Simple basis problem

May 2010
20
1
I have dosed off in class way too many times and I need help with a few problems from my pre-test:

#2
Let T: M_22 -> M_22 be defined by


\(\displaystyle T \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & a-d \\ c+b & d \end{bmatrix}
\)

Determine whether
\(\displaystyle \begin{bmatrix} 5 & 3 \\ -1 & 4 \end{bmatrix}\) is in R(T).
Find a basis for R(T).


A simple hint in the right direction would be greatly appreciated.

Advathanksnce!
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
I have dosed off in class way too many times and I need help with a few problems from my pre-test:

#2
Let T: M_22 -> M_22 be defined by


\(\displaystyle T \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & a-d \\ c+b & d \end{bmatrix}
\)

Determine whether
\(\displaystyle \begin{bmatrix} 5 & 3 \\ -1 & 4 \end{bmatrix}\) is in R(T).
Find a basis for R(T).


A simple hint in the right direction would be greatly appreciated.

Advathanksnce!
\(\displaystyle \begin{bmatrix} 5 & 3 \\ -1 & 4 \end{bmatrix}\)

\(\displaystyle a=5\), \(\displaystyle d=4\), \(\displaystyle a-d=5-4=1\)

If a=5 and d=4, then the matrix has to be of the form
\(\displaystyle \begin{bmatrix} 5 & 1 \\ b+c & 4 \end{bmatrix}\); therefore, \(\displaystyle \begin{bmatrix} 5 & 3 \\ -1 & 4 \end{bmatrix}\notin R(T)\)

Basis:
\(\displaystyle \begin{bmatrix} a & a-d \\ c+b & d \end{bmatrix}\rightarrow a\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}+b\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}+c\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}+d\begin{bmatrix} 0 & -1 \\ 0 & 1 \end{bmatrix}\)

\(\displaystyle \left(\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & -1 \\ 0 & 1 \end{bmatrix}\right)\)