# Determinant of linear transform with ordered basis is independent of the choice

• Jan 27th 2011, 07:16 PM
lllll
Determinant of linear transform with ordered basis is independent of the choice
Given a basis $\beta$ for $V$ define $\det(T) = \det([T]_\beta)$. Prove that the definition is independent of choice of an ordered basis for V (i.e. if $\beta$ and $\gamma$ are ordered basis for $V$ then $\det([T]_\beta) = \det([T]_\gamma)$).

I don't have much in the form of a proof. I tried working out an example but it got me nowhere. So what I have as an example is:

let $T(a,b) = (a+b,a-2b)$ and $\beta=\{(1,0), (0,1)\} \ \gamma = \{(1,2),(1,-1)\}$ then under the transformation we have

$[T]_\beta = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}$ where $\det[T]_\beta = -3$

and for

$[T]_\gamma= \begin{pmatrix} 0 & 3 \\ 1 & -1 \end{pmatrix}$ where $\det[T]_\gamma= -3$

where $[T]_\gamma= \begin{pmatrix} 0 & 3 \\ 1 & -1 \end{pmatrix}$ was obtained by $[T]_\gamma^\beta= \begin{pmatrix} 3 & -3 \\ 0 & 3 \end{pmatrix}$ and $[\hat{T}] = \frac{1}{3} (a+b,2a-b)$ which gives the desired result.

Now I'm thinking it has something with the eigenvalues and corresponding eigenvectors but that got nowhere, since I'm not to sure on how to proceeds.
• Jan 27th 2011, 07:47 PM
roninpro
Actually, it is possible to show that if you have two matrix representations of a linear transformation, then they are similar. That is, if $A$ and $B$ are both representations of $T$, then $A=SBS^{-1}$ for some invertible matrix $S$. Then, $\det(B)=\det(SAS^{-1})=\det(S)\det(A)\det(S^{-1})=\det(A)$. This proves invariance of the determinant with respect to a basis.