# Thread: Prove A is digagonalizablie if...

1. ## Prove A is digagonalizablie if...

A is an nxn matrix with two distinct eiganvalues lambda 1 and lambda 2

the dimension of the eigenspace for lambda 1 is n-1

prove that A is diagonalizable
.................................................. ..................................
so i know the algebraic multiplicity for the first eiganvalue is greater then n-1

and greater than 1 for the second...

and for A to diagonalizeable i need the chrasteric polynomial to split...witch it does and also for the eiganspace's dimensions to match they're multiplicty?...if that is correct how to do write it out?

2. Originally Posted by mtlchris
A is an nxn matrix with two distinct eiganvalues lambda 1 and lambda 2

the dimension of the eigenspace for lambda 1 is n-1

prove that A is diagonalizable
.................................................. ..................................
so i know the algebraic multiplicity for the first eiganvalue is greater then n-1

and greater than 1 for the second...

and for A to diagonalizeable i need the chrasteric polynomial to split...witch it does and also for the eiganspace's dimensions to match they're multiplicty?...if that is correct how to do write it out?
"for A to diagonalizeable i need the chrasteric polynomial to split...witch it does" ..... how do you say that?

As you said,
"so i know the algebraic multiplicity for the first eiganvalue is greater then n-1......and greater than 1 for the second..."

The algebraic multiplicity of the first eigenvalue cannot be $n$ because the second eigenvalues algebraic multiplicity is at least 1. Thus the algebraic multiplicity of the first eigenvalue is exactly $n-1$. This means the algebraic multiplicity and the geom. multiplicity of the second eigenvalue is exactly 1 and thus the matrix is diagonalizable.

3. I say that the characteristic polynomial splits because A has two distinct eigenvalues so the characteristic polynomial has to be a product of linear factors...right??