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.
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