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"
 A: 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  and the elements of . 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 $|_k−|$ 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 $(−)v=0$ ,$v \neq 0$ and each component of $v$ is polynomial in  and the elements of .
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,  is a simple root. This implies null space of  − 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...
A: 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 $\lambda$.
