0
$\begingroup$

I have just started learning about eigenvectors and eigenvalues, and encountered the following (unjustified) statement.

$ (A - \lambda I)\mathbf{v} = 0$. If the matrix $ (A - \lambda I) $ were invertible, then the solution would simply be $\mathbf{ v } = 0$. But this is not allowed by definition.

I'm curious as to why it is "not allowed by definition"? I would appreciate it if someone could elaborate on the reasoning behind this.

Thank you.

$\endgroup$
3
  • 1
    $\begingroup$ I assume this discussion takes places briefly after a definition like this: A non-zero vector $v$ is an eigenvalue of $A$ with eigenvalue $\lambda$ if $(A-\lambda I)v=0$. Or maybe the definition says $Av=\lambda v$ but the other form is discussed later. If this is the case and $v$ is an eigenvector, then it is non-zero by definition. But to judge, more context is needed. What is said before this statement? $\endgroup$ – Joonas Ilmavirta Nov 17 '16 at 13:35
  • 1
    $\begingroup$ We only allow a vector to be an eigenvector if it is non-zero. Otherwise the zero vector would be an eigenvector for every eigenvalue. And it contributes nothing to a basis of eigenvectors. $\endgroup$ – Joe Johnson 126 Nov 17 '16 at 13:35
  • $\begingroup$ @JoonasIlmavirta You are correct. I understand now. You both have my thanks. $\endgroup$ – The Pointer Nov 17 '16 at 13:38
1
$\begingroup$

The eigenvalue equation can also be written as $Av=\lambda v $, so we are looking for vectors $v$, on which the operator $A$ acts by scalar multiplication. These vectors usually say some interesting things about the nature of the operator $A$, as well as helps with solving problems. After all, many linear operator-related problems are far easier if the operator is diagonal, and the diagonalization of operators are done via finding the eigenvalues/eigenvectors.

The thing is, if $v=0$, then the equation $Av=\lambda v$ is always satisfied, regardless of what $A$ is. Therefore, although this is a solution of the equation, it is not an interesting solution, since the whole point of solving an eigenvalue equation is to understand the structure of the operator $A$. If something is true for all $A$s, then it must contain no information about specific $A$s.

$\endgroup$
0
1
$\begingroup$

We exclude the zero vector as it is a trivial solution to the problem. It's a solution, but not an interesting or useful one.

$\endgroup$
1
$\begingroup$

As others have said, zero being an eigenvector does create some problems, particularly in the case that every eigenvector has a unique eigenvalue (if 0 were an eigenvector, its eigenvalue would be arbitrary.

However, as others have failed to mention, 0 also creates problems when it is not an eigenvector, particularly in the case of the eigenspace. An eigenspace is the set of all eigenvectors of a specific eigenvalue plus the zero vector. This can be avoided using the current definition by defining the eigenspace to be the null space of the transformation minus the identity matrix scaled by the eigenvalue, but the problem still exists as a matter of intuition.

I would highly advise you read this Math SE question and in particular its top answer.

$\endgroup$

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.