What is the difference between $A^TA$ and $A^2$?

Question

What is the difference between squaring a matrix by multiplying it by its transpose and squaring it by multiplying it by itself. When I am asked to square a matrix, which method is preferred?

Answer 1

In general we have $A^TA \ne A^2.$

Example: $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.$

Answer 2

Matrix multiplication fails in some cases for example: Take matrices $A=\begin{bmatrix}1&0&1\\0&1&0\end{bmatrix}$, $ A^T=\begin{bmatrix}1&0\\0&1\\1&0\end{bmatrix}$.You can find $A.A^T$ but you cannot find $A^2$?

Answer 3

In the special case that $A$ is an invertible square matrix, we have $$A^T A = A^2 \iff (A^T A) A^{-1} = (A^2)A^{-1} \iff A^T = A \iff \text{$A$ is symmetric} $$ So, for example, $A=\pmatrix{0 & 1 \\ -1 & 1}$ is invertible (because its determinant is not zero) and not symmetric (obviously), hence $A^T A \ne A^2$ for this example (which an easy calculation shows).

For another example, $A = \pmatrix{1.49 & 98.07 \\ 3.6 & -47.458}$ is also invertible (because its determinant is obviously negative) and not symmetric (obviously), hence $A^T \ne A^2$ (which a hard calculation shows).