Rank of "Characteristic Matrix": $A - \lambda I$ Matrix Analysis Horn, Johnson 1st ed. Problem: 1.2.6

If $\lambda \in \sigma(A)$ $(\lambda$ is an eigenvalue of $A)$, for $A \in \mathbb{C}^{n \times n}$, has multiplicity $1$ as a root of the characteristic polynomial $p_{A}(t) = \det(A - \lambda I)$, show that $\text{rank}(A - \lambda I) = n - 1$.

The hint suggests this route:
In a previous problem we have shown that
$$\frac{d}{dt}p_{A}(t) = \sum\limits_{i=1}^{n}p_{A_{i}}(t)$$
where $A$ is the matrix obtained by the deleting the $i$-th row and column of $A$. Since $\lambda$ is a single root of $p_{A}$, $p_{A}'(\lambda) \neq 0$ (correct? is there another reason this is true?). This implies that at least one of the characteristic polynomials of the principal submatricies of size $n-1$ is nonzero at $\lambda$. 
I'm now to conclude that one of these principle submatrices is nonsingular. I can't see why though.
 A: Lemma: Let $A  \in \mathbb{C}^{n \times n}$ and $\lambda$ be an eigenvalue for $A$. Let $\operatorname{mult}(\lambda)$ denote the multiplicity of $\lambda$ as a root of $\chi_A$. Then $\dim\ker (A-\lambda I)\leq \operatorname{mult}(\lambda)$.
Here this implies $\dim\ker (A-\lambda I)\leq 1$ and since $\lambda$ is an eigenvalue we also have $\dim\ker (A-\lambda I)\geq 1$, hence $\operatorname{rank}(A-\lambda I)=n-1$ by the null-rank theorem.
Proof of the lemma: Let $F$ be a vector space such that $\ker (A-\lambda I) \oplus F = \mathbb C^n$. Let $p=\dim \ker (A-\lambda I).$
In this new decomposition, $A$ has the form
$$
\left[
\begin{array}{c|c}
\lambda I_p & B \\
\hline
0 & C
\end{array}
\right]
$$
that is to say there exists some invertible $P$ such that $A=P^{-1}\left[
\begin{array}{c|c}
\lambda I_p & B \\
\hline
0 & C
\end{array}
\right]P$ 
hence $\chi_A(t) = (\lambda-t)^p\det(C-tI_{n-p})=(\lambda-t)^pQ(t)$ where $Q$ is some polynomial. That implies $$p\leq \operatorname{mult}(\lambda)$$ie $$\dim\ker (A-\lambda I)\leq \operatorname{mult}(\lambda)$$
A: When you substitute $\lambda$ for $t$, the left-hand side is nonzero; thus at least one of the summands in the right hand side is different from zero.
Therefore, $\lambda$ is not an eigenvalue of $A_i$, for some $i$, which means the matrix $A_i-\lambda I$ is invertible. Such a matrix is obtained from $A-\lambda I$ by removing the $i$-th row and column, so $A-\lambda I$ has $n-1$ linearly independent columns.
A: The algebraic multiplicity of eigenvalue $\lambda$ is $1$ so the geometric multiplicity will be $\le1$. Let $v$ be the eigenvector corresponding to $\lambda$, then $(A-\lambda I)v=0$ gives only one linearly independent eigenvector as its solution which in turn implies that $rank(A-\lambda I)=n-1$.
A: let $A$ be a $n\times n$ matrix, and if $c$ is an eigenvalue of the characteristic equation, then $|A-cI|$ would be $0$ and hence it won't be a full rank matrix (rank $< n$)


*

*In fact, if it is a simple eigenvalue i.e. its algebraic multiplicity is 1, same as geometric multiplicity, then the rank would be $n-1$.

*If it has algebraic multiplicity > $1$ and $\operatorname{dim}\big(\operatorname{Ker} (A-cI)\big)=g$ (geometric multiplicity), then the rank would be $n-g$.( According to the rank nullity theorem which says that $\text{nullity} + \text{rank} = n$, here nullity is $g$, so we have rank of the characteristic equation $= n-g$)
