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I understand that each linear operator on a finite dimemsional vector space can be represented as a matrix with respect to a given basis. For example, the linear operator on $R^2$, $ T\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}x_1 \\ x_1 + x_2 \end{bmatrix}$ with respect to the basis B ={$\begin{bmatrix}1\\2\end{bmatrix}$, $\begin{bmatrix}1\\1\end{bmatrix}$} is matrix A = $\begin{bmatrix}2&1\\-1&0\end{bmatrix}$. I am trying to go in the other direction, that is, given matrix A and basis B, how do I derive operator T?

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  • $\begingroup$ Try multiplying the given matrix by a column vector (the column vector containing the coordinates of an arbitrary vector in the given basis). $\endgroup$ – Larara Jun 9 '16 at 22:57
  • $\begingroup$ ok, but that doesn't give me the same operator T that I started with $\endgroup$ – David Warren Katz Jun 9 '16 at 23:06
  • $\begingroup$ Possibly (depending on the given basis) you'll have to apply the change of basis matrix besides the matrix of the operator $A$... $\endgroup$ – Larara Jun 9 '16 at 23:14
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If you are asking what "$T$" is [the matrix with respect to the standard basis], then solving $$\begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} =\begin{bmatrix}x_1\\x_1+x_2\end{bmatrix}$$ shows $$T = \begin{bmatrix}1&0\\1&1\end{bmatrix}.$$

More generally, if $B$ is the matrix whose columns are your basis elements and $A$ is the matrix of $T$ with respect to basis $B$, then $T=BAB^{-1}$.


If you are asking more generally how a matrix $A$ with respect to an arbitrary basis $B=\{v_1,\ldots,v_n\}$ defines an operator on $\mathbb{R}^n$, note that by the definition of a basis, any $v$ can be written uniquely in the form $v=c_1 v_1 + \cdots + c_n v_n$ for some scalars $c_1,\ldots,c_n$. Then $$[Tv]_B = A \begin{bmatrix}c_1 \\ \vdots \\ c_n\end{bmatrix}.$$

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Suppose T is the transformation in the cannonical basis and $T_B$ is transformation in the new basis. And $B$ takes a vector in the canonical basis to the new basis.

$T\mathbf v = \mathbf w\\ B\mathbf v = \mathbf v_{B}\\ T_{B}\mathbf v_{B} = \mathbf w_{B}\\ T_{B}B\mathbf v = B\mathbf w\\ B^{-1}T_{B}B\mathbf v = \mathbf w\\ B^{-1}T_{B}B = T\\ T_{B} = BTB^{-1}$

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