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I know that for a matrix $A$:

If $A^TA = A$ then $A=A^2$

but is it if and only if? I mean:

is this true that "If $A=A^2$ then $A^TA = A$"?

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The answer is no.

Consider $A = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}$. We have

$$A^2 = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = A$$ but $$A^TA = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \ne A$$

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What about of $A=\begin{bmatrix} 1&1\\0&0\end{bmatrix}\large{?}$

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Since $$\det (A)^2 = \det (A)\det (A) = \det (A^2)= \det (A) \implies \det (A)\in\{0,1\}$$

So if $\det (A)= 1$ then exsist $A^{-1}$ so $A = I$ and the answer is yes.

If $\det(A)=0$ then examples show that answer is negative.

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