1. ## [SOLVED] Diagonalization

if there is a 3x3 matrix which has exactly 3 distinct eigenvalues why must it be diagonalizable?

2. Originally Posted by jkeatin
Suppose now that A is some other 3x3 matrix which has exactly 3 distinct eigenvalues. why A must be diagonalizable.
well, it depends how much you know about diagonalizable matrices. i'll assume you know nothing but this definition that an $n \times n$ matrix $A$ with entries in a field $\mathbb{F}$ is diagonalizable iff $\mathbb{F}^n$ has a

basis $\mathfrak{B}$ such that $[T]_{\mathfrak{B}}$ is a diagonal matrix, where as usual $T: \mathbb{F}^n \longrightarrow \mathbb{F}^n$ is defined by $T(\bold{x})=A\bold{x}.$

so suppose an $n \times n$ matrix $A$ with entries in a field $\mathbb{F}$ has $n$ distinct eigenvalues in $\mathbb{F}$, say $\lambda_i, \ 1 \leq i \leq n.$ let $A\bold{x}_i=\lambda_i \bold{x}_i$ with $\bold{0} \neq \bold{x}_i \in \mathbb{F}^n.$ then $\mathfrak{B}=\{\bold{x}_1, \cdots, \bold{x}_n \}$ is a basis for $\mathbb{F}^n. \ \ \ \ \ \ \color{red}(*)$

now consider the linear map $T: \mathbb{F}^n \longrightarrow \mathbb{F}^n$ defined by $T(\bold{x})=A \bold{x}.$ then $T(\bold{x_i})=\lambda_i \bold{x}_i$ and thus: $[T]_{\mathfrak{B}}=\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ . & . & \cdots & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}. \ \Box$

$\color{red}(*)$ to see this we only need to show that $\bold{x}_1, \cdots, \bold{x}_n$ are linearly independent. so suppose $c_{i_1}\bold{x}_{i_1} + \cdots + c_{i_k}\bold{x}_{i_k}=\bold{0},$ where not all of the coefficients are 0 and $k$ is minimal. call it (1). taking $T$

of both sides of (1) gives us: $c_{i_1}T(\bold{x}_{i_1}) + \cdots + c_{i_k}T(\bold{x}_{i_k})=\bold{0}$ and hence $c_{i_1}\lambda_{i_1}\bold{x}_{i_1} + \cdots + c_{i_k}\lambda_{i_k}\bold{x}_{i_k}=\bold{0}.$ call this one (2). now multiply (1) by $\lambda_{i_1}$ and subtract the result from (2) to get:

$c_{i_2}(\lambda_{i_2}-\lambda_{i_1})\bold{x}_{i_2} + \cdots + c_{i_k}(\lambda_{i_k}-\lambda_{i_1})\bold{x}_{i_k}=\bold{0},$ which contradicts the minimality of $k$ unless $c_{i_2}(\lambda_{i_2}-\lambda_{i_1}) = \cdots = c_{i_k}(\lambda_{i_k}-\lambda_{i_1})=0.$ since $\lambda_i$ are distinct, we must have $c_{i_2}= \cdots = c_{i_k}=0,$

and thus $c_{i_1}=0$ by (1). contradiction! thus the elements of $\mathfrak{B}$ are linearly independent. Q.E.D.

3. so it basically has to be linearly independent? meaning 3 independent eigenvectors?