# When do we have linearly dependent eigenvectors

Suppose we have a $$3 \times 3$$ real invertible matrix $$A$$. If $$A$$ has two distinct eigenvalues, so one of them has multiplicity $$2$$. Is it possible to have only two linearly independent eigenvectors? i.e., is it possible that the eigenvectors corresponding to the eigenvalue with multiplicity $$2$$ are linearly dependent?

• What do you think? May 19, 2020 at 21:53

## 2 Answers

What about $$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$ ?

• Could you explain what the idea behind that is? How do we construct this kind of matrix? May 19, 2020 at 22:04
• I took a 2x2 matrix with only one eigenvalue that is not the identity, so it has only one eigenspace of dimension 1. I completed in an orthogonal basis so that it is an invertible matrix May 19, 2020 at 22:06

$$A=\begin{bmatrix}2&0&0\\0&1&1\\0&0&1\end{bmatrix}$$ For the eigenvalue 1,which has algebraic multiplicity 2, the dimension of the eigenspace is 1. Any eigenvector for eigenvalue 1 is a multiple of $$\begin{bmatrix}0\\1\\0\end{bmatrix}.$$

• Could you explain what the idea behind that is? How do we construct this kind of matrix? May 19, 2020 at 22:06