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Sending each point $(x_1, x_2)$ in the plane $\mathbb{R}^2$ to the point $x_2$ in $\mathbb{R}^1$ is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^1$. What is the corresponding matrix?

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    The dimension of the required matrix is what by what? – 2017-02-16
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    It should be 1x2 – 2017-02-16
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    Right, so an unknown 1x2 matrix would take what form? – 2017-02-16

2 Answers 2

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Let $\mathcal{B}_{\mathrm{can}, \, \mathbb{R}^2} = (e_1, e_2)$ be the canonical basis of $\mathbb{R}^2$, with $e_1 = (1,0)$ and $e_2 = (0,1)$, and let $\mathcal{B}_{\mathrm{can}, \, \mathbb{R}} = (1)$ be the canonical basis of $\mathbb{R}$. Also, let $p_2$ be the linear mapping defined by: $$ \forall (x,y) \in \mathbb{R}^2, \; p_2\big( (x,y) \big) = y. $$ Then: $$ p_2(e_1) = 0 \quad \text{and} \quad p_2(e_2) = 1.$$ Therefore, the matrix of $p_2$ with respect to the pair of basis $\big( \mathcal{B}_{\mathrm{can}, \, \mathbb{R}^2}, \mathcal{B}_{\mathrm{can}, \, \mathbb{R}})$ is given by :

$$ \mathrm{Mat}\big( p_2, \mathcal{B}_{\mathrm{can}, \, \mathbb{R}^2}, \mathcal{B}_{\mathrm{can}, \mathbb{R}} \big) = \begin{pmatrix} 0 & 1 \end{pmatrix}. $$

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$$x_2 = \begin{pmatrix}a &b\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=ax_1+bx_2.$$

Now you should be able to solve on your own.