How can I show that for matrix $A$ , $A^t A $ is not equal to $ A A^t $ in general?

Question

How can I show that for matrix $A$ , $A^t A \neq A A^t $

$A^t$ means the transpose of $A$.

That is the entire question and I have no idea how to begin... please help!

Answer 1

Consider $e_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. In general any counterexample to a matrix problem that works in dimension $2$ works in higher dimensions by extending a matrix appropriately, so look at the $2$ dimensional case first! Note that $e_{12}^t=e_{21}$ and that $e_{12}e_{21}=e_{11}$ while $e_{21}e_{12}=e_{22}$. In general it is not a bad idea to look at the elementary matrices $e_{ij}=(\delta_{ij})$, since they have simple rules for computation, like

$$e_{ij}e_{kl}=\delta_{jk}e_{il}\\ e_{ij}^t=e_{ji}$$

Answer 2

Let $$A = \left[ \begin{array}{ c c } 1 & a \\ b & 1 \end{array} \right] $$

Now,

$$A\cdot A^T = \left[ \begin{array}{ c c } 1 & a \\ b & 1 \end{array} \right] \cdot \left[ \begin{array}{ c c } 1 & b \\ a & 1 \end{array} \right] = \left[ \begin{array}{ c c } 1+a^2 & a+b \\ a+b & b^2+1 \end{array} \right] $$

Whereas

$$ A^T \cdot A = \left[ \begin{array}{ c c } 1 & b \\ a & 1 \end{array} \right] \cdot \left[ \begin{array}{ c c } 1 & a\\ b & 1 \end{array} \right] = \left[ \begin{array}{ c c } 1+b^2 & a+b \\ a+b & a^2+1 \end{array} \right] $$

Thus, $A^T \cdot A \neq A\cdot A^T$ in general.