# Determine the kernel and range. HELP???

## Determine the kernel and range of the transformation defined by the matrix 6 4 3 2 . (Enter your answers as a comma-separated list. Enter each vector in the form (x1, x2, ...). Use r for any arbitrary scalar.) Show that dim ker(T) + dim range(T) = dim domain(T). dim ker(T) + dim range(T) = dim domain(T) right double arrow implies

Jun 28, 2018

color(red)("range"(T) = {((2a),(a))| a in RR}),qquad dim"range"(T)=1
color(red)("ker"(T) = {((2t),(-3t))| t in RR}),qquad dim"ker"(T)=1

#### Explanation:

The transformation takes the vector $\left(\begin{matrix}x \\ y\end{matrix}\right)$ to the vector $\left(\begin{matrix}{x}^{'} \\ {y}^{'}\end{matrix}\right)$ where

$\left(\begin{matrix}{x}^{'} \\ {y}^{'}\end{matrix}\right) = \left(\begin{matrix}6 & 4 \\ 3 & 2\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) \implies$

${x}^{'} = 6 x + 4 y$
${y}^{'} = 3 x + 2 y$

The range
It is easy to see that ${x}^{'} = 2 {y}^{'}$, so that vectors in the range of the map must be of the form $\left(\begin{matrix}2 a \\ a\end{matrix}\right) , a \in \mathbb{R}$. Conversely, if a vector is of the form $\left(\begin{matrix}2 a \\ a\end{matrix}\right)$ with $a \in \mathbb{R}$, then any of the infinite vectors $\left(\begin{matrix}x \\ y\end{matrix}\right)$ with $3 x + 2 y = a$ will map into it. Thus

$\textcolor{red}{\text{range} \left(T\right) = \left\{\left(\begin{matrix}2 a \\ a\end{matrix}\right) | a \in \mathbb{R}\right\}}$

It is easy to see that all the elements of $\text{range} \left(T\right)$ are multiples of the single vector $\left(\begin{matrix}2 \\ 1\end{matrix}\right)$. The single element set $\left\{\left(\begin{matrix}2 \\ 1\end{matrix}\right)\right\}$ is thus a spanning set for $\text{range} \left(T\right)$. Since it is obviously linearly independent, this set provides a basis for $\text{range} \left(T\right)$. Hence we have

$\textcolor{red}{\text{dim"\ "range} \left(T\right) = 1}$

The Kernel

The kernel of the transformation is defined by

$\text{ker} \left(T\right) \equiv \left\{v \in {\mathbb{R}}^{2} | T v = 0\right\}$

So, if $v = \left(\begin{matrix}x \\ y\end{matrix}\right) \in \text{ker} \left(T\right)$, we have

$6 x + 4 y = 0$
$3 x + 2 y = 0$

Which will be satisfied by $x = 2 t , \setminus y = - 3 t$ for any $t \in \mathbb{R}$. So

$\textcolor{red}{\text{ker} \left(T\right) = \left\{\left(\begin{matrix}2 t \\ - 3 t\end{matrix}\right) | t \in \mathbb{R}\right\}}$

It is obvious that $\left\{\left(\begin{matrix}2 \\ - 3\end{matrix}\right)\right\}$ is a basis for $\text{ker} \left(T\right)$. Hence

$\textcolor{red}{\text{dim"\ "ket} \left(T\right) = 1}$

Rank-nullity theorem

Since $\text{domain} \left(T\right) = {\mathbb{R}}^{2}$, we have $\dim \text{domain} \left(T\right) = 2$

Thus the rank-nullity theorem

$\dim \text{range"(T)+dim"ker"(T) = dim"domain} \left(T\right)$

is verified by $1 + 1 = 2$