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Thread: Vector space of all 2x2 matrices

  1. #1
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    Vector space of all 2x2 matrices

    Hi,

    Consider the vector space M22 of all 2 2 matrices with real number entries. Let T be the linear transformation T : M22 > M22 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 M22 and write down the dimension of M22.
    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?


    Please help!
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  2. #2
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    Quote Originally Posted by Pixel View Post
    Hi,

    Consider the vector space M22 of all 2 2 matrices with real number entries. Let T be the linear transformation T : M22 > M22 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 M22 and write down the dimension of M22.
    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?


    Please help!
    The column vectors of the transformation are lin dep.; therefore, the rank will be 1. Dim=rank+nullity.
    Basis
    $\displaystyle \begin{bmatrix}
    1 & 1\\
    0 & 0
    \end{bmatrix},\begin{bmatrix}
    0 & 0\\
    1 & 1
    \end{bmatrix}$
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  3. #3
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    Quote Originally Posted by Pixel View Post
    Hi,

    Consider the vector space M22 of all 2 2 matrices with real number entries. Let T be the linear transformation T : M22 > M22 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 M22 and write down the dimension of M22.
    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 $\displaystyle \begin{bmatrix}a & 0 \\ c & 0\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$


    Please help!
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  4. #4
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    So for part b you just multiply the 4 standard bases matrices by the transition matrix and then put the resulting matrices in terms of the standard basis? and these will be columns of a 4x4 matrix?
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