1. ## eigenvectors

if det(A-xI) = x^2 - 6x + 9 must there exist two linearly independent eigenvectors? or can just the one eigenvector for the double root of 3 be enough?

2. Originally Posted by ktcyper03
if det(A-xI) = x^2 - 6x + 9 must there exist two linearly independent eigenvectors? or can just the one eigenvector for the double root of 3 be enough?
$x_1=x_2=-3$

You will have two eigenvalues and one eigenvector.

3. Since you get one eigenvalue with an algebraic multiplicity of two, you will AT MOST get two eigenvectors, so really you may have one or two vectors for the basis of your eigenspace.

4. why must the number of eigenvectors be <= 2 though?

5. The eigenvectors will be at most two because when you put your eigenvalue back in the $A-\lambda I$ matrix, you will reduce it down to find the numbers of vectors that form a basis for your eigenspace.

Eigenspace = $\left \{ \vec{a} \vert (A-\lambda I)\vec{a}=\vec{0} \right \}$

6. Originally Posted by dwsmith
$x_1=x_2=-3$

You will have two eigenvalues and one eigenvector.
The matrix $\begin{bmatrix}-3 & 0 \\ 0 & -3\end{bmatrix}$ has -3 as a double eigenvalue and both [1 0] and [0 1] as eigenvectors.

$\begin{bmatrix}-3 & 1 \\ 0 & -3\end{bmatrix}$, on the other hand, has -3 as a double eigenvalue and only [1 0] as an eigenvector.

By the way, you shouldn't say a matrix "has one eigenvector" or "has two eigenvectors". Every matrix has an infinite number of eigenvectors corresponding to each eigenvalue. The question here is whether there are one or two independent eigenvectors.