# Can a real 2 by 2 matrix have one eigenvalue with geometric multiplicity 2?

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.

• What about the identity matrix and its eigen values/vectors? – ITA Jun 3 '20 at 9:34
• you are missing a very basic and fundamental law of the universe: you cannot divide by zero – Exodd Jun 3 '20 at 9:34
• yes, sorry, I was thinking of $c\neq0$ of course – PepeToro Jun 3 '20 at 9:53

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