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I am trying to verify the final statement in the following picture. The bases in example 1 are $\beta = \{(1,1), (1, -1) \}$ and $\beta' = \{(2,4), (3, 1) \}$

What exactly is $[T]_{\beta}$? From my understanding it's the identity matrix of space generated by T with respect to the basis $\beta$.

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Taken from the 4th edition Linear Algebra by Stephen H. Friedberg, Arnold J. Insel, Lawrence E. Spence. Page 128

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2 Answers 2

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$\begin{equation} \begin{split} [T]_\beta &= \beta^{-1} T \beta \\ &=\begin{bmatrix} 1 && 1 \\ 1 && -1 \end{bmatrix}^{-1} \begin{bmatrix} 3 && -1 \\ 1 && 3 \end{bmatrix} \begin{bmatrix} 1 && 1 \\ 1 && - 1\end{bmatrix} \\ &= -\frac{1}{2} \begin{bmatrix} -1 && -1 \\ -1 && 1 \end{bmatrix} \begin{bmatrix} 2 && 4 \\ 4 && -2 \end{bmatrix} \\ &= \begin{bmatrix} 3 && 1 \\ -1 && 3 \end{bmatrix} \end{split} \end{equation} $

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  • $\begingroup$ Thank you so much for this! And I gather that the middle matrix T is the basis matrix for the linear operator T? $\endgroup$
    – Malcolm
    Feb 12 at 17:39
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    $\begingroup$ @Malcolm Yes, it is. $\endgroup$ Feb 12 at 17:54
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The $[T]_{\beta}$ denotes the matrix associated with the linear transformation $T$ with respect to the ordered basis $\beta$, and satisfies $[Tv]_{\beta}=[T]_{\beta}[v]_{\beta}$ for all vectors $v \in \mathbb{R^2}$. If $\beta_1=(1,1)$ and $\beta_2=(1,-1)$, then $[T]_{\beta}=[A_1,A_2]$ where $A_i=[T\beta_i]_{\beta}, i=1,2.$ You can find the column vectors $A_i$ of the matrix $[T]_{\beta}$ by evaluating the basis vectors $\beta_i$ with respect to $T$. The result is $T\beta_1=3\beta_1-\beta_2$ and $T\beta_2=\beta_1+3\beta_2$.

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