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

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

Given a basis $\displaystyle \beta$ for $\displaystyle V$ define $\displaystyle \det(T) = \det([T]_\beta)$. Prove that the definition is independent of choice of an ordered basis for V (i.e. if $\displaystyle \beta$ and $\displaystyle \gamma$ are ordered basis for $\displaystyle V$ then $\displaystyle \det([T]_\beta) = \det([T]_\gamma)$).

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

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

and for

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

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

2. Actually, it is possible to show that if you have two matrix representations of a linear transformation, then they are similar. That is, if $\displaystyle A$ and $\displaystyle B$ are both representations of $\displaystyle T$, then $\displaystyle A=SBS^{-1}$ for some invertible matrix $\displaystyle S$. Then, $\displaystyle \det(B)=\det(SAS^{-1})=\det(S)\det(A)\det(S^{-1})=\det(A)$. This proves invariance of the determinant with respect to a basis.

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### Determinant is linear transformatio

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