Question

Suppose that A is an $\mathit{n}\mathbf{\times }\mathit{n}$ matrix where n is a positive integer. Show that$\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$ is symmetric.

Short answer

It is proved that$\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$ is symmetric matrix.

Step-by-step solution

Step 1:

Given: A is a $n\times n$-matrix with n a positive integer.

To prove: $\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$is symmetric

Step 2:

$\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$is symmetric if the matrix is found to be the same as its transpose.

Use$\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}{\mathbf{)}}^{\mathbf{t}}\mathbf{=}{\mathit{A}}^{\mathbf{t}}\mathbf{+}{\mathit{B}}^{\mathbf{t}}$

$\mathbf{(}\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}{\mathbf{)}}^{\mathbf{t}}\mathbf{=}{\mathit{A}}^{\mathbf{t}}\mathbf{+}\mathbf{(}{\mathit{A}}^{\mathbf{t}}{\mathbf{)}}^{\mathbf{t}}$

Step 3:

By taking the result from exercise 16:$\mathbf{(}{\mathit{A}}^{\mathbf{t}}{\mathbf{)}}^{\mathbf{t}}\mathbf{=}\mathit{A}$

$\mathbf{\Rightarrow }\mathbf{(}\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}{\mathbf{)}}^{\mathbf{t}}\mathbf{=}{\mathit{A}}^{\mathbf{t}}\mathbf{+}\mathit{A}$

Step 4:

Use commutative property of matrices (see exercise 8 ):

$\mathbf{(}\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}{\mathbf{)}}^{\mathbf{t}}\mathbf{=}\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$

Therefore, it is found that transposed matrix$\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$ of is the matrix $\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$itself and thus the matrix$\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{t}}$ is symmetric.