Question

Why do only non-zero eigenvalues have eigenvectors in the image?

Answer 1

I'm not sure what your question means. Which image? But to get the ball rolling I will point out that zero eigenvalues only occur if the matrix is singular. In other words,
\[\left( A - \lambda I \right)x = 0\ = Ax = 0\] has solutions.

Conversely, if the matrix ONLY has non-zero eigenvalues, then that matrix is invertible.

Answer 2

The image of the matrix is another term for the column space.

Eigenvectors with zero eigenvalues lie in the null space of the matrix.

Answer 3

Okay, thanks to fwizbang for defining the image of a matrix = the column space of a matrix, I am now in a position to thoroughly (I hope) demonstrate why only non-zero eigenvalues have eigenvectors in the image(column space of A).

Eigenvectors are defined as vectors whose direction does not change when multiplied by a matrix, A. That is for a matrix A with eigenvalue = lambda and eigenvector = x:\[Ax = \lambda x\] then
\[Ax - \lambda x = 0\]and
\[(A - \lambda I) x = 0\]
so, if the eigenvalue = 0, then the eigenvector, x, is in the nullspace. That is,
\[Ax = 0\]

On the other hand, if\[\lambda \neq 0\]
then the eigenvector, x, clearly is in the column space. Because equations of the general form Ax = b, are solvable only when b is in the image (column space) of A, and b is in the image of the eigenvector, because it is merely the scalar eigenvalue multiplied times the eigenvector.

Answer 4

@datanewb what eigenvectors means parallel or perpendicular vectors

Answer 5

An eigenvector, x, for a sqaure matrix, A, is any vector that when multiplied by the matrix A, comes out parallel to itself.

So, in equation form, that looks like: \[Ax = \lambda x\]

Take the very simple square matrix : \[ \left[\begin{matrix}1&1\\1&1\end{matrix}\right] \] it has two eigenvectors. One of which is in the column space (parallel to the column space)
\[x_{1} = \left[\begin{matrix}1\\1\end{matrix}\right] \\
\left[\begin{matrix}1&1\\1&1\end{matrix}\right] \left[\begin{matrix}1\\1\end{matrix}\right] = \left[\begin{matrix}2\\2\end{matrix}\right] \\\left[\begin{matrix}1\\1\end{matrix}\right] \|\left[\begin{matrix}2\\2\end{matrix}\right] \]
For the other eigenvector, it is in the nullspace (for this specific example). The nullspace consists of vectors which are perpendicular to the column space.

\[ x_{2} = \left[\begin{matrix}1\\-1\end{matrix}\right] \\
\left[\begin{matrix}1&1\\1&1\end{matrix}\right] \left[\begin{matrix}1\\-1\end{matrix}\right] = \left[\begin{matrix}0\\0\end{matrix}\right] \]