Let $A$ be a real $d \times d$ matrix, and suppose that $\text{rank}(A)=d-1$.

Is there a nice expression (say in terms of matrix operations) for the adjugate of $A$?

Recall the adjugate matrix $\text{adj} A$ satisfies

$$ A \cdot \text{adj} A=\det A \cdot \text{Id},$$

so when $A$ is invertible, we get $\text{adj} A=\det A \cdot A^{-1}$.

I am wondering if there is a similar expression for $\text{adj} A$ when $\text{rank}(A)=d-1$.

(If $\text{rank}(A)<d-1$, $\text{adj}A=0$).

Ideally, I would like something involving standard matrix operations; I tried using generalized inverses but this failed.

If it helps, here is a description of the diagonal case:

If $A=\text{diag}(0,\lambda_2,\dots,\lambda_d)$ then $\text{adj} A=\text{diag}(\Pi_{i=2}^d \lambda_i,0,\dots,0)$.


This is probably not what you are looking for but your description generalizes for arbitrary matrices. In the language of operators, assume $T \colon \mathbb{F}^d \rightarrow \mathbb{F}^d$ is an operator of rank $d - 1$ and choose some basis $e_1,\dots,e_d$ of $\mathbb{F}^d$ such that $Te_1 = 0$. Set $V = \operatorname{span} \{ e_2, \dots, e_d \}$ and let $\iota \colon V \hookrightarrow \mathbb{F}^d$ be the inclusion and $p \colon \mathbb{F}^d \rightarrow V$ be the projection (with respect to the chosen basis).

Then with respect to the basis $e_1,\dots,e_d$, the operator $\operatorname{adj}(T)$ is diagonal with $\operatorname{adj}(T)(e_1) = \det(p \circ T \circ \iota) e_1$ and $$\operatorname{adj}(T)(e_2) = \dots = \operatorname{adj}(T)(e_d) = 0. $$ When $T$ is diagonal with respect to the basis $e_1,\dots,e_d$, we recover your description.

In general, if $A$ has rank $d - 1$ then $\operatorname{adj} A$ has rank one and each column of $\operatorname{adj} A$ belongs to the kernel of $A$ (by the identity $A \cdot \operatorname{adj} A = \det(A) I$). So to write down a "more explicit" formula for $\operatorname{adj} A$ (other than the formula which involves the determinants of the minors) you would have to somehow write down a formula for an element in the kernel of $A$ which I assume will necessarily involve some sort of determinant.

  • $\begingroup$ The rows are multiples of the kernel of $A^t$, so the matrix is $k×( ker( A ))(ker(A^t))^t$ for some number $k$ $\endgroup$ – Empy2 Sep 19 '18 at 14:37

Let $A\in\mathbb{M}_n(K)$ and let $p_A$ be its characteristic polynomial. We have $p_A(X)=Xq(X)+(-1)^n\det(A)$ for some polynomial $q\in K[X]$, so that $$(-1)^{n-1}\det(A)I_n=Aq(A).$$

Multiply by $\text{adj}(A)$ on the left to get


Now consider $\det(A), q(A), \text{adj}(A)$ as polynomials in the entries of $A$, in $K[X_1,\ldots,X_{n^2}]$. Since $\det$ is a nonzero element in an integral domain, we can cancel it and get $$\text{adj}(A)=(-1)^{n-1}q(A).$$

Therefore $$\text{adj}(A)=(-1)^{n-1}\frac{p_A(X)-(-1)^n\det(A)}X\left\rvert_{X=A}\right.$$ is the formula we are seeking. In the special case that $\text{rank}(A)=n-1$, we have $\det(A)=0$ and $$\text{adj}(A)=(-1)^{n-1}\frac{p_A(X)}X\left\rvert_{X=A}\right., \ \text{rank}(A)\leq n-1.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.