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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.

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    Null space of $A-\lambda I$ is one dimensional. – 2017-01-05
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    What is $p_{A-λI}(t)$? Since $p_A(t)=\det(A-λI)$, I understand that $$p_{A-λI}(t)=\det(A-λI-λΙ)=\det(A-2λI)$$ Is this what you want? – 2017-01-05
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    It should be correct now. Thanks for the help. Sorry. – 2017-01-05

3 Answers 3

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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 $rank(A-\lambda I)=n-1$.

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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 $$\dim\ker (A-\lambda I)\leq \operatorname{mult}(\lambda)$$

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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.

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    After thinking about it for more than a second, it's pretty obvious. Since $\lambda$ is not an eigenvalue of $A_{i}$, $\det{A_{i} - \lambda I} \neq 0$, which implies invertibility. – 2017-01-05