# Linear Algebra - Showing a Matrix is not Diagonalizable

## Homework Statement

Show that the matrix A is not diagonalizable. Explain your reasoning.

A=\begin{bmatrix}k&0\\0&k\\\end{bmatrix}

## Homework Equations

Important theorem: A nxn matrix is diagonalizable if and only if it has n linearly independent eigenvectors.

## The Attempt at a Solution

Since A is triangular, the eigenvalues are the entries on the main diagonal. In this case the only eigenvalue is k. So then I solve for B = (λI - A) where λ=k, which turns out to equal a 2x2 zero matrix. Then I solve Bx = 0 to try and find the eigenvectors. Here is where I think i'm going wrong. Since B is the zero matrix, I believe that x1 = t, x2 = s, where t and s are any real number. So I find that the vector x is equal to t(1,0) + s(0,1) which would indicate to me that the matrix has two linearly independent eigenvectors and should be diagonalizable.

But this is the opposite of which I wished to prove! So clearly my thinking must be wrong. The solution for this problem says that a basis for the eigenspace is simply {(0,0)}, so since A does not have two linearly independent eigenvectors, it does not satisfy the theorem I have above and cannot be diagonalizable.

My problem is that I don't see how you can say from Bx=0 where B is the 2x2 zero matrix that x can only be (0,0) and thus the basis for the eigenspace is only {(0,0)}. Can't x be any vector?

Thank you.

EDIT: Another point. Isn't this matrix also symmetric (A=Atranspose)? Then shouldn't it be diagonalizable (it's already diagonal anyways)?

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