# 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

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.