1. ## Linear Algebra help

Suppose that A is a square matrix and it has two distinct eigenvalues, lamda 1 and lamda 2. and that dim(E_lamda1)=n-1. Prove that A is diagonalizable.

2. Suppose that A is a square matrix and it has two distinct eigenvalues, lamda 1 and lamda 2. and that dim(E_lamda1)=n-1. Prove that A is diagonalizable.

GIVEN INFORMATION:

A is a square matrix ( n x n ).
A has two distinct eigenvalues, $\displaystyle \lambda_{1}$ and $\displaystyle \lambda_{2}$.
$\displaystyle dim(E_{\lambda_{1}}) = n-1$

Proof:

For an n x n matrix, the dimensions of the eigenspaces must add up to n for the matrix to be diagonalizable. We are given $\displaystyle dim(E_{\lambda_{1}}) = n-1$. Furthermore, we know that $\displaystyle dim(E_{\lambda_{2}}) \geq 1$ and $\displaystyle dim(E_{\lambda_{1}}) + dim(E_{\lambda_{1}}) \leq n$. Thus, $\displaystyle dim(E_{\lambda_{2}}) = 1$.

So, finally, $\displaystyle dim(E_{\lambda_{1}}) + dim(E_{\lambda_{2}}) = (n-1) + 1 = n$. We have proven that A is diagonalizable.

-Andy