Linear transformation ker and image

Question

Let $\varphi\colon \mathbb{R}^4 \rightarrow \mathbb{R}^3$ be described by $\varphi(X)=AX$ where $A=\begin{pmatrix} 3 & 2 & 1 & 3 \\ 1 & 1 & 1 & 1 \\ 2 & 1 & 0 & 1 \end{pmatrix} $ . Find base vectors of $\ker \varphi$ and $\operatorname{Im} \varphi$. In my opinion those vectors will be $[-1,1,0,0]$ and $[-1,0,1,0]$ for kernel and $[3,1,2]$ and $[3,1,1] $ for image. Am I correct?

Answer 1

Start with the definitions: $$\ker\varphi:=\{X\in\mathbb R^4:\varphi(X)=0\},$$ that is, the set of all $(x_1,x_2,x_3,x_4)$ such that $$ \begin{bmatrix} 3 & 2 & 1 & 3 \\ 1 & 1 & 1 & 1 \\ 2 & 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix}. $$

Can you find the solution set of this system? If so, you will be able to write down a basis for $\ker\varphi$. Note that row reducing yields $$ \begin{bmatrix} 3 & 2 & 1 & 3 \\ 1 & 1 & 1 & 1 \\ 2 & 1 & 0 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$ so \begin{align} x_1-x_3&=0,\\ x_2+2x_3&=0,\\ x_4&=0 \end{align} and thus $x_4=0$, $x_2=-2x_3$, $x_1=x_3$ and $x_3$ is free. Hence a basis for $\ker\varphi$ is $$ \left\{\begin{bmatrix} 1\\-2\\1\\0\end{bmatrix}\right\}.$$

As for $\text{Im}\,\varphi$, again go to the definition $$ \text{Im}\,\varphi:=\{Y\in\mathbb{R}^3:\varphi(X)=Y\text{ for some }x\in\mathbb{R}^4\}.$$ So look at all possible outputs of $\varphi$ acting on an $X$ after row reducing: \begin{align} \begin{bmatrix} 3 & 2 & 1 & 3 \\ 1 & 1 & 1 & 1 \\ 2 & 1 & 0 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. \end{align} Since you have leading ones in columns 1, 2, and 4, use columns 1, 2, and 4 of the original matrix as the basis for $\text{Im}\,\varphi$.

Answer 2

Reducing your matrix to RREF gives $\;$ $R=\begin{bmatrix}1&0&-1&0\\0&1&2&0\\0&0&0&1\end{bmatrix}$;

Now solve $Rx=0$ to get a basis for the kernel, and

take the columns in A corresponding to the leading 1's in R to get a basis for the image.