The matrix is given as:

$A=\begin{bmatrix} 0 & 1 & 1 \\0 & 0 & 4 \\ 0 & 0 & 3 \end{bmatrix}$

So the matrix has eigenvalues of $0$ ,$0$,and $3$.

The matrix has a free variable for $x_1$ so there are only $2$ linear independent eigenvectors. So this matrix is not diagonalizable.

What conditions would be necessary for $A$ to be diagonalizable?

Is it simply all $3$ eigenvectors must be linearly independent? Or perhaps the opposite?

  • 2
    $\begingroup$ A $n\times n$ matrix is diagonalizable iff it has $n$ independent eigenvectors. $\endgroup$ Dec 12, 2014 at 19:53

2 Answers 2


As you remarked correctly, the eigenvalues, with multiplicity, are $0,0,3$.

A matrix is diagonalizable if and only of for each eigenvalue the dimension of the eigenspace is equal to the multiplicity of the eigenvalue.

For the eigenvalue $3$ this is trivially true as its multiplicity is only one and you can certainly find one nonzero eigenvector associated to it.

For the eigenvector $0$ however you would need to find $2$ linearly indepedent eigenvectors Yet as you said, indirectly, the eigenspace associated to $0$ is the space generated by $(1,0,0)$. Its dimension is thus one and you cannot find two independent eigenvectors for the eigenvalue $0$.

The matrix is thus not diagonalizable.

Necessary conditions for diagonalizable include:

  • There exists a basis of eigenvectors, which in you case would mean there exist $3$ linearly independent eigenvectors. (This is almost what you said, but note that I said "there exists" and not "the three eigenvectors" since there are infinitely many eigenvectors.)

  • For each eigenvalue the dimension of the respective eigenspace is equal to the multiplicity of the eigenvalue. (As mentioned at the start.)

  • $\begingroup$ Ok so essentially the number of linearly independent eigenvectors for an eigenvalue must match the multiplicity of the eigenvalue? $\endgroup$ Dec 12, 2014 at 20:19
  • $\begingroup$ Yes, exactly this, and you need this for each eigenvalue. Also recall that the multiplicity of the eigenvalue is always an upper bound for the dimension of the eigenspace, but the dimension of the eigenspace can be strictly smaller. However, since it cannot be less than 1, you do not have do check anything for eigenvalues with multiplicity 1. $\endgroup$
    – quid
    Dec 13, 2014 at 1:10

A matrix $A$ is diagonalisable with distinct eigenvalues $\lambda_1,\ldots,\lambda_k$ (or a subset thereof) if and only if the product $(A-\lambda_1I)\ldots(A-\lambda_kI)$ is zero. In your example clearly $0,3$ are the only eigenvalues, but $$ A(A-3I)=\begin{pmatrix}0&-3&4\\0&0&0\\0&0&0\end{pmatrix} $$ is nonzero, so $A$ is not diagonalisable.

  • $\begingroup$ Can you supply a proof of this? $\endgroup$ May 16, 2017 at 12:16
  • $\begingroup$ Proof of what part? $\endgroup$ May 16, 2017 at 14:01
  • $\begingroup$ The general statement that A is diag iff said product is zero. This result saves the effort to find dim of eigenspaces to check diagonalizability. $\endgroup$ May 16, 2017 at 14:05
  • 1
    $\begingroup$ Well, if $A$ is diagonalisable then each factor in the product kills the corresponding eigenspace, so the product is zero everywhere. And if the product is zero then $A$ is annihilated by a polynomial that is split with simple roots (the $\lambda_i$ occurring in the product) and that is known to imply diagonalisability. $\endgroup$ May 16, 2017 at 14:09

You must log in to answer this question.

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