Diagonalizability of 3x3 Matrices with 3 Distinct Eigenvalues

[jkeatin]

Homework Statement

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

Homework Equations

The Attempt at a Solution

 

[Vid]

How is diagolization related to the dimension of the eigenspace? Why would distinct eigenvalues let you satisfy this dimension criteria?

 

[jkeatin]

well since its a 3x3 matrix, the most eigenvalues it could have is 3? so since there is 3 unique eigenvalues, then it is definitely diagonalizable?

 

[Vid]

You need two theorems to show this is true and I hinted at them in my post.

 

[HallsofIvy]

I think it is better to think in terms of linear transformations- any linear transformation can be represented as a matrix in a given basis: Apply the linear tranformation to each of the basis vectors in turn, the write the result as a linear combination of the basis vectors- the coefficients are the columns of the matrix.

The three distinct eigenvalues must have 3 independent eigenvectors. Using those eigenvectors as a basis for the vector space, the linear operator is represented by a diagonal matrix with the eigenvalues on the diagonal.

Equivalently, if A is a matrix with three distinct eigenvalues, B is the matrix having those three eigenvectors as columns, then B^-1AB is the diagonal matrix having the eigenvalues on the diagonal.

By the way, having three independent eigenvectors is a necessary condition for a matrix to be diagonalizable. Having three distinct eigenvalues is not necessary.

 

[jkeatin]

so what if a 3x3 matrix only has two eigenvalues, does that mean its not able to be diagonalized?

 

[Vid]

No, reread what Halls wrote. One of the eigenvalues could have a two dimensional eigenspace.

 

[jkeatin]

Thanks guys, I think its making sense, we will see on finals week lol.