# Is adjoint of singular matrix singular? What would be its rank?

Let $$A$$ be a square and singular matrix of order $$n$$.

Is $$\operatorname{adj}(A)$$ necessarily singular? What would be the rank of $$\operatorname{adj}(A)$$?

• The title is a bit misleading, adjoint (most commonly meaning the complex conjugate of the tranpose) should be replaced by adjunct (the transpose of the comatrix).
– Max
Mar 18, 2020 at 6:25

## 2 Answers

No, the adjugate of a singular matrix can be non-singular. But it happens only for the $$1\times 1$$ zero matrix.

Here is a complete classification, referring to this answer and this one. One always has $$\def\adj{\operatorname{adj}}A \cdot \adj(A) = \det(A) I_n.$$

• If $$A$$ has rank$$~n$$, then it is invertible, and so is $$\det(A)$$, and $$\adj(A)=\det(A)A^{-1}$$ is invertible too, and has rank$$~n$$.

• If $$A$$ has rank$$~n-1$$ then at least one $$(n-1)\times(n-1)$$ minor is nonzero, and so $$\adj(A)\neq0$$. On the other hand by the given relation the image of $$\adj(A)$$ is contained in the kernel of $$A$$ which has dimension$$~1$$ by rank-nullity; it follows that $$\adj(A)$$ has rank$$~1$$ in this case.

• If $$A$$ has rank$${} then all $$(n-1)\times(n-1)$$ minors are equal to zero, and so $$\adj(A)$$ has rank$$~0$$.

The cases where $$\adj(A)$$ has rank$$~n$$ are the first case for any$$~n$$, and the second case for $$n=1$$. (And, I would be inclined to say, the last case for $$n=0$$; but that of course cannot happen at all.) So the only case where $$A$$ is singular but $$\adj(A)$$ is not, is the case $$A=(0)$$ (with $$n=1$$).

• If $A$ has rank $n-1$, how do we prove that there exists at least one cofactor matrix of $A$ with non zero determinant? That is, why is there some $i, j$ such that $A_{ij} \neq 0$?
– Anu
Feb 16, 2018 at 12:27
• Because one characterisation of the rank is that it is the largest size of a nonzero minor (with the $0\times0$ minor always being $1$, this is well defined). To see that such a nonzero minor indeed exists, let $r$ be the rank; select $r$ linearly independent columns (possible because $r$ is the column rank), which form a sub-matrix still of rank $r$, then select from it $r$ linearly independent rows (possible because $r$ is the row rank), they from a non-singular $r\times r$ sub-matrix. Feb 20, 2018 at 6:21
• How the image of $adj(A)$ lies in the Kernal of $A$? Can you please explain?
– user464147
May 28, 2019 at 5:09
• @unknownx A matrix equation $P\cdot Q=0$ means that the image of$~Q$ is contained in the kernel of$~P$. This is an instance of that general fact, with $Q=\adj(A)$ and $P=A$. May 28, 2019 at 5:29
• for the case where $A$ has rank $n-1$ you're assuming $n \ge 2$ ?
– BCLC
Oct 23, 2021 at 18:11

Hint : $$A \cdot\mbox{adj}(A)=\det(A) I$$

If $$\det(A)=0$$, we get $$A \cdot\mbox{adj}(A)=0$$. Can $$\mbox{adj}(A)$$ be invertible?

For the rank, if you are familiar with linear transformations, prove that the above relation implies that the image of the transformation defined by $$A$$ must be in the kernel of the transformation defined by $$\mbox{adj}(A)$$. This yields an inequality of ranks.

• If $|A| =0$ then $|A \cdot \text{adj}(A)| = |A| \cdot | \text{adj}(A)| =0$. Then I don't think it says anything about $|\text{adj}(A)|$? So how do we conclude if it is invertible or not? Feb 7, 2022 at 21:01
• @William you are making the problem too complicated by looking at determinants. You know that $A \cdot\mbox{adj}(A)=0$.. Now, if $AB=0$ and $B$ is invertible, what does this tell you about $A$? Feb 8, 2022 at 1:12
• That $A=0$ is the zero matrix? I don't follow? Feb 8, 2022 at 8:43
• Oh if $A=0$ then $\text{adj}(A)=0$ hence non invertible. If $A ≠0$ then $\text{adj}(A)$ can't be invertible because that would imply $A=0$, hence a contradiction. Is that it? Have I done it? Feb 8, 2022 at 8:45
• @William Exactly. Feb 8, 2022 at 14:52