I was asked this question in an exam.

Let $A$ be a square matrix.

  1. $A^TA=A \implies A^2=A$, true or false?
  2. $A^2=A \implies A^TA=A$, true or false?

I rewrote the equations as $(A^T-I)A=0$ and $(A-I)A=0$, but I am unsure how to proceed. I also tried to consider it in terms of columns and rows, A^2=A means that the dot product of row i and column j equals $A_{ij}$, but that doesn't get me anywhere. I know that if I assume $A$ to be symmetric, both statements are true. My hunch would be that 1 is false, 2 is true.

  • $\begingroup$ If the matrix is invertible, then it's the identity. Do you know whether the class of matrices that either of this relations holds is bigger than that? $\endgroup$ – tst Mar 9 at 3:54

Note that $A^TA$ is a symmetric matrix, so if $A^TA=A$, then $A$ is symmetric. Consequently, $A^T=A$ which implies that $A^2=A^TA=A$. Conclusion: (1) is true.

(2) is false. Here's a counterexample $$A= \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} $$ Then $A^2=A$, but $A^TA\neq A$ (otherwise $A$ would be symmetric).

  • $\begingroup$ Thanks for the answer. I missed the fact that $A^TA$ is symmetric... XD $\endgroup$ – eatfood Mar 9 at 8:25
  • $\begingroup$ No problem. Yeah noting that that matrix is symmetric does help a lot with the reasoning. $\endgroup$ – Stefan Lafon Mar 9 at 15:30

(1) is true:

If $A^{t}A=A$, then $A^{t}=(A^{t}A)^{t}=A^{t}A^{tt}=A^{t}A=A$. (Here, we have used the properties that $A^{tt}=A$ and $(AB)^{t}=B^{t}A^{t}$). It follows that $A^{2}=AA=A^{t}A=A$.

(2) is false:

Let $A=\begin{pmatrix}1 & 0\\ a & 0 \end{pmatrix}$, where $a$ can be any non-zero number. Then $A^{2}=A$. Now $A^{t}=\begin{pmatrix}1 & a\\ 0 & 0 \end{pmatrix}$, so $A^{t}A=\begin{pmatrix}1+a^{2} & 0\\ 0 & 0 \end{pmatrix}\neq A$.


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