Re: Is this matrix diagonalizable?

On 30 sep, 12:24, Tonico <Tonic...@xxxxxxxxx> wrote:
On Sep 30, 7:10 pm, "Carl R." <solrac...@xxxxxxxxxxx> wrote:

Hello

Let A = [1 3 1 2 ]
[0 -1 1 3]
[0 0 2 5]
[0 0 0 -2]

So I found that we have two repeated eigenvalues, namely lambda = 2.
Therefore among the 4 eigenvectors there are two eigenvectors which
have the same value, hence they can't
be linearly independent because a vector is a multiple of itself.
Therefore A is not diagonalizable. Is this correct?
Is there another way to see it?

Thanks in advance

********************************************************
The matrix is given in ujpper triangular form, and thus its
eigenvalues are the elements on its main diagonal, namely -1, 1, -2, 2
==> four different eigenvectors, one for each eigenvalue, will be lin.
indep0endent and thus will conmform a basis of the vector space ==>
the matrix is diagonalizable.
I don't know why you think there're "repeated eigenvalues"....

Regards
Tonio

Thanks guys! I found my mistake.


.

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