Prove that a linear operator $T$ on a finite-dimensional vector space is invertible if and only if zero is not an eigenvalue of $T$.

Definition: Let $T$ be a linear operator on a vector space $V$. A nonzero vector $v \in V$ in an eigenvector of $T$ if there exists a scalar $\lambda$ such that $T(v)= \lambda v$. The scalar $\lambda$ is called an eigenvalue.

Let $A$ be in $M_{n,n}(F)$. A nonzero vector $v\in F^n$ is an eigenvector of $A$ if $v$ is an eigenvector of $L_{A}$. The scalar $\lambda$ is called the eigenvalue of $A$.

Proof: $\Rightarrow$ Let $T$ be a finite linear operator, and $T(v)=Av=\lambda v$ for $A$ to be a $M_{n,n}(F)$ matrix. If $T(v)$ is invertible, then $T(T^{-1})=(Av)(Av)^{-1}=(\lambda v)(\lambda v)^{-1}=I_n$. That means $\lambda v$ is nonzero.

$\Leftarrow$ If zero is not an eigenvalue of $T$, that means $\lambda v=Av \neq0$, then $det(Av)$ $\neq$ $0$. Hence $T$ is invertible.

I know this is a crappy work, but this is all I can think about.

  • 1
    $\begingroup$ Do you know that the determinant of $T$ is the product of the eigenvalues of $T$? $\endgroup$
    – Jan
    Feb 20, 2020 at 5:28
  • $\begingroup$ Yes I am aware of that. Can you say more about that? $\endgroup$
    – neveryield
    Feb 20, 2020 at 5:30
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    $\begingroup$ Perhaps in the first part you should prove (unless it was done previously) that $\lambda^{-1}$ is the inverse eigenvalue $\endgroup$
    – IAmNoOne
    Feb 20, 2020 at 5:32
  • 1
    $\begingroup$ If all eigenvalues are non zero then $\det T$ is non zero. If any eigenvalue is zero then $\det T = 0$. $\endgroup$
    – copper.hat
    Feb 20, 2020 at 5:34
  • $\begingroup$ @IAmNoOne Yeah, I should do so, or it violates definition of invertible. $\endgroup$
    – neveryield
    Feb 20, 2020 at 5:36

2 Answers 2


It is not entirely true say that $\textsf{T}(v) = Av$ for some $A\in\textsf{M}_{n \times n}(F)$ since we don't know that $v$ is a column vector that can be multiplied with $A$, $v$ could be anything!

Now, let me help you with the proof. For the $(\Rightarrow)$ direction, prove the contrapositive, that is, prove that if $0$ is an eigenvalue of $\textsf{T}$, then $\textsf{T}$ cannot be invertible (recall that, $\textsf{T}$ is one-to-one if and only if the only vector that it sends to $\textbf0$ is the same $\textbf0$).

For $(\Leftarrow)$, the same idea, if $\textsf{T}$ is not invertible, then $\textsf{T}$ is not one-to-one, and then there exists a non-zero $v\in\textsf{V}$ with $\textsf{T}(v) = \textbf0$, but this means that $v$ is an eigenvector of $\textsf{T}$ with eigenvalue $\lambda=0$.


Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of $T$. Then

$$\det(T) = \prod_{i = 1}^n \lambda_i.$$

Assume $T$ is invertible. Then $\det(T) \neq 0$. Now use the fact that in a field a product is zero if and only if at least one factor is zero. Hence, $\lambda_i \neq 0$ for all $i$. This also proves the backwards direction.


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