Show that \(A^T A \neq A A^T\) in general. (Proof and demonstration.)

Proof:

Let \(A\) be an \(m \times n\) matrix: \[ A= \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \\ \end{bmatrix} \] \[ A A^T = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \\ \end{bmatrix} \begin{bmatrix} a_{11} & a_{21} & \ldots & a_{m1} \\ a_{12} & a_{22} & \ldots & a_{m2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \ldots & a_{mn} \\ \end{bmatrix} = \begin{bmatrix} (a_{11})^2 + (a_{21})^2 + ...+ (a_{m1})^2 & a_{11}a_{21} + a_{12}a_{22} + ...+a_{1n}a_{2n} & \ldots & a_{11}a_{m1} + a_{12}a_{m2} +...+a_{1n}a_{mn} \\ a_{21}a_{11} + a_{22}a_{12}+...+a_{2n}a_{1n} & (a_{21})^2+(a_{22})^2+...+(a_{2n})^2 & \ldots & a_{21}a_{m1} + a_{22}a_{m2} +...+a_{2n}a_{mn} \\ \vdots & \vdots & \ddots &\vdots \\ a_{m1}a_{11} + a_{m2}a_{12} +...+ a_{mn}a_{n1} & a_{m1}a_{21} + a_{m2}a_{22}+...+a_{mn}a{2n} & \ldots & (a_{m1})^2 + (a_{m2})^2+...+(a_{mn})^2 \end{bmatrix} \] \[ A^TA = \begin{bmatrix} a_{11} & a_{21} & \ldots & a_{m1} \\ a_{12} & a_{22} & \ldots & a_{m2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & \ldots & a_{mn} \\ \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn} \\ \end{bmatrix} = \begin{bmatrix} (a_{11})^2+(a_{12})^2+...(a_{m1})^2 & a_{11}a_{12} + a_{21}a_{22}+...+a_{m1}a_{m2} & \ldots & a_{11}a_{1n}+a_{21}a_{2n}+...+a_{m1}a_{mn} \\ a_{12}a_{11}+a_{22}a_{21}+...+a_{m2}a_{m1} & (a_{12})^2+(a_{22})^2+...+(a_{2n})^2 & \ldots & a_{12}a_{1n} + a_{22}a{2n} +...+ a_{m2}a_{mn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n}a_{11}+a_{2n}a_{21}+...+a_{mn}a{m1} & a_{1n}a_{12}+a_{2n}a_{22}+...+a_{mn}a_{m2} & \ldots & (a_{1n})^2 + (a_{2n})^2 +...+ (a_{mn})^2 \end{bmatrix} \]

\[ \begin{bmatrix} (a_{11})^2 + (a_{21})^2 + ...+ (a_{m1})^2 & a_{11}a_{21} + a_{12}a_{22} + ...+a_{1n}a_{2n} & \ldots & a_{11}a_{m1} + a_{12}a_{m2} +...+a_{1n}a_{mn} \\ a_{21}a_{11} + a_{22}a_{12}+...+a_{2n}a_{1n} & (a_{21})^2+(a_{22})^2+...+(a_{2n})^2 & \ldots & a_{21}a_{m1} + a_{22}a_{m2} +...+a_{2n}a_{mn} \\ \vdots & \vdots & \ddots &\vdots \\ a_{m1}a_{11} + a_{m2}a_{12} +...+ a_{mn}a_{n1} & a_{m1}a_{21} + a_{m2}a_{22}+...+a_{mn}a{2n} & \ldots & (a_{m1})^2 + (a_{m2})^2+...+(a_{mn})^2 \end{bmatrix} \neq \begin{bmatrix} (a_{11})^2+(a_{12})^2+...(a_{m1})^2 & a_{11}a_{12} + a_{21}a_{22}+...+a_{m1}a_{m2} & \ldots & a_{11}a_{1n}+a_{21}a_{2n}+...+a_{m1}a_{mn} \\ a_{12}a_{11}+a_{22}a_{21}+...+a_{m2}a_{m1} & (a_{12})^2+(a_{22})^2+...+(a_{2n})^2 & \ldots & a_{12}a_{1n} + a_{22}a{2n} +...+ a_{m2}a_{mn} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1n}a_{11}+a_{2n}a_{21}+...+a_{mn}a{m1} & a_{1n}a_{12}+a_{2n}a_{22}+...+a_{mn}a_{m2} & \ldots & (a_{1n})^2 + (a_{2n})^2 +...+ (a_{mn})^2 \end{bmatrix} \]

Therefore,

\(A^T A \neq A A^T\)

For a special type of square matrix A, we get \(A^T A = AA^T\). Under what conditions could this be true? :

The conditions under which we get \(A^TA = AA^T\) are when every element that is not contained on the diagonal of the square matrix is = 0.

This causes \(A = A^T\) and thus \(A^TA = AA^T\)

Proof

Let \(A\) be an \(n \times n\) matrix with the only non-zero elements contained on the diagonal:

\[ \begin{bmatrix} a_{11} & 0 & \ldots & 0 \\ 0 & a_{22} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & a_{nn} \end{bmatrix} \]

\[ A^T = \begin{bmatrix} a_{11} & 0 & \ldots & 0 \\ 0 & a_{22} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & a_{nn} \end{bmatrix} \]

\[ A^TA = \begin{bmatrix} (a_{11})^2 & 0 & \ldots & 0 \\ 0 & (a_{22})^2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & (a_{nn})^2 \end{bmatrix} \]

\[ AA^T = \begin{bmatrix} (a_{11})^2 & 0 & \ldots & 0 \\ 0 & (a_{22})^2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & (a_{nn})^2 \end{bmatrix} \]

\[ \begin{bmatrix} (a_{11})^2 & 0 & \ldots & 0 \\ 0 & (a_{22})^2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & (a_{nn})^2 \end{bmatrix} = \begin{bmatrix} (a_{11})^2 & 0 & \ldots & 0 \\ 0 & (a_{22})^2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & (a_{nn})^2 \end{bmatrix} \]

Therefore the condition for square matrices where \(A^TA = AA^T\) is when the only non-zero elements are contained on the diagonal of the matrix.