Respuesta :
Answer:
Ker(T) has basis [tex]\emptyset[/tex]
Range(T) has basis [tex]\left\{\left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \right\}.[/tex]
Step-by-step explanation:
If the matrix you wanted to represent is [tex]A=\left(\begin{array}{cc}2&1\\3&4\end{array}\right)[/tex], then:
[tex]ker(T)=\left\{{\bf x}\in \mathbb{R}^{2}: A{\bf x}=0\right\}[/tex].
So, to find ker(T) you must solve the homogenous equation
[tex] 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)[/tex]
Using Gauss elimination we obtain the simpler equivalent system
[tex]\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)[/tex]
Then, we have that
[tex] x=0,y=0 [/tex].
We have that [tex] ker(T)=\{\left(\begin{array}{c}0&0\end{array}\right)\}[/tex]. On this case we say that the basis is the empty set [tex]\emptyset[/tex].
The range of [tex]T[/tex] is the set of vectors of the form
[tex]\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)[/tex]
So,
[tex]Range(T)=\langle \left(\begin{array}{c}2&3\end{array}\right), \left(\begin{array}{c}1&4\end{array}\right) \rangle.[/tex] Where the angle brakets denotes the span.
