I have this question:

Show that the matrix $$A = \begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix}$$ is not diagonalizable.

So is the general strategy is

  1. To Find the eigenvectors and then
  2. Show that the matrix of eigenvectors is not invertible? If they are invertible, then it has a unique solution to ($\lambda \bf {I - A)x = 0}$ which would imply that they are linearly independent. If it's linearly independent, then it would be diagonalizable?

I'm following this theorem

Condition for Diagonalization
A $n \times n$ matrix is diagonalizable iff it has $n$ linearly independent eigenvectors.

So I have to find the eigenvalue first, which is $2$ because the $2$ is on the diagonal of this matrix in a triangular matrix, using this theorem.

Eigenvalues of Triangular Matrices
If $A$ is a $n \times n$ triangular matrix, then its eigenvalues are the entries on its main diagonal.

Solving for $\lambda {\bf I - A}$:

$$\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$

Since this matrix is not invertible, it is not diagonalizable. Is this right?

This is the proof that I'm relying on:

enter image description here


2 Answers 2


What you wrote is not correct.

If $\lambda$ is an eigenvalue of $A$, then $A-\lambda I$ is never invertible, no matter if $A$ is diagonalizable or not.

For example, your "proof" can be used to prove that the matrix $I$ (the identity matrix) is also not diagonalizable, which is clearly absurd.

A matrix is diagonalizable if there exists an invertible matrix $P$ such that $A=P^{-1}DP$ and $D$ is a diagonal matrix. You have probably proven in classes (or a textbook you are learning from did) that this happens if and only if there exits $n$ linearly dependent eigenvectors for the matrix.

This is what you should start from. In your case, the matrix has only one eigenvalue, and that eigenvalue has only one eigenvector (up to, of course, scaling), and from that, we can conclude the matrix is not diagonalizable.

  • $\begingroup$ I updated my post with the proof that I'm using. I don't really follow @5xum....the proof says that the matrix P could be invertible no? $\endgroup$ Sep 25, 2018 at 13:25
  • $\begingroup$ @KittyCapital Which part of my answer do you not follow? $\endgroup$
    – 5xum
    Sep 25, 2018 at 13:26
  • 2
    $\begingroup$ @KittyCapital The only thing you have proven is that $A-\lambda I$ is non-invertible, which is always true if $\lambda$ is an eigenvector. In fact, the kernel vectors of $A-\lambda I$ are exactly eigenvectors of $A$. However, in your case, you can only get one linearly independent eigenvector, not two, and therefore, the matrix is not diagonalizable. $\endgroup$
    – 5xum
    Sep 25, 2018 at 13:28
  • $\begingroup$ omg I accidentally set P = $A - \lambda * I$ even though all that produces is an eigenvector which is single as you said. My mistake! $\endgroup$ Sep 25, 2018 at 13:41
  • 1
    $\begingroup$ @KittyCapital $P$ is the matrix whose columns are all eigenvectors of $A$! $\endgroup$
    – 5xum
    Sep 25, 2018 at 13:42

Diagonalizable $\iff$ every eigenvalue has same algebraic multiplicity and geometric multiplicity. $A$ has an eigenvalue $\lambda=2$ with algebraic multiplicity 2, however, since $A-\lambda I$ has rank 1, thus its geometric multiplicity $=\dim \{N(A- \lambda I)\} = 2 - 1=1 \neq 2$ by rank-nullity theorem. Thus not diagonalizable.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .