# Thread: Computing Eigenvectors

1. ## Computing Eigenvectors

Hey guys, just got into computing eigenvectors and values just last week, and found that this thing in my homework, typically the ones on the homework were easy, but this one threw me off, and im not sure why.

Find the Eigenvalues and Eigenvectors of $C= \left(\begin{array}{ccc}1&1&0\\0&1&0\\0&0&1\end{ar ray}\right)$ if you can, diagonalize C as well.

So my work is basically this, I computed the eigenvalue of c=1 with algebraic multiplicity = 3, stuck it into B-cI, where c is the eigenvalue coresponding to B, and I is the identity matrix, and got a reduced matrix with with wierd results...i mean can an eigenvector be the zero vector? If anyone could show me how they got their eigenvectors as a result would be greatly helpful.

2. I found two eigenvectors : $\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $\begin{pmatrix}0\\0\\1\end{pmatrix}$ which are linearly independent.
You can't diagonalize $C$ because the dimension of the eigenspace is two.

3. is it because you got those two vectors because x1 and x3 are free variables since it reduced?

4. Originally Posted by Dave2718
is it because you got those two vectors because x1 and x3 are free variables since it reduced?

Who knows what you meant by "...are free variables because it reduced"......but it is so since when he/she substitutes $\lambda=1$ instead of x in the char. pol. $\det(xI-A)$ and forms the corresponding homogeneous system, she/he only gets one equation: $x_2=0\Longrightarrow$ the system's solution (i.e., the eigenspace of the only eigenvalue 1) is the subspace of all vectors whose second coordinate is zero, and he/she gave you a basis for it.
Since this basis is not a basis for the whole space, the matrix isn't diagonalizable.

Tonio