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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.

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  • $\begingroup$ 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. $\endgroup$
    – Charlie Vanaret
    Commented 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). $\endgroup$
    – DanLewis3264
    Commented Dec 14, 2020 at 23:14
  • $\begingroup$ A singular matrix is a matrix that is non-invertible. $\endgroup$
    – amWhy
    Commented 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
    Commented 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
    Commented Dec 14, 2020 at 23:24

1 Answer 1

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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 $\lambda$.

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