Eigenvalue Q: What if ##\lambda=0##?

[matematikuvol]

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

And what if $\lambda=0$?

 

[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.

 

[matematikuvol]

Is there some easy way to see that?

 

[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.

 

[homeomorphic]

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

But if A is non-singular, then X=A−10=0.

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

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.