# checking if matrix is diagonalizable

• Mar 29th 2011, 11:25 AM
Jskid
checking if matrix is diagonalizable
My textbook gives two theorems that are used to determine if a matrix is diagonilable
1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

So if I have characteristic polynomial with two roots that are the same (e.g. $\lambda=1$ but it has n linearly independent eigenvectors, which one takes presidents? Or should the two always agree and this means I made a mistake?
• Mar 29th 2011, 11:30 AM
tonio
Quote:

Originally Posted by Jskid
My textbook gives two theorems that are used to determine if a matrix is diagonilable
1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

So if I have characteristic polynomial with two roots that are the same http://mathhelpforum.com/advanced-algebra/176204-checking-if-matrix-diagonalizable.html#post634879\" rel=\"nofollow\">
My textbook gives two theorems that are used to determine if a matrix is diagonilable
1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

So if I have characteristic polynomial with two roots that are the same (e.g. $\lambda=1$ but it has n linearly independent eigenvectors, which one takes presidents? Or should the two always agree and this means I made a mistake?

the 2nd statement is sufficent but not required. It is just saying the if you have no repeated eigenvalues and using the fact that each eigenspace must have dimention at least 1 you will always get n linearly independent eigenvectors.

The problem that can occur is if you have a repeated eigenvalue say of multiplicity 2 the eigenspace for that eigenvalue may only have 1 linearly indpendant eigenvector. So the matrix is not diagonalizable. But if you do get two linearly independant eigenvectors then the matrix would be diagonalizable.
• Mar 29th 2011, 11:51 AM
Jskid
Quote:

Originally Posted by TheEmptySet
The problem that can occur is if you have a repeated eigenvalue say of multiplicity 2 the eigenspace for that eigenvalue may only have 1 linearly indpendant eigenvector. So the matrix is not diagonalizable. But if you do get two linearly independant eigenvectors then the matrix would be diagonalizable.

That's what happened and confused me.