# Thread: determinant of T linear operator

1. ## determinant of T linear operator

Let T be a linear operator on a finite-dimensional vector space V, and let β be an ordered basis for V, we define det(T) = det ( [T]β )

i) 1) Prove that det(T – tIv) = det ( [T]β – tI ) for any scalar t and any ordered basis β
ii) 2) Prove that if det (T) ≠ 0 , then T is invertible

2. Originally Posted by mystic
Let T be a linear operator on a finite-dimensional vector space V, and let β be an ordered basis for V, we define det(T) = det ( [T]β )

i) 1) Prove that det(T – tIv) = det ( [T]β – tI ) for any scalar t and any ordered basis β
ii) 2) Prove that if det (T) ≠ 0 , then T is invertible
I'm not quite sure what you mean for the first one. For the second I assume you're allowed to use that the invertibility of a matrix is equivalent to invertible determinant. If so, then do the following. Let $V$ be an $F$-space. Then, there exists an isomorphism $\phi:V\to F^{\dim_F V}$ afforded by the choice of the ordered basis $\]beta$. Suppose then that $\det\left([T]_\beta\right)\ne 0$ then for every $v\in V$ one has that $\phi(v)\in F^{\dim_F V}$ and since $[T]_\beta$ is invertible one has that $[T]_\beta^{-1}$ exists. Prove then that $T\left(\phi^{-1}\left([T]_{\beta}^{-1}(\phi(v))\right)=v$. Thus, $T$ is surjective and the conclusion follows since $\dim_F V<\infty$

3. it seems to me the first problem boils down to showing that [tIv]β = I, the matrix whose entries are the kronecker delta, no matter what β is, since if so, you can subtract the matrices on the RHS. you may need to invoke the isomorphism between Mn(F) and the elements of End(V) (you need β to exhibit this isomorphism, but the fact that they are isomorphic doesn't depend on β).

4. The problem defines $\det(T)=\det([T_{\beta}])$ . In order to see if $\det (T)$ is well defined we need to prove that the right side does not depends on the chosen basis. Indeed, if $\beta'$ is another basis of $V$ then, $[T]_{\beta}$ and $[T]_{\beta'}$ are similar matrices so, $\det([T]_{\beta})=\det([T]_{\beta'})$ .

Now, fixed $\beta$ an using the standard isomorphism between $\textrm{End}_{\mathbb{K}}(V)$ and $\mathbb{K}^{n\times n}$ we have $[T-tI_V]_{\beta}=[T]_{\beta}-tI_n$ .