Question

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

Answer 1

Answer:

Ker(T) has basis $\emptyset$

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

Step-by-step 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.