How to prove $\tilde{A}=O$ if $\mathrm{rank}(A) \le n - 2$

Question

In the final exam of the linear algebra course I took, the following problem was asked and I wasn't able to solve it:

Let $n \ge 2$ and $A$ be a $n \times n$ matrix. If $\mathrm{rank}(A) \le n - 2$, then prove $\tilde{A} =O$.

In this course $\tilde{A}$ denotes the adjugate matrix of $A$. I suspect that using $\ker{A} \ge 2$ by rank-nullity theorem works well, but I don't know how to prove it. I'm glad for anyone to tell me any hints or any solutions.

Answer 1

The proof depends on your definition of adjugate.

• If adjugate is defined using minors, note that each element of $\tilde A$ is an $(n-1)$-rowed minor of $A$.
• If adjugate is defined via characteristic polynomial, let the characteristic polynomial of $A$ be $x^mq(x)$, where $m\ge2$ and $q$ is a polynomial of degree $n-m$ such that $q(0)\ne0$. Since $\dim\ker A\ge2$, the ascending chain $\ker A\subseteq\ker A^2\subseteq\cdots$ must have been stabilised before the exponent of $A$ reaches $m$. Therefore $\ker A^{m-1}=\ker A^m$ and in turn, $A^{m-1}q(A)=0$. It follows that $\tilde{A}=(-1)^{n-1}A^{m-1}q(A)=0$.
• If adjugate is defined using exterior product, pick any two linearly independent vectors $x_1,x_2\in\ker(A)$ and let $\{x_1,x_2,\ldots,x_n\}$ be a basis of the ambient vector space $V$. For any $v\in V$, note that $$ \tilde{A}v\wedge\bigwedge_{i\in[n]\setminus\{j\}} x_i =v\wedge\bigwedge_{i\in[n]\setminus\{j\}} Ax_i=0 $$ for $j=1,2,\ldots,n$.