# Is it true that $A^TA=A \implies A^2=A$? Is the converse true?

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.

• 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? – 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).

• Thanks for the answer. I missed the fact that $A^TA$ is symmetric... XD – eatfood Mar 9 at 8:25
• No problem. Yeah noting that that matrix is symmetric does help a lot with the reasoning. – 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 $$A^{t}A=\begin{pmatrix}1+a^{2} & 0\\ 0 & 0 \end{pmatrix}\neq A$$.