# Finding Eigenvector from a 3x3 matrix

Given the constant matrix

A = $$\begin{bmatrix}4&-1&0\\3&1&-1\\1&0&1\end{bmatrix}$$

Find the fundamental matrix.

After finding the eigenvalue $$\lambda$$=2 with multiplicity 3, then trying to find the eigenvector corresponding to the eigenvalue, I end up with matrix $$\begin{bmatrix}2&-1&0\\3&-1&-1\\1&0&-1\end{bmatrix}$$, then after trying to find the corresponding eigenvector I end up with u=(1,2z,z), which I obtained from letting x=1 and subtracting the first and second rows with appropriate multiplication of 3$$R_1$$ and -2$$R_2$$ and adding the rows. Also, the reduced matrix has the form

$$\begin{bmatrix}1&0&-1\\0&1&-2\\0&0&0\end{bmatrix}$$

Then the first eigenvector u is u=(1,2,1). Nonetheless, since the eigenvalue has muliplicity 2 shouldn't there be two more eigenvectors in terms of some variable t?

• Note that $$A-2I=\begin{bmatrix}2&-1&0\\3&-1&-1\\1&0&\color{#f00}{-3}\end{bmatrix}.$$ – awllower Oct 20 '18 at 8:07
• Are you sure that you wrote the right matrix. I ask this because $2$ is not an eigenvalue of the one that you mentioned (in fact, it has no rational eigenvalues). Besides, what is a “fundamental matrix”? – José Carlos Santos Oct 20 '18 at 8:10
• I apologize I have fixed the matrix, it seems I accidentally added a negative to the element (3,3) it should be just 1, NOT -1 – lastgunslinger Oct 20 '18 at 8:20
• Eigenvector is $u=(z,2z,z)$ or: any non-zero multiple of $(1,2,1).$ – user376343 Oct 20 '18 at 9:03
• The algebraic multiplicity of the eigenvalue $2$ is $3$, but the geometric multiplicity is $1$. – Math1000 Oct 20 '18 at 10:37