Prove that the adjugate of $A$ is diagonalizable for every $A$ with the same properties

Question

Let $A = (a_{ij})\in \mathbb C^{3x3}$ such that

$$\det(A) = \det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \det \begin{pmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{pmatrix} = 0 \; \text{and } \det \begin{pmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \neq 0$$

Then, knowing that for every $B \in \mathbb C^{nxn}$ such that $\operatorname{rank}(B) = 1$, $$B \text{ is diagonalizable} \Leftrightarrow \operatorname{trace}(B) \neq 0$$ prove that $\operatorname{adj}(A)$ is diagonalizable.

This is what I got so far. Let $\operatorname{adj}(A) = (b_{ij})$:

• The minors that have determinant 0 are entries in the diagonal of $\operatorname{adj}(A)$, meaning that $b_{22} = b_{33} = 0$

• The trace of the adjugate is then $b_{11} = \det \begin{pmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \neq 0$

I now want to prove that the adjugate has rank 1.

• Using that $\det(A) = 0$, writing the determinant of A along all 3 columns I get the following system of equations

$$\begin{cases} a_{13}b_{31} + a_{23}b_{32} = 0 \\ a_{12}b_{21} + a_{32}b_{23} = 0 \\ a_{11}b_{11} + a_{21}b_{12} + a_{31}b_{13} = 0 \end{cases} $$

Which can be read as "$\operatorname{adj}(A) \cdot A$" has all $0$ in the diagonal and thus its trace is $0$

• Doing the same but along all 3 rows of the matrix yields a similar system of equations that says the same about $A \cdot \operatorname{adj}(A)$.

With ALL of this taken into account, I'm stuck trying to prove that the adjugate has rank 1. Its rank is most certainly not 3 (though I don't know how to prove it, it probably has to do with $A$ being singular), but right now I'm more interested in proving that its rank is not 2.

Any tips in how to proceed are appreciated.

Answer 1

We can prove a stronger statement:

• if $A\in M_3(\mathbb C)$ is singular and $\operatorname{adj}(A)$ has a nonzero trace, then $\operatorname{adj}(A)$ is diagonalisable.

Since $A$ is singular, the rank of its adjugate is at most $1$. As $B=\operatorname{adj}(A)$ has nonzero trace, its rank must be $1$. Let $B=uv^T$. Then $v^Tu=\operatorname{tr}(B)\ne0$. Let $\{y,z\}$ be a basis of the subspace $\{x\in\mathbb C^3:u^Tx=0\}$ and let $$ P=\pmatrix{\frac{v^T}{v^Tu}\\ y\\ z}. $$ Then $P$ is nonsingular and $Pu=e_1=(1,0,0)^T$. Therefore $PBP^{-1}=(Pu)(v^TP^{-1})=e_1\left((v^Tu)e_1^T\right)$ is a diagonal matrix, i.e. $B$ is diagonalisable.

Answer 2

The answer is pretty straightforward once you spend 2 hours re-reading your notes to remember the identity $A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = \operatorname{det}(A) \cdot I_n$

From this identity we get $\operatorname{adj}(A) \cdot A = 0$ since $A$ is singular. In particular it can be rewritten as $$\operatorname{adj}(A) \cdot (A_1 \; A_2\; A_3) = (0 \;0 \;0)$$

Meaning that $\operatorname{Span}\{A_1, A_2, A_3\} \subset \operatorname{ker}(\operatorname{adj}(A))$. Since $A$ has a nonsingular $2\times 2$ minor, its rank is $2$ and then the rank of the adjugate is 1 (since it's not the zero matrix)