This question is being asked not because textbooks do not provide for an explanation about this. They do but, I really can't seem to grasp it very well. I know matrices but honestly my professor has not defined what a singular matrix means.
" To be of any use, the nullspace of $A− \lambda I$ must contain vectors other than zero."
I also do not have an idea of what nullspace is. I have tried searching about it but would also like to ask for those concepts here as it might be explained better.
I get that with eigenvalues and eigenvectors, we don't want the eigenvector $v$ in $Av = \lambda v$ to be zero because that would just result into a useless solution. But I don't get where all the talk about it being singular came from. I do understand from a definition I read that a singular matrix has a determinant of zero, which led to why $|A− \lambda I| = 0$ came and from there I can do the solutions.
I have read about the derivation of the equation $|A− \lambda I| = 0$ and the only part I don't get is how it was concluded that $A− \lambda I$ has to be singular. I am aware that probably I don't have a good grasp of the definitions and if I did I would understand why it led to becoming singular. The idea is probably everywhere on the Internet, I just do not manage to get it or find a good enough reference for me to do so. But that's why I'm asking here, probably a good explanation or a reference with a good explanation will be mentioned. Thank you.
Best Answer
Start with $Av=\lambda v$. Equivalently, we could write $Av=\lambda Iv$, since $Iv=v$. Then we can bring both terms to the left to get $Av-\lambda Iv=0$, or equivalently, $$(A-\lambda I)v=0.$$ This means that $v$ is in the nullspace of the matrix $A-\lambda I$ (by definition, any vector $x$ such that $Mx=0$ is in the nullspace of matrix $M$). Any matrix with a non-trivial nullspace is singular.
We can also work backward. If $A-\lambda I$ is singular matrix, then we know that it must have at least one vector $v$ in its nullspace (other than the zero vector). This vector satisfies $$(A-\lambda I)v=0\implies Av-\lambda I v=0\implies Av=\lambda Iv=\lambda v.$$ Therefore, if $A-\lambda I$ is singular, any vector in its nullspace is an eigenvector of $A$ with eigenvalue $\lambda$.