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

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.

• What can you say about the determinant of $A$? And how is that related to the adjugate matrix? Mar 20 at 4:13
• Hint: since $\operatorname{rank}(A) < n - 1$, any choice of $n - 1$ row vectors from $A$ are linearly dependent. Moreover, the application of any linear map to these vectors will still keep them linearly dependent (e.g. linear maps that eliminate a coordinate). Mar 20 at 4:14

• 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 $$j=1,2,\ldots,n$$.