Diagonalizability with matrices?

Question

Let $A$ be a $n\times n$ matrix. Assume that $A$ has two eigenvalues $\lambda_1$ and $\lambda_2$, and assume that the eigenspace corresponding to $\lambda_1$ is $n-1$ dimensional, i.e. $A-\lambda_1I$ has an $(n-1)$-dimensional null space. Prove that $A$ is diagonalizable.

So I know that since the eigenspace that corresponds to the first eigenvalue is of dimension $n-1$. I assume that the eigenvector of the second eigenvalue will be independent from all the vectors in the first eigenspace, but how do I go about proving that?

For some basic information about writing math at this site see e.g. here, here, here and here. — AlexR, Apr 15 '15 at 07:27
The usual method for proving linear independence. You set up a linear combination, assume it's zero, and deduce, somehow, that the coefficients are all zero. In this case, at some point you'll have to use the fact that those vectors are eigenvectors. — Gerry Myerson, Apr 15 '15 at 07:27
Eigenspaces associated to different eigenvalues are always in direct sum. You may have a look at this question. — Jean-Claude Arbaut, Apr 15 '15 at 07:29

score 5 · Answer 1 · answered Apr 15 '15 at 07:36

The key point is that eigenvectors related to different eigenvalues are always linearly independent. Suppose $v_2\neq 0$ is related to $\lambda_2$, i.e. $Av_2=\lambda_2v_2$. Then it is impossible that $v_2$ is in the null space of $\lambda_1$, because in this case we would have also $Av_2=\lambda_1v_2$, and then $\lambda_1v_2=\lambda_2v_2$, which is impossible if $\lambda_1\neq \lambda_2$ and $v_2\neq 0$.

AlexR · Answer 2 · 2015-04-15T08:54:14.123

First of all, it is implied (and essential) that $\lambda_1 \ne \lambda_2$. Now chose a set of basis vectors $\{u_i\}_{i=1}^{n-1}$ for $\mathrm{Eig}_{\lambda_1}(A)$ (it has dimension $n-1$ by assumption). If an eigenvector $v$ of $\lambda_2$ was linearly dependent of them, then it is inside $\mathrm{Eig}_{\lambda_1}(A)$ and thus $v = \sum_{i=1}^{n-1} \alpha_i u_i$ so $$\lambda_2 v = Av = \sum_{i=1}^{n-1} \alpha_i Au_i = \sum_{i=1}^{n-1} \alpha_i \lambda_1 u_i = \lambda_1 v \Rightarrow \lambda_1 = \lambda_2\quad \huge\unicode{x21af}$$ Now to complete the assignment, nothing else is needed by dimension arguments. We've already found a diagonalisation of $A$:

Let $V = \pmatrix{u_1&\cdots&u_{n-1}&v}\in\mathbb K^{n\times n}$. Then $$A = V \pmatrix{\lambda_1 & & & \\ & \ddots & & \\ & & \lambda_1 &\\ & & & \lambda_2} V^{-1}$$

Bravo. This is the constructive proof. – Vim Apr 15 '15 at 08:05 — Vim, Apr 15 '15 at 08:05

Diagonalizability with matrices?

2 Answers2