# Is there a nice expression for the adjugate of a matrix with almost full degree?

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.

• The rows are multiples of the kernel of $A^t$, so the matrix is $k×( ker( A ))(ker(A^t))^t$ for some number $k$ – 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

$$(-1)^{n-1}\det(A)\text{adj}(A)=\det(A)q(A).$$

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 $$\text{adj}(A)=(-1)^{n-1}\frac{p_A(X)}X\left\rvert_{X=A}\right., \ \text{rank}(A)\leq n-1.$$