Linear mapping/Dimension formula/R/011022134246/Example

We consider the linear mapping

$\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{4},{\begin{pmatrix}x\\y\\z\end{pmatrix}}\longmapsto M{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}y+z\\2y+2z\\x+3y+4z\\2x+4y+6z\end{pmatrix}},$

given by the matrix

${}M={\begin{pmatrix}0&1&1\\0&2&2\\1&3&4\\2&4&6\end{pmatrix}}\,.$

To determine the kernel, we have to solve the homogeneous linear system

${}{\begin{pmatrix}y+z\\2y+2z\\x+3y+4z\\2x+4y+6z\end{pmatrix}}={\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}\,.$

The solution space is

${}L={\left\{s{\begin{pmatrix}1\\1\\-1\end{pmatrix}}\mid s\in \mathbb {R} \right\}}\,,$

and this is the kernel of ${}\varphi$. The kernel has dimension one, therefore the dimension of the image is ${}2$, due to the dimension formula.