Solve the equation $A=\text{adj}(A)$ in $\mathbf{M}_n(\mathbb{R})$.
Where $\text{adj}(A)$ is the adjugate of $A$ (tranpose of cofactor matrix).
Here is what I know:
-
$A^2=\det(A)I_n$
-
$\text{rank(A)} \lt n-1 \implies \text{rank}(\text{adj}(A)) = 0$
Here is what I have done so far:
We have 3 cases:
-
If $det(A) = 0$
Then $A^2=0$, thus $\text{rank}(A)\leq2-1=1$.
This implies $\text{rank(A)} = \text{rank}(\text{adj}(A)) \leq 1$
So $\text{rank}(A) = 0$, that is, $A=0$
-
If $det(A) > 0$
Then $\left(X-\sqrt{\det A}\right)\left(X+\sqrt{\det A}\right)$ has $A$ as a root.
Thus $A$ is diagonalizable (since its minimal polynomial has simple roots) and its eigenvalues are a subset of $\{\sqrt{\det A}, -\sqrt{\det A} \}$.
Thus $A=P^{-1}
\begin{bmatrix}
\pm\sqrt{\det A} & &\\
& \ddots &\\
& & \pm\sqrt{\det A}
\end{bmatrix}
P$Which we can verify as a solution to our equation.
-
If $det(A) < 0$
Then $A^2=\det(A)I_n$.
Thus $B^2=-I_n$ with $B=\frac{1}{\sqrt{-\det A}}A$
How do I solve for $B$?
Best Answer
We consider three possibilities according to the rank of $A$:
In short, $A=\operatorname{adj}(A)$ if and only if $A=0$ or $A$ is similar to $I_{n-k}\oplus-I_k$ for some even integer $k$.
Two comments to your solution attempt: