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## Solution to Linear Algebra Done Wrong Exercise 1.3.3

For each linear transformation below find its matrix

a) $T:\mathbb R^2\to \mathbb R^3$ defined by $$T(x,y)^T=(x+2y,2x-5y,7y)^T;$$

Solution: We have $$T\begin{pmatrix}x\\ y\end{pmatrix}=x\begin{pmatrix} 1\\ 2\\ 0\end{pmatrix}+y\begin{pmatrix} 2\\ -5\\ 7\end{pmatrix},$$ hence the matrix is $\begin{pmatrix} 1 & 2\\ 2 & -5\\ 0 & 7\end{pmatrix}$.

b) $T:\mathbb R^4\to \mathbb R^3$ defined by $$T(x_1,x_2,x_3,x_4)^T=(x_1+x_2+x_3+x_4,x_2-x_4,x_1+3x_2+6x_4)^T;$$

Solution: We have $$T\begin{pmatrix}x_1\\ x_2\\ x_3\\ x_4\end{pmatrix}=x_1\begin{pmatrix} 1\\ 0\\ 1\end{pmatrix}+x_2\begin{pmatrix} 1\\ 1\\ 3\end{pmatrix}+x_3\begin{pmatrix} 1\\ 0\\ 0\end{pmatrix}+x_4\begin{pmatrix} 1\\ -1\\ 6\end{pmatrix},$$ hence the matrix is $\begin{pmatrix} 1 & 1 & 1 & 1\\ 0 & 1 & 0 & -1\\ 1 & 3 & 0 & 6\end{pmatrix}$.

c) $T:\mathbb P_n\to \mathbb P_n$, $Tf(t)=f’(t)$ (find the matrix with respect to the standard basis $1,t,t^2,\dots,t^n$);

Solution: We have $$T\begin{pmatrix} 1 \\ t \\ t^2 \\ \vdots \\ t^{n-1} \\ t^n\end{pmatrix}=\begin{pmatrix} 0 \\ 1 \\ 2t \\ \vdots \\ (n-1)t^{n-2} \\ nt^{n-1}\end{pmatrix},$$namely $$T\begin{pmatrix} 1 \\ t \\ t^2 \\ \vdots \\ t^{n-1} \\ t^n\end{pmatrix}=1\begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \\0\end{pmatrix}+t\begin{pmatrix} 0 \\ 0 \\ 2 \\ \vdots \\ 0 \\ 0\end{pmatrix}+\cdots+t^{n-2}\begin{pmatrix} 0 \\0 \\ 0 \\ \vdots \\ n-1 \\ 0\end{pmatrix}+t^{n-1}\begin{pmatrix} 0 \\ 0\\0 \\ \vdots \\ 0 \\ n\end{pmatrix}+t^n\begin{pmatrix} 0 \\0 \\ 0 \\ \vdots \\ 0 \\ 0\end{pmatrix}.$$Therefore, the matrix is $\begin{pmatrix} 0 & 0 &0 &\cdots & 0 & 0 & 0 \\1 & 0 &0 &\cdots & 0 & 0 & 0 \\ 0 & 2 &0 &\cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots \\ 0 & 0 &0 &\cdots & n-1 & 0 & 0 \\ 0 & 0 &0 &\cdots & 0 & n & 0 \end{pmatrix}$.

d) $T:\mathbb P_n\to \mathbb P_n$, $Tf(t)=2f(t)+3f’(t)-4f’’(t)$ (find the matrix with respect to the standard basis $1,t,t^2,\dots,t^n$);

Solution: Similar to c), the matrix is $$\begin{pmatrix} 2 & 0 &0 &\cdots & 0 & 0 & 0 \\3 & 2 &0 &\cdots & 0 & 0 & 0 \\ -8 & 6 &2 &\cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots \\ 0 & 0 &0 &\cdots & 3(n-1) & 2 & 0 \\ 0 & 0 &0 &\cdots & -4n(n-1) & 3n & 2 \end{pmatrix}.$$