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.

  • $\begingroup$ 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. $\endgroup$
    – cvanaret
    Dec 14, 2020 at 23:14
  • $\begingroup$ 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). $\endgroup$ Dec 14, 2020 at 23:14
  • $\begingroup$ A singular matrix is a matrix that is non-invertible. $\endgroup$
    – amWhy
    Dec 14, 2020 at 23:21
  • $\begingroup$ 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. $\endgroup$
    – AndroidV11
    Dec 14, 2020 at 23:21
  • $\begingroup$ 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. $\endgroup$
    – amWhy
    Dec 14, 2020 at 23:24

1 Answer 1


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 $\lambda$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.