Given the a real matrix $A=\begin{bmatrix} a & b \\ c& d\end{bmatrix}$, we assume that it has only one real eigenvalue $\lambda$. I am wondering if it is possible that the eigenvalue $\lambda$ has geometric multiplicity 2, but it seems like it is not possible.

Let $v=\begin{bmatrix} v_1 \\ v_2\end{bmatrix}$. When I solve the usual equation $(\lambda I-A)v=0$, because of the dimension of course, I only obtain one condition for the eigenvector, namely $v_1=\frac{(\lambda-d)}{c}v_2$, which would indicate that there is only one eigenvector and would not be possible to have 2 linearly independent eigenvectors of the repeated eigenvalue $\lambda$. Perhaps this is a very trivial observation for real $2\times 2$ matrices? Am I missing something very silly?

For posterity after the comments: ... with $c\neq0$ indeed is not possible.

  • 1
    $\begingroup$ What about the identity matrix and its eigen values/vectors? $\endgroup$
    – ITA
    Jun 3, 2020 at 9:34
  • $\begingroup$ you are missing a very basic and fundamental law of the universe: you cannot divide by zero $\endgroup$
    – Exodd
    Jun 3, 2020 at 9:34
  • $\begingroup$ yes, sorry, I was thinking of $c\neq0$ of course $\endgroup$
    – PepeToro
    Jun 3, 2020 at 9:53

1 Answer 1


Your construction of the eigenvector assumes $c$ is non-zero. A matrix with full geometric multiplicity is diagonalizable, so the only such matrix is $\lambda I$.


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