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Prove if $\operatorname{adj}(A) =0$, then $\operatorname{rank}A \leq n-2$

My initial idea is to setup an induction. Base case $A \in M_{2×2}$ which obviously implies that $A = [O]_{2×2} \rightarrow \operatorname{rank}A \leq 0 = n-2$. However, I can't seem to reason how assuming it worked in the $M_{n\times n}$ case allows for the $M_{(n+1)×(n+1)}$ case to work....

Bernard
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A. Radek Martinez
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2 Answers2

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I don’t think that it require to use induction here. For avoid possible confusion, let’s just separate the case when $n=2$ and $n \neq 2$.

Proof. Let $n \in \Bbb{N}$ such that $n \geq 2$ and $A \in \mathcal{M}_{n}(\Bbb{K})$ and suppose that $\text{adj}(A) = \mathcal{O}_n$. If $n = 2$, then (as you well pointed out) $A = \mathcal{O}_2$. Since $\text{rank}\mathcal{O}_2 = 0$ and $0 \leq 0 = 2 - 2$, the proposition if true. Suppose now that, $n > 2$. Since, $\text{adj}A = \mathcal{O}_{n}$, we know that $\det{A} = 0$. So, $A$ is not invertible, which means that $\text{rank}A < n$, i.e., $\text{rank}A \leq n-1$. Suppose that $\text{rank}A = n-1$. This means that there is one row that is a linear combination of the remaking rows. Then, if we consider the matrix of size $n-1 \times n$ that we obtain by eliminating that row, we get a matrix that has rank equal to $n-1$. This means that we also have a column in this $n \times 1$ that is a linear combination of the remaining columns. So, the matrix of size $n-1 \times n-1$ that we obtain by further eliminating this column has rank equal to $n-1$. This means that this matrix is invertible. Hence, its determinant is not null. But this implies that one of the entries of $\text{adj}A$ is not zero, which is a contradiction. Therefore, the rank of $A$ cannot be $n-1$. Therefore, we conclude that $\text{rank}A \leq n-2$. $\square$

Air Mike
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  • There are a few problems with your answer. First, you don't explain your notation: what do you mean by $A(i|j)$? Second, you seem to assume that if $A$ is equivalent to $B$ and $B$ has a non-zero adjoint, then $A$ has a non-zero adjoint. It is not clear how one would justify this – Ben Grossmann Mar 21 '21 at 05:03
  • @BenGrossmann thank you for the feedback. I already edited my question so it becomes more clear (I would appreciate if you could take a look and give some feedback) – Air Mike Mar 21 '21 at 09:54
  • What I was saying was that in that matrix we can eliminate a row $i$ and a column $j$ such that the remaking matrix of size $n-1 \times n-1$ would be invertible. When I wrote $A(i|j)$ that means the matrix of size $n - 1 \times n -1$ that we obtain by eliminating the row $i$ and column $j$ – Air Mike Mar 21 '21 at 09:56
  • Unfortunately I forgot to mention what that means. Once again, thank you for your feedback. I really appreciate it. – Air Mike Mar 21 '21 at 09:56
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One approach is to use the notion of "determinantal rank", as explained in this post. In particular: if the adjoint of $A$ is zero, then all size $n-1$ submatrices of $A$ have determinant zero which means that $A$ has rank at most $n-2$.

For a direct proof: we are given that $\operatorname{adj}(A) = 0$. Suppose for the purpose of contradiction that $\operatorname{rank}(A) \geq n-1$. It follows that we may select $n-1$ linearly independent rows of $A$. This submatrix has rank $n-1$, which means that we may select $n-1$ linearly independent columns. These columns form a square, size $(n-1)$ submatrix of $A$.

Because this submatrix is invertible, its determinant is non-zero. However, this means that one of the entries of $\operatorname{adj}(A)$ is non-zero, contradicting our premise.

Ben Grossmann
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