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Let phi(A):R^2->R^2 be a linear operator given by multiplication matrix

A =

(1 1
01)

Find a matrix that represents this operator with respect to the basis (1,0) and (-1,1).

Any help would be much appreciated because I'm a bit confused as to what I should be doing here.

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  • $\begingroup$ what is trial ??? $\endgroup$
    – user476275
    Mar 4 '18 at 19:49
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Express $(1,0) = u_1$ and $(-1,1) = u_2$ in the canonical basis which is $((1,0),(0,1)) = (e_1, e_2)$.

As you know $f(e_1) = e_1$ and $f(e_2) = e_1+e_2$, you'll be able to express $f(u_1)$ and $f(u_2)$ in $(u_1,u_2)$

Indeed, $u_1 = e_1$ and $u_2 = -e_1 + e_2$.

Hence $f(u_1) = f(e_1) = e_1 = u_1$

And $f(u_2) = f(-e_1 + e_2) = -f(e_1) + f(e_2) = -e_1 + (e_1 + e_2) = e_2 = u_1 +u_2$

So $A$ is the wanted matrix.

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For any linear transformation $\phi\colon \mathbb{R}^2 \to \mathbb{R}^2$, and basis $(v_1,v_2)$ of $\mathbb{R}^2$, there exist constants $b_{ij}$ such that \begin{align*} \phi(v_1) &= b_{11}v_1 + b_{12}v_2 \\ \phi(v_2) &= b_{21}v_1 + b_{22}v_2 \end{align*}

Those coefficients $b_{ij}$ make up a matrix $B$, which is the matrix that represents $\phi$ with respect to $(v_1,v_2)$.

You have been given $\phi$, $v_1$, and $v_2$; you should be finding $b_{ij}$.

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You are trying to find a new matrix that encodes inputs and outputs in terms of $(1,0)$ and $(-1,1)$ instead of the standard basis, so you should start by writing relationships between the standard basis and $(1,0)$ and $(-1,1)$, i.e. $(1,0)=1 \cdot (1,0)$ and $(0,1) =1 \cdot (1,0) + 1 \cdot (-1,1)$. Since you want your inputs to be of the form $[x,y]$ where $[x,y]$ represents $x \cdot (1,0) + y \cdot (-1,1)$, you should find a matrix to change inputs in this form to the standard basis. Similarly, you should find a matrix that then takes outputs in the standard basis to their coordinates in the $\{ (1,0), (-1,1) \}$ basis.

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