Question

T(v) = Av represents the linear transformation T. Find a basis for the kernel of T and the range of T.

A = [2 1]
[3 4]

Answer 1

Ker(T) has basis
`\emptyset`

Range(T) has basis
`\left\{\left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \right\}.`

Explanation:

If the matrix you wanted to represent is
`A=\left(\begin{array}{cc}2&1\\3&4\end{array}\right)`, then:

`ker(T)=\left\{{\bf x}\in \mathbb{R}^(2): A{\bf x}=0\right\}`.

So, to find ker(T) you must solve the homogenous equation

`A{\bf x}=\left(\begin{array}{cc}2&1\\3&4\end{array}\right) \left(\begin{array}{c} x&y \end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)`

Using Gauss elimination we obtain the simpler equivalent system

`\left(\begin{array}{cc} -6&3\\0&5\end{array}\right) \left(\begin{array}{c} x&y\end{array}\right)=\left(\begin{array}{c}0&0\end{array}\right)`

Then, we have that

`x=0,y=0`.

We have that
`ker(T)=\{\left(\begin{array}{c}0&0\end{array}\right)\}`. On this case we say that the basis is the empty set
`\emptyset`.

The range of
`T` is the set of vectors of the form

`\left(\begin{array}{c} \alpha &\beta\end{array}\right)=\left(\begin{array}{cc}2&1\\3&4\end{array}\right)\left(\begin{array}{c}x&y\end{array}\right)=x\left(\begin{array}{c}2&3\end{array}\right)+y\left(\begin{array}{c}1&4\end{array}\right)`

So,

`Range(T)=\langle \left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \rangle.` Where the angle brakets denotes the span.