) Verify that this S is skew-symmetric when A = [4 3]
[2 1]

(e) Prove that if A is a square matrix with complex elements, then H = 1/2 (A + A^*) is hermitian.

(f) Verify that this H is hermitian when A = [1+i 4-3i]

[2-i 3+2i]

Any help with any of these questions would be greatly appreciated!

2. ## Transpose of a matrix

Hello mr_motivator
Originally Posted by mr_motivator
(a) Prove that the transpose of the sum of two matrices A and B is the sum of their transposes.

(b) Verify this is so when
A = [1 2 3] and B = [4 6 8]
[4 5 6] [5 7 9]

(a) Denote the elements in row $\displaystyle i$, column $\displaystyle j$ of the matrices $\displaystyle A$ and $\displaystyle B$ by $\displaystyle a_{i,j}$ and $\displaystyle b_{i, j}$. In the matrix sum $\displaystyle A+B$, the element in row $\displaystyle i$, column $\displaystyle j$ is therefore $\displaystyle a_{i,j} + b_{i, j}$.

This is then the element in row $\displaystyle j$, column $\displaystyle i$ of $\displaystyle (A+B)^T$, which is the sum of elements in row $\displaystyle j$, column $\displaystyle i$ of $\displaystyle A^T$ and $\displaystyle B^T$.

This is true for all valid $\displaystyle i$ and $\displaystyle j$. Hence $\displaystyle (A+B)^T=A^T +B^T$.

(b) is a simple matter of checking this out.
$\displaystyle A+B = \begin{pmatrix}5 & 8 & 11\\9 & 12 & 15\end{pmatrix}$

$\displaystyle \Rightarrow (A+B)^T =\begin{pmatrix}5 & 9\\8 & 12\\11 & 15\end{pmatrix}$

= etc...

3. (c) Prove that if A is a square matrix then S = 1/2 (A -A^t) is skew-symmetric.
You have to prove: $\displaystyle S^T = -S$

Use these 3 properties:
• $\displaystyle (cA)^T = cA^T$
• $\displaystyle (A+B)^T = A^T + B^T$
• $\displaystyle \left(A^T\right)^T = A$

So:
\displaystyle \begin{aligned} S^{T} & = \left[\frac{1}{2} \left(A - A^{T}\right)\right]^{T} \\ & = \frac{1}{2}\left(A - A^T\right)^{T} \\ & = \cdots \end{aligned}

I'm sure you can finish off.

For (d), just show that: $\displaystyle A^T = -A$

4. ## Hermitian Matrices

Hello mr_motivator
Originally Posted by mr_motivator
(e) Prove that if A is a square matrix with complex elements, then H = 1/2 (A + A^*) is hermitian.

(f) Verify that this H is hermitian when A = [1+i 4-3i]

[2-i 3+2i]
A matrix is Hermitian if its transpose is also its conjugate. In other words, if $\displaystyle (a_{i,j})^* = a_{j, i}$ for all valid $\displaystyle i, j$. We have to prove that

$\displaystyle \tfrac{1}{2}(A + (A^T)^*)$ is Hermitian.

Now suppose in matrix $\displaystyle A, a_{i, j}$ is written $\displaystyle p + iq$ and $\displaystyle a_{j,i}$ is written $\displaystyle r + is$. Then the element in row $\displaystyle i$, column $\displaystyle j$ of $\displaystyle \tfrac{1}{2}(A + (A^T)^*)$ is

$\displaystyle \tfrac{1}{2}(a_{i,j} + (a_{j,i})^*) = \tfrac{1}{2}[(p+iq)+(r-is)]= \tfrac{1}{2}[(p+r) +(q-s)i]$
(1)

and the element in row $\displaystyle j$, column $\displaystyle i$ of $\displaystyle \tfrac{1}{2}(A + (A^T)^*)$ is

$\displaystyle \tfrac{1}{2}(a_{j,i} + (a_{i,j})*) = \tfrac{1}{2}[(r+is)+(p-iq)]= \tfrac{1}{2}[(p+r) +(s-q)i]$

which is the conjugate of the element in (1).

This then is the required proof.

I think you can complete part (f) now.