Distinct Characteristic Roots - Bellman

Problem: Show that if $\lambda$ is a simple root (multiplicity 1) of $A$, a characteristic vector associated with $\lambda$ can always be taken to be a vector whose components are polynomial in $\lambda$ and the elements of $A$.

Attempt & Question: From an earlier exercise, we showed that if $\lambda$ is a simple root, then at least one of the determinants $|A_k-\lambda I|$ is nonzero where $A_k$ is the matrix $A$ excluding the $k$th row and column (we are hinted to use this fact). From here I'm confused what to do - I know that $|A_k-\lambda I|$ is polynomial of degree $N-1$ with respect to $\lambda$, but how does this imply the components of the characteristic vector are polynomial in $\lambda$ and the elements of $A$. Any help for clarifying the goal and/or tactics to get there is appreciated.

Source: Richard Bellman - "Introduction to Matrix Analysis"

Let $B$ be the adjugate matrix of $A-\lambda I$. Then we have $(A-\lambda I)B=det(A-\lambda I)I=0$. The components of B are polynomial in 𝜆 and the elements of 𝐴. So if we show $B$ is not a zero matrix then some non-zero column vector of $B$ has meet the desired property for $(A-\lambda I)B=0$

We already know $|𝐴_k−𝜆𝐼|$ is nonzero for some $k$. This elements equals to $(k,k)$-th elements of B ($B$ is the adjugate matrix of $A-\lambda I$ !). Let v be the $k$-th column vector of $B$, then $(𝐴−𝜆𝐼)v=0$ ,$v \neq 0$ and each component of $v$ is polynomial in 𝜆 and the elements of 𝐴.

Another proof

It suffice to show $B \neq 0$.

We use a easy but useful property of adjugate matrix. Let $S$ be a $n$ by $n$ matrix and $T$ the adjugate matrix. Then

(1)$rank(S)\leq n-2$ then $T=0$ ($rank(T)=0$).

(2)$rank(S) = n-1$ then $rank(T)=1$

(3)$rank(S)=n$ then $rank(T)=n$

proof of (1) : Any $n-1$-by-$n-1$ submatrix of $S$ has rank less than $n-2$ so the adjugate matrix of S is zero matrix.

proof of (2) : For some $n$ by $n-1$ submatrix W has rank $n-1$ so some $n-1$ by $n-1$ submatrix $W'$ has rank $n-1$ so $det(W')\neq0$. Because the components of $S$ are detarminant of the $n-1$ by $n-1$ submatrix (times power of -1), the adjugate matrix T is greater than zero. On the other hand $ST=0$. This means the image of $T$ is contained in null space of $S$. Then by "dimension formura" $ rank(S)+ rank(\text{null space of $S$})=n$ ,we have $rank(T)=1$.

proof of (3): Trivial

Well, 𝜆 is a simple root. This implies null space of 𝐴−𝜆𝐼 has dimension $1$ and then $rank(A-\lambda I)=n-1$. By (2), the adjugate matrix $B$ of $A-\lambda I$ has rank $1$, i.e.　$B$ is not a zero matrix.

Even if $\lambda$ is not simple root,if the dimension of eigenspace for $\lambda$ is equal to $1$ then same property holds because the dimension of null space is $n-1$. This is a little generalization.

sorry for my poor English...

If you know that the characteristic polynomial of $A$ is $p(\mu)=(\mu-\lambda)q(\mu)$, where $q$ has no root $\lambda$, then we know that there is a vector $x$ such that $q(A)x \ne 0$. From these facts it follows that $$ q(A)x\ne 0,\;\; (A-\lambda I)q(A)x = 0, $$ which gives a non-trivial eigenvector $q(A)x \ne 0$ with eigenvalue $\lambda$.