# In computing eigenvalues, why does $A− \lambda I$ have to be a singular matrix?

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.

• From $Ax = \lambda x$ for a nonzero $x$, we get $(A - \lambda I)x = 0$. Since $x$ is nonzero, the nullspace of the matrix $A - \lambda I$ contains a nontrivial element $x$ and thus $A - \lambda I$ is singular. Dec 14, 2020 at 23:14
• Is that not simply the definition of eigenvalue? Eigenvalues are, for me, roots of the characteristic polynomial $\chi(x) = \det(A - xI)$. How are you defining eigenvalue? The nullspace of a linear map is its kernel (i.e. the set of elements of the domain that map to zero). Dec 14, 2020 at 23:14
• A singular matrix is a matrix that is non-invertible. Dec 14, 2020 at 23:21
• I understand that x is nonzero, but I don't get the part where the nullspace of the matrix contains a nontrivial element. I know it's really stupid, but maybe someone could dumb it down better for me to understand. That's the only part I don't get. Dec 14, 2020 at 23:21
• What do you mean by $x$? And a non-invertible matrix $X$ is a matrix for which there is no matrix $Y$ such that $XY= YX= I$. There are many many non-invertible matrices that contain a nontrivial element. Dec 14, 2020 at 23:24

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 $$\lambda$$.