# 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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## Answers and Replies

lanedance
Homework Helper
I would go with your EDIT, this is already a diagonal matrix.

Are you sure the matrix is written correctly?

Thank you for replying. Yes, it is written correctly.

There must be some idea I'm missing here in solving the problem. I have trouble believing the book is incorrect on this, since I'm usually wrong on such things.