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I know how to find the coordinate vector respect to a vector matrix. However, in my textbook, I see the following problem, which asks to find the coordinate vector of M respect to a matrix basis:

$M=\begin{bmatrix}1 & 1\\0& 1\end{bmatrix}\in Mat(2,2,\mathbb{Q})$

$\mathcal{B} _ { 1 } = \left\{ A = \left[ \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right] , B = \left[ \begin{array} { c c } { 0 } & { 1 } \\ { 0 } & { 0 } \end{array} \right] , C = \left[ \begin{array} { c c } { 0 } & { 0 } \\ { 1 } & { 0 } \end{array} \right] , D = \left[ \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { - 1 } \end{array} \right] \right\}$

How can I proceed? Can I explicit the matrix components as column vectors?

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Start by setting up the equation you want to solve:

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = a\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + b\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + c\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + d \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}.$$

Next, simplify:

$$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a + d & b \\ c & a - d \end{bmatrix}.$$

Then turn into a system of equations:

\begin{align*} 1 &= a + d \\ 1 &= b \\ 0 &= c \\ 1 &= a - d. \end{align*}

Finally, solve!

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  • $\begingroup$ General approach is OK but $(\cdot)_{21}=-c$, not $c$. $\endgroup$ – obareey Jan 12 '19 at 17:41
  • $\begingroup$ @obareey Actually, I just put down the wrong matrix, but thanks for letting me know. $\endgroup$ – Theo Bendit Jan 12 '19 at 22:45

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