# Why is this “not allowed by definition”? Eigenvectors and Eigenvalues

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.

• 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? – Joonas Ilmavirta Nov 17 '16 at 13:35
• 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. – Joe Johnson 126 Nov 17 '16 at 13:35
• @JoonasIlmavirta You are correct. I understand now. You both have my thanks. – The Pointer Nov 17 '16 at 13:38

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.