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

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.

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)

It is known, that $$\operatorname{adj} {(AB)} = \operatorname{adj} {(B)} \operatorname{adj} {(A)}$$ for any two $$n \times n$$ matrices $$A$$ and $$B$$.

Suppose that $$A = PDP^{-1}$$ for some diagonal matrix $$D$$ and some invertible matrix $$P$$. Then $$\operatorname{adj} {(A)} = \operatorname{adj} {(PDP^{-1})} = \operatorname{adj} {(P^{-1})} \operatorname{adj} {(D)} \operatorname{adj} {(P)}.$$ Since $$I = \operatorname{adj} {(I)} = \operatorname{adj} {(P^{-1} P)} = \operatorname{adj} {(P)} \operatorname{adj} {(P^{-1})},$$ we conclude that $$\operatorname{adj} {(A)}$$ is similar to the diagonal matrix $$\operatorname{adj} {(D)}$$.

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.