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Find all real square matrix $A$ which $$A=\operatorname{adj}(A)$$

My thoughts:

We have the well-known equation $$ A\operatorname{adj}(A)=\operatorname{det}(A)I_n$$

Since $A=\operatorname{adj}(A)$, it comes to $$ A^2=\operatorname{det}(A)I_n$$

So,

  • if $\operatorname{det}(A)=0$, then $A^2=O$.
  • if $\operatorname{det}(A)\neq0$, then we get (if $n\neq2$) $$(\operatorname{det}(A))^2=(\operatorname{det}(A))^n$$ $$ \operatorname{det}(A)=\pm1$$ $$ A^2=\pm I_n$$ In this case, $A$ seems like a involutory matrix.

How to characterize the matrics mentioned above? Or what further properties should $A$ follows so we can completely characterize $A$?

Thanks for your help!

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    $\begingroup$ Interestingly: in the $2 \times 2$ case, all non-zero matrices satisfying $A^2 = 0$ seem to satisfy $\operatorname{adj}(A) = -A$ (let's call that a conjecture). For larger matrices: if $A$ is non-zero and non-invertible, then the adjugate of $A$ has rank strictly less than that of $A$. All together, it seems that we never have $A = \operatorname{adj}(A)$ when $A$ is non-zero and non-invertible. $\endgroup$ – Omnomnomnom Oct 24 '17 at 15:33
  • $\begingroup$ In fact, as a proof of my "conjecture": it suffices to note that for an invertible $S$, we have $$ \operatorname{adj}(SJS^{-1}) = \operatorname{adj}(S)\operatorname{adj}(J)\operatorname{adj}(S^{-1}) = \operatorname{adj}(S)(-J)\operatorname{adj}(S^{-1}) $$ where $$ J = \pmatrix{0&1\\0&0} $$ $\endgroup$ – Omnomnomnom Oct 24 '17 at 15:36
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The first question to answer: which matrices satisfy $A^2 = I$ and $\det(A) = 1$?

It suffices to note that since $A^2 - I = 0$, the minimal polynomial of $A$ must divide $x^2 - 1 = (x-1)(x + 1)$. Hence, $A$ must be diagonalizable with eigenvalues equal to $\pm 1$. Moreover, since the product of eigenvalues is equal to $\det(A)$ which is equal to $1$, the $-1$ eigenvalue must have even multiplicity. All together, we can characterize these matrices as those of the form $$ A = S \pmatrix{-I_{2k}&0\\0&I_{n - 2k}}S^{-1}, \quad k = 0,1,\dots,\lfloor n/2 \rfloor $$ where $I_k$ denotes the $k \times k$ identity matrix and the matrices here are $n \times n$.

The second question: which matrices satisfy $A^2 = -I$ and $\det(A) = -1$? In fact, there are no such real matrices. Because $A^2 + I = 0$, the (complex) eigenvalues of $A$ must solve $x^2 + 1 = 0$, which is to say that the eigenvalues of $A$ are $\pm i$. Because $A$ is a real matrix, its complex eigenvalues come in conjugate pairs. Thus, its determinant must have the form $\det(A) = (-i)^k(i)^k = 1 \neq -1$.

An alternative explanation: the minimal polynomial $x^2 + 1$ of $A$ is an irreducible polynomial over $\Bbb R$, so the characteristic polynomial must have the form $\det(xI - A) = (x^2 + 1)^k$. Thus, we find that $A$ has even size, and that $\det(A) = \det(0I - A) = (0^2 + 1)^k = 1$.

As I note in the comments above, there are no solutions satisfying $A^2 = 0$, and in fact no non-invertible solutions other than $A = 0$.

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$A^2=I_n$, let $U=\{x:A(x)=x\}, V=\{x:A(x)=-x\}$. $U\cap V=\{0\}$, for every $x$, you can write $x={1\over 2}(A(x)+x)+{1\over 2}(x-A(x))$, ${1\over 2}(x+A(x))\in U$ and ${1\over 2}(x-A(x))\in V$.

Then take any $U,V$ supplementary subspaces and define $A_{\mid U}=I_U$ and $A_{\mid V}=-Id_V$.

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