How can I prove that if a matrix $A$ is symmetric so is its transponse $A^T$

Question

How can I prove that if a matrix $A$ is symmetric so is its transponse $A^T$.

I know that this claim is true because $A = A^T$ but how can I prove it formally?

And are there two symmetric matrices such that their marix product is not symmetric.

Thanks

Answer 1

The definition of a symmetric $n \times n$ matrix $A = (a_{ij})$ is $$ a_{ij} = a_{ji} \quad (i, j \in \{ 1, \dotsc, n \}) \quad (*) $$

The definition of the transpose $A^T = (b_{ij})$ of a $m \times n$ matrix $A = (a_{ij})$ is $$ b_{ij} = a_{ji} \quad (i \in \{ 1, \dotsc, n \}, j \in \{ 1, \dotsc, m \}) \quad (**) $$

For a symmetric matrix $A$, we have $m = n$, and for the elements of $A^T$ we have $$ \underbrace{ b_{ij} \stackrel{(**)}{=} a_{ji} \stackrel{(*)}{=} a_{ij} \stackrel{(**)}{=} b_{ji} }_{(*)} $$ so $A^T$ is symmetric.

Answer 2

Hint (almost answer):

By the very, very definition of "transpose of matrix", convince yourself that transposing twice gets us back to the original: $\;(A^t)^t=A\;$

Example for the second part:

$$\begin{pmatrix}0&1\\1&1\end{pmatrix}\cdot\begin{pmatrix}1&0\\0&\!-1\end{pmatrix}=\ldots$$

Answer 3

if a matrix $A$ is symmetric then $aij = aji$; if $B = A^T$ then $bij = aji$

since $bij = aji = aij$ then $B = A^T = A$

Answer 4

Note that $$ (A^T)^T=A^T. $$

(I substituted $A^T=A$.)