Write down the matrix that belongs T in the standard basis of R^3

Question

Let be $T: \mathbb R^3$ a linear transformations which transforms $\vec i$ into $\vec j$, vector $\vec j$ into $\vec i$ + $\vec j$ and 2*$\vec k$ into 4* $\vec i$. Write down the matrix that belongs T in the standard basis of $\mathbb R^3$ .

My notes show me that I have to write down the columns are $\vec i$, $\vec j$, and 2* $\vec k$, and rows as the three standard vectors $\vec i$, $\vec j$ and $\vec k$. When filling up the coresponding matrix I get the result

$$\begin{bmatrix} 0 & 1 & 4 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$ That is apparently wrong but I don't know why. I am using as a template assignments like tranformation from $T(\vec a) = \vec a - \vec b$ which gives me the vector $$\begin{bmatrix} 1 & x & x \\ -1 & x & x \\ 0 & x & x \\ \end{bmatrix}$$

($x \rightarrow b-c$ and $c-a$ in corresponding columns.)

Answer 1

Since $T(2\vec k)=4\vec i$, then $T(\vec k)=2\vec i$ and therefore your matrix should be$$\begin{bmatrix}0&1&2\\1&1&0\\0&0&0\end{bmatrix}.$$

Answer 2

Recall $\vec{i}=(1,0,0)$, $\vec{j}=(0,1,0)$ and $\vec{k}=(0,0,1)$. Then $$\left\{\begin{aligned}&T(\vec{i})=\vec{j}\\&T(\vec{j})=\vec{i}+\vec{j}\\&T(2\vec{k})=4\vec{i}\end{aligned}\right.\iff \left\{\begin{aligned}&T(1,0,0)=\color{green}{(0,1,0)}\\&T(0,1,0)=(1,0,0)+(0,1,0)=\color{green}{(1,1,0)}\\&T(2(0,0,1))\underset{\text{linearity}}{=}2T(0,0,1)=4(1,0,0)\to T(0,0,1)=\color{red}{(2,0,0)}\end{aligned}\right.$$ Thus the matrix is $$\begin{pmatrix}0&1&\color{red}2\\1&1&0\\0&0&0\end{pmatrix}.$$