# possible Eigenvector basis

I am learning to form bases with eigenvalues and want to make sure I understand. I was given the matrix $C= \begin{pmatrix} 6 & -3 & -1 \\ 11 & -6 & -3 \\ -10 & 7 & 5 \end{pmatrix}$. I found an eigenvector $\begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$ for $\lambda =1$. I also found eigenvector $\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ for $\lambda=2$. Now knowing $\lambda=1$ and $\lambda=2$ are the only eigenvalues of C, is it possible to get a basis of $\mathbb{R}^3$ consisting of eigenvectors of C?

I don't think it's possible since for both eigenvectors I was given one degree of freedom, thus there was "only one" vector (quotes because I understand there isn't only one). Am I understanding this correctly or is it possible?

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## 2 Answers

You are correct: it is not always possible to find a basis of eigenvectors. As you have discovered, the matrix only has two eigenvectors.

Since $2$ is an eigenvalue of multiplicity $2$, you may well wonder why there is only one eigenvector associated with it. The answer is that there is a vector $v$ so that $(C-2I)v \neq 0$, but $(C-2I)^2v=0$; this is called a generalised eigenvector. If we include generalised eigenvectors, more generally defined as vectors for which $(C-\lambda I)^n v = 0$ for any $n>0$, then we can form a basis; this is the idea behind the Jordan Normal Form.

No it is not possible to form a basis of $\Bbb R^3$, since you only have $2$ vectors, and $3$ are required to do this, since this is a $3$-dimensional space. So yes you're completely correct.