# Prove that $T$ is invertible if and only if $0$ is not an eigenvalue of $T$

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.

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

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.
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 $$\lambda=0$$.