# Thread: Vector space of all 2x2 matrices

1. ## Vector space of all 2x2 matrices

Hi,

Consider the vector space M2×2 of all 2 × 2 matrices with real number entries. Let T be the linear transformation T : M2×2 > M2×2 given by:

T( $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$) = $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$ = $\begin{bmatrix} b & b \\ d & d \end{bmatrix}$

(a) Write down the standard basis S for M2×2 and write down the dimension of M2×2.
I found that S={(1,0),(0,1)} and the dimension of M2x2 is 4 (2x2=4)
(b) Write down the image under T of each element of S and hence write down the matrix of T with respect to S.
Not sure?
(c) Find a basis (consisting of 2 × 2 matrices) for the image of T and hence write down the rank and nullity of T.
Not sure?
(d) Find a basis (consisting of 2 × 2 matrices) for the nullspace of T.

Not sure?

2. Originally Posted by Pixel
Hi,

Consider the vector space M2×2 of all 2 × 2 matrices with real number entries. Let T be the linear transformation T : M2×2 > M2×2 given by:

T( $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$) = $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$ = $\begin{bmatrix} b & b \\ d & d \end{bmatrix}$

(a) Write down the standard basis S for M2×2 and write down the dimension of M2×2.
I found that S={(1,0),(0,1)} and the dimension of M2x2 is 4 (2x2=4)
(b) Write down the image under T of each element of S and hence write down the matrix of T with respect to S.
Not sure?
(c) Find a basis (consisting of 2 × 2 matrices) for the image of T and hence write down the rank and nullity of T.
Not sure?
(d) Find a basis (consisting of 2 × 2 matrices) for the nullspace of T.

Not sure?

The column vectors of the transformation are lin dep.; therefore, the rank will be 1. Dim=rank+nullity.
Basis
$\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix},\begin{bmatrix}
0 & 0\\
1 & 1
\end{bmatrix}$

3. Originally Posted by Pixel
Hi,

Consider the vector space M2×2 of all 2 × 2 matrices with real number entries. Let T be the linear transformation T : M2×2 > M2×2 given by:

T( $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$) = $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ $\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$ = $\begin{bmatrix} b & b \\ d & d \end{bmatrix}$

(a) Write down the standard basis S for M2×2 and write down the dimension of M2×2.
I found that S={(1,0),(0,1)} and the dimension of M2x2 is 4 (2x2=4)
Did you think about this? If your basis contained only a single matrix, then the dimension would be 1, not 4. The "standard basis" for M2x2 is $\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$, $\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$, and $\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$. That should have been one of the first things you learned.

(b) Write down the image under T of each element of S and hence write down the matrix of T with respect to S.
Not sure?

$T\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$ which is just the sum of 0 times each of the basis matrices. The first column is of the matrix representation of T is all 0s, $\begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}$.

$T\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}= \begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}$ $= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ so the second column is $\begin{bmatrix}1 \\ 1 \\ 0 \\ 0\end{bmatrix}$.
Now, try the next two columns.

(c) Find a basis (consisting of 2 × 2 matrices) for the image of T and hence write down the rank and nullity of T.
Not sure?

Any vector in the image of T can be written as $T\begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix}b & b \\ d & c\end{bmatrix}$ $= \begin{bmatrix}b & b \\ 0 & 0\end{bmatrix}+ \begin{bmatrix}0 & 0 \\ d & d\end{bmatrix}$ $= b\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}+ d\begin{bmatrix} 0 & 0 \\ 1 & 1\end{bmatrix}$. Get the point?

(d) Find a basis (consisting of 2 × 2 matrices) for the nullspace of T.
Not sure?
Any $T\begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix} b & b \\ d & d\end{bmatrix}$ will be 0 (and so the matrix $\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ in the null space of T) if and only if b= 0 and d= 0. That leaves $\begin{bmatrix}a & 0 \\ c & 0\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$