I'm learning for a test and tried to calculate the eigenvalue(s) and eigenspace(s) of the matrix

$$A=\begin{pmatrix} 3 & -1\\ 1 & 1 \end{pmatrix}$$

whose characteristic polynomial is

$$p_A(\lambda) = \lambda^2 -4\lambda+4$$

Calculating the eigenvalue(s),

$$\lambda_{1,2} = \frac{4}{2} \pm \sqrt{\left(-\frac{4}{2}\right)^2 -4} = \color{red}{2 \pm \sqrt{0}}$$

Do we have two eigenvalues, namely, $\lambda_1 = 2$ and $\lambda_2=2$ or do we just have one eigenvalue $\lambda_1 = 2$?

I would say we just have one eigenvalue because it's the same value.


It has one eigenvalue $2$ with (algebraic) multiplicity $2$ (because it occurs twice as a root of the characteristic polynomial).

There is only a one-dimensional eigenspace though, generated by $\begin{pmatrix}1\\1\end{pmatrix}$, as you can check, so the matrix cannot be diagonalised. This is the so-called geometric multiplicity of the eigenvalue.

  • $\begingroup$ I thought the multiplicity of the eigenvalue was the dimensionality of the eigenspace? Thus only one? $\endgroup$ – Oscar Lanzi May 26 '18 at 12:10
  • 1
    $\begingroup$ There is a geometric and an algebraic multiplicity. @OscarLanzi, e.g. see here $\endgroup$ – Henno Brandsma May 26 '18 at 12:11

The matrix has just one eigenvalue, $\lambda=2$.

We can also ask whether it has one or two independent eigenvectors associated with this eigenvalue. Let $x$ and $y$ be the components of any eigenvector, then the eigenvector equations are



Both equations have only the solution $y=x$ so there is just one linearly independent eigenvector. This, not the degeneracy of roots in the characteristic equation, governs the geometric multiplicity of the eigenvalue. Thus $2$ is the lone eigenvalue with geometric multiplicity $1$.


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