# Thread: Base of Eigenvectors

1. ## Base of Eigenvectors

I started with the above matrix and found that the eigenvalues were 2 and 4.

When I was trying to calculate the base for for lambda=4, I came up with (0,1,0)^T and (0,0,1)^T. What bothers me is that the algebraic multiplicity in this case is 1, right? But the geometric multiplicity appears to be 2. This can't be because the algebraic multiplicity must be greater than or equal to the geometric multiplicity.

So where have I gone wrong in my reasoning?

2. ## Re: Base of Eigenvectors

Oh nevermind, I see where the mistake is. The base should be (1,0,0)^T. As x2 and x3=0.

3. ## Re: Base of Eigenvectors

For the benefit of others, then, when in doubt go back to your basic definitions: Saying that 4 is an eigenvalue for matrix A means there exist some vector v such that Av= 4v. Here, that gives
$\begin{bmatrix}4 & -2 & -1 \\ 0 & 2 & -1 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ $= \begin{bmatrix}4x- 2y- z \\ 2y- z \\ 2z\end{bmatrix}$ $= \begin{bmatrix}4x \\ 4y \\ 4z\end{bmatrix}$
so we have the three equations 4x- 2y- z= 4x, 2y- z= 4y, and 2z= 4z. Because the original matrix was "triangular" those are very easy to solve- the last equation is the same as "2z= 0" and is true only if z= 0. Then the second equation is 2y- z= 2y= 4y which is the same as 2y= 0 so y= 0. Putting y= z= 0 into the first equation, 4x= 4x which is satisfied for all x. Yes, that gives $\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$

Of course, for the double eigenvalue, 2, we must have
$\begin{bmatrix}4 & -2 & -1 \\ 0 & 2 & -1 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ $= \begin{bmatrix}4x- 2y- z \\ 2y- z \\ 2z\end{bmatrix}$ $= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}$
which gives the equations 4x- 2y- z= 2x, 2y- z= 2y, and 2z= 2z. But now the second equation reduces to -z= 0 which says that z= 0 but y can be anything. That makes the final equation 4x- 2y= 2x or 2y= 2x. An eigenvector corresponding to eigenvalue 2 is $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$. Note that even though the "algebraic multiplicity" of eigenvalue 2 is 2, its "geometric multiplicity" is only 1. That's possible. The geometric multiplicity is always "less than or equal to" the algebraic multiplicity.