# Vector space of all 2x2 matrices

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

(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?

#### dwsmith

MHF Hall of Honor
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($$\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$) = $$\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ $$\displaystyle \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$$ = $$\displaystyle \begin{bmatrix} b & b \\ d & d \end{bmatrix}$$

(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
$$\displaystyle \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0\\ 1 & 1 \end{bmatrix}$$

#### HallsofIvy

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

(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 $$\displaystyle \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$$, $$\displaystyle \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$$, $$\displaystyle \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$$, and $$\displaystyle \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$. 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?

$$\displaystyle T\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$ 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, $$\displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$.

$$\displaystyle T\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}= \begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}$$$$\displaystyle = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$$ so the second column is $$\displaystyle \begin{bmatrix}1 \\ 1 \\ 0 \\ 0\end{bmatrix}$$.
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 $$\displaystyle T\begin{bmatrix}a & b \\ c & d\end{bmatrix}= \begin{bmatrix}b & b \\ d & c\end{bmatrix}$$$$\displaystyle = \begin{bmatrix}b & b \\ 0 & 0\end{bmatrix}+ \begin{bmatrix}0 & 0 \\ d & d\end{bmatrix}$$$$\displaystyle = b\begin{bmatrix}1 & 1 \\ 0 & 0\end{bmatrix}+ d\begin{bmatrix} 0 & 0 \\ 1 & 1\end{bmatrix}$$. Get the point?

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