Eigenvalue question

1. Oct 31, 2012

matematikuvol

If $\hat{A}\vec{X}=\lambda\vec{X}$ then $\hat{A}^{-1}\vec{X}=\frac{1}{\lambda}\vec{X}$

And what if $\lambda=0$?

2. Oct 31, 2012

micromass

Staff Emeritus
0 can never be an eigenvalue of an invertible matrix. So if $\lambda=0$, then $\hat{A}$ is not invertible, so $\hat{A}^{-1}$ makes no sense.

3. Nov 1, 2012

matematikuvol

Is there some easy way to see that?

4. Nov 1, 2012

AlephZero

If $X$ is an eigenvector of $A$ with eigenvalue zero, then $AX = 0$ and $X \ne 0$.

But if $A$ is non-singular, then $X = A^{-1}0 = 0$.

For any matrix $A$, only one of the above can be true.

5. Nov 1, 2012

homeomorphic

To put it less formally (but perhaps less transparently if you haven't gone far in linear algebra yet), in order for a matrix to have an inverse, it has to be associated to a one to one linear transformation. If you can hit a vector, v, with a linear map T and kill it (make it zero), then you hit λv with T and kill it for any scalar λ, so the map is not one to one because the pre-image of 0 (or any other vector) contains at least one parameter worth of stuff.