I'm working on this problem:
Given $A$ be an $n\times n$ complex matrix with $\operatorname{rank} A = n-1$, how to find its adjugate matrix?
I have solved the case when (algebraic) multiplicity of eigenvalue $0$ is $1$.
Definition
The adjugate of a matrix $A = (a_{ij})_{n\times n}$, denoted $\operatorname{adj} A$, is $(M_{ji})_{n\times n}$, where $M_{ij}$ is the determinant of the submatrix of $A$ with row $j$ and column $i$ removed, multiplied by $(-1)^{i+j}$.
The conjugate transpose of an $m\times n$ matrix $A$ is denoted $A^*$.
Pre conditions
Assume a Jordan form decomposition of $A$ is given:
$$
A = MJM^{-1},
$$
which means all eigenvalues of $A$, plus their multiplities, the Jordan blocks, and the corresponding Jordan basis are known.
It would be welcomed if you need to add new conditions to answer this question.
My Solution
As $\operatorname{rank} A = n-1$, there exists $p\neq 0$ such that $Ap=0$, and if $Ax=0$ then $x=kp$ for some $k\in \mathbb{C}$.
For $A^*$, there also exists $q\neq 0$ such that $A^*q=0$, and if $A^*y=0$ then $y= k'q$ for some $k'\in \mathbb{C}$.
Note that $\operatorname{rank} \operatorname{adj} A = 1$, so there exists nonzero $x, y\in \mathbb{C}^n$ such that $\operatorname{adj} A = xy^*$.
As $(\operatorname{adj} A)A = A(\operatorname{adj} A) = (\det A)I = 0$, we have
$$
x(A^*y)^* = (Ax)y^* = 0.
$$
Then $x,y\neq 0$ implies $A^*y = Ax=0$, which means there exists a scalar $k\in \mathbb{C}$ such that
$$
\operatorname{adj} A = k(pq^*).
$$
Set all eigenvalues of $A$ be $\lambda_1,\dots,\lambda_n$, where $\lambda_1=0$, which are repeated by their multiplicities. A Jordan form of $A$ is
$$
J = \begin{pmatrix}
J_1 & &\\
& \ddots &\\
& & J_m
\end{pmatrix}
$$
where $J_1 = (0)$ and $J_k$ is the Jordan block with the same nonzero entries
on its diagonal and $1$'s on its upper subdiagonal.
If
$$
J = \begin{pmatrix}
\lambda & 1 & &\\
& \ddots & \ddots & \\
& & \ddots & 1\\
& & & \lambda
\end{pmatrix}
$$
is a Jordan block with $\lambda\neq 0$, then its inverse is $J^{-1} = (k_{ij})$, where
\begin{equation}
k_{ij} = \begin{cases}
0,&\quad\text{if $i>j$},\\
\lambda^{-1},&\quad\text{if $i=j$},\\
(-1)^{i+j}\lambda^{i-j+1},&\quad\text{if $i<j$}.
\end{cases}\tag{1}
\end{equation}
For matrix $A$, there exists a matrix $M$ whose first column is $p$, such that
$$
A = MJM^{-1}.
$$
By the continuity of matrices, there exists $\delta>0$ such that
$$
\det(A+\epsilon I)\neq 0 \quad \text{for all $\epsilon \in (0,\delta)$}.
$$
Then
\begin{equation}\begin{split}
\operatorname{adj}(A+\epsilon I) &= \det(A+\epsilon)(A+\epsilon I)^{-1}\\
&= \prod_{i=1}^n{(\lambda_i+\epsilon)}(M(J+\epsilon I)^{-1}M^{-1})\\
&= M\begin{pmatrix}
\prod_{i=2}^n(\lambda_i+\epsilon) & & & \\
& \prod_{i=1}^n(\lambda_i+\epsilon)(J_2+\epsilon I)^{-1} & &\\
& &\ddots & \\
& & & \prod_{i=1}^n(\lambda_i+\epsilon)(J_k+\epsilon I)^{-1}
\end{pmatrix}M^{-1}.
\end{split}\tag{2}\end{equation}
By the structure of $J_k^{-1}$ with $k\geq 2$, when $\epsilon \to 0$, $\prod_{i=1}^n(\lambda_i+\epsilon)(J_k+\epsilon I)^{-1}\to 0$.
This means
\begin{equation}
\operatorname{adj}(A) = \lim_{\epsilon\to 0}\operatorname{adj} (A+\epsilon I) = M\begin{pmatrix}
\lambda^* & & &\\
& 0 & &\\
& &\ddots &\\
& & & 0
\end{pmatrix}M^{-1},\tag{3}
\end{equation}
where $\lambda^* = \prod_{i=2}^n \lambda_i$.
For $\operatorname{adj} A = kpq^*$, the nonzero eigenvalue of $\operatorname{adj} A$ is $k(q^*p)$. From $(1)$, the nonzero eigenvalue of $\operatorname{adj} A$ is $\lambda^*$, so
$$
k = (q^*p)^{-1}\prod_{i=2}^n\lambda_i,
$$
and
$$
\operatorname{adj} A = \prod_{i=2}^n \lambda_i \frac{pq^*}{q^*p}.\tag(4)
$$
Problem remained
When the mutiplicity of eigenvalue $0$ is not $1$, $J_1\neq (0)$, and $\prod_{i=1}^n(\lambda_i+\epsilon)(J_1+\epsilon I)^{-1}$ diverges when $\epsilon \to 0$, so the method above seem to fail.
At the same time, I'm also not sure whether equation $(3)$ works. Are there any algebraic method solving this problem?
Processes (updating)
-
Equation (4) seems to be true for arbitrary matrices by numerical checking; sometimes $(-1)$ should be multiplied. I think that it depends on whether $d$, the multiplicity of eigenvalue $0$, is even or odd. (It seems to be $(-1)^{d+1}$.)
-
If $A = MJM^{-1}$ with multiplicity of eigenvalue $0$ be $d$, then $\operatorname{adj}(A) = kME_{1d}M^{-1}$, where $E_{1d}$ is an $n\times n$ matrices with $(1,d)$ entry be $1$, and $0$ elsewhere.
Best Answer
I haven't examined the details of your approach, but it seems overly complicated. Here's another approach. First note that $\mathop{\mathrm{adj}}(A)$ is characterized by the identity $$\det(Ax_1,\ldots,Ax_{n-1},x_n)=\det(x_1,\ldots,x_{n-1},\mathop{\mathrm{adj}}(A)x_n)\tag{1}$$ If $\mathop{\mathrm{rank}}A=n-1$, then $\mathop{\mathrm{im}}\mathop{\mathrm{adj}}(A)=\ker(A)$. Indeed, on the one hand $$A\mathop{\mathrm{adj}}(A)x=\det(A)x=0x=0$$ so $\mathop{\mathrm{im}}\mathop{\mathrm{adj}}(A)\subseteq\ker(A)$. On the other hand we can choose a basis $v_1,\ldots,v_n$ with $\langle v_n\rangle=\ker(A)$ and choose $v$ so that $Av_1,\ldots,Av_{n-1},v$ is a basis. Plugging these values into (1) shows that $\mathop{\mathrm{adj}}(A)v\ne0$ so $0\ne\mathop{\mathrm{im}}\mathop{\mathrm{adj}}(A)=\ker(A)$.
It follows that $\mathop{\mathrm{adj}}(A)=f\otimes v_n$ where $f$ is a linear functional. If we scale so that $\det(v_1,\ldots,v_n)=1$, then by (1) $$f(x)=\det(Av_1,\ldots,Av_{n-1},x)\tag{2}$$ You can plug the basis vectors in (2) to compute an explicit matrix representation for $\mathop{\mathrm{adj}}(A)$.