# Diagonalizable matrix

• April 5th 2010, 01:37 PM
temaire
Diagonalizable matrix
Are the following matrices diagonalizable:

http://img693.imageshack.us/img693/4198/91350081.jpg

I think neither are diagonalizable because they don't have at least two linearly independent eigenvectors. Am I correct?
• April 5th 2010, 02:34 PM
harish21
Quote:

Originally Posted by temaire
Are the following matrices diagonalizable:

http://img693.imageshack.us/img693/4198/91350081.jpg

I think neither are diagonalizable because they don't have at least two linearly independent eigenvectors. Am I correct?

To diagonalize a matrix, I would first find the eigenvectors. The three eigenvectors $v_1, v_2, v_3$ will form the three columns(c1,c2,c3) of the matrix $P$.

If P is invertible , then the matrix given by $P^{-1} A P = D$ is the diagonal matrix
• April 5th 2010, 02:41 PM
temaire
Quote:

Originally Posted by harish21
To diagonalize a matrix, I would first find the eigenvectors. The three eigenvectors $v_1, v_2, v_3$ will form the three columns(c1,c2,c3) of the matrix $P$.

If P is invertible , then the matrix given by $P^{-1} A P = D$ is the diagonal matrix

I have found the eigenvectors for both matrices. However, I only found one eigenvector for each matrix, which means that the two matrices do not have at least two linearly independent eigenvectors, which means that they are not diagonalizable. Is this correct?
• April 5th 2010, 04:32 PM
harish21
Quote:

Originally Posted by temaire
I have found the eigenvectors for both matrices. However, I only found one eigenvector for each matrix, which means that the two matrices do not have at least two linearly independent eigenvectors, which means that they are not diagonalizable. Is this correct?

A $n \times n$ matrix is diagonalizable only if it has n linearly independent eigenvectors.

So your matrix is not diagonalizable!