relation eigenvalue and adj(A-λI)

Question

Let $A$ be a matrix in $\mathbb C^{n×n}$, let $λ$ be an eigenvalue of $A$ with eigenvector $x$.

Why is there some $y \in \mathbb C^n$ such that $adj(A−λI)=x{y^*}$?

Answer 1

The question basically asks: given a singular matrix $B\in F^{n,n}$, $F$ any field, with $Bx=0$, $x\ne 0$, then there is $y\in F^n$ such that $$ adj(B) = x y^T. $$ This implies that in this case $adj(B)$ has rank at most one, with a very special structure.

First, assume that rank of $B$ is less than $n-1$. This implies $adj(B)=0$, as all minors of size $(n-1)\times (n-1)$ do not have full rank.

Second, assume $rank(B)=n-1$. Then there are invertible matrices $P,Q$ such that $$ B = P \pmatrix{ I_{n-1} & 0 \\ 0 & 0 } Q, $$ with the property $adj(AB)=adj(B)adj(A)$, we have $$ adj(B) = adj(Q) adj\left(\pmatrix{ I_{n-1} & 0 \\ 0 & 0 }\right) adj(P) = adj(Q) \pmatrix{ 0_{n-1}& 0 \\ 0 & I_1 } adj(P), $$ which implies that $rank(adj(B))=1$. Hence there are vectors $a,b\in F^n$, $b\ne 0$, such that $$ adj(B)=ab^T. $$ Since $B$ is singular, it holds $Badj(B)=0$, which implies $$ (Ba)b^T=0. $$ As $b\ne0$, this implies $Ba=0$, hence $a$ is in the null space of $B$, which is by assumption spanned by $x$. Then $a = cx$ with some $c\in K$. This proves $$ B = ab^T = cxb^T = x (cb)^T, $$ which was the claim.

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This is a rather lengthy answer. Is anyone aware of a more elegant approach?

Answer 2

As daw points out, your question boils down to proving that if $B$ is a singular matrix and $Bx=0$ for some nonzero vector $x$, then $\renewcommand{\adj}{\operatorname{adj}}\adj(B)=xv^T$ for some vector $v$. User "1015" (then "julien") has actually explained why this is the case in another thread. Essentially, this is because the rank of $\adj(B)$ is at most $1$ when $B$ is singular:

• If $\renewcommand{\rank}{\operatorname{rank}}\rank(B)\le n-2$, every minor of $B$ of size $n-1$ will be zero; therefore $\adj(B)=0=x0^T$.
• If $\renewcommand{\rank}{\operatorname{rank}}\rank(B)= n-1$, then $B$ has a one-dimensional null space. Therefore the null space is the linear span of $x$. However, since $B\adj(B)=\det(B)I=0$, it follows that every column of $\adj(B)$ is in the null space of $B$, and in turn a scalar multiple of $x$. Hence $\adj(B)=xv^T$ for some vector $v$.

The above argument does not depend on the ground field. Hence transpose is used instead of conjugate transpose. In the case where the ground field is $\mathbb C$, simply put $y=\bar{v}$.