# Proving Transformation defined as a matrix is linear

I came across a problem thatwhile doing some review that states:
Consider the transformation $\textit{T}$:$\mathbb{R}^2\rightarrow\mathbb{R}^2$ defined by the matrix
$$\begin{pmatrix} 2 & 0 \\ 1 & 3 \\ \end{pmatrix}$$ Prove that the transformation is linear, state the kernel of T and find the image of the vector (1, 2) under this transformation.
For showing the linear transformation, I wrote in a more familiar fashion: T($x_1$, $x_2$) = (2$x_1$, $x_1$+$3x_2$). However, I am not stuck as to how to define the vectors/matrices for showing closure under addition and scalar multiplication. Any advice/solutions?

Let $u,v \in \mathbb R^2$ and $\alpha \in \mathbb R$. If we prove that $T(\alpha u + v) = \alpha T(u)+T(v)$ then $T$ its a linear transformation.

Let $u = (u_1, u_2)$ and $v = (v_1,v_2)$ for any $u_1,u_2,v_1, v_2, \alpha \in \mathbb R$ then: $$\alpha u + v = \alpha(u_1,u_2)+(v_1,v_2)=(\alpha u_1+v_1,\alpha u_2 + v_2)$$

We calculate $T(\alpha u + v)$: $$T(\alpha u + v)=T(\alpha u_1+v_1,\alpha u_2 + v_2)$$

$$=(2(\alpha u_1+v_1),(\alpha u_1+v_1)+3(\alpha u_2 + v_2))$$

$$=(2 \alpha u_1 + 2v_1,(\alpha u_1+3 \alpha u_2)+(v_1+3v_2))$$

$$=(2\alpha u_1,\alpha u_1+3 \alpha u_2)+(2v_1,v_1+3v_2)$$

$$=\alpha (2u_1, u_1+3u_2)+(2v_1,v_1+3v_2)=\alpha T(u)+T(v)$$

So $T$ is a linear transformation.

Forthe kernel of $T$, you only need to do Gauss elimination on the matrix and then find the free variables. In this case after doing Gauss elimination we find that

$$\begin{pmatrix} 1 & 3 \\ 0 & -6 \\ \end{pmatrix}$$

This means that $ker(T) = \{0 \}$

You can calculate the image of $(1,2)$ directly simply put $T(1,2)$ or make the matrix multiplication: $$\begin{pmatrix} 1 & 3 \\ 0 & -6 \\ \end{pmatrix} \begin{pmatrix} 1\\ 2 \end{pmatrix}$$

I hope it helps.