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$T: M_{22} \mapsto M_{22}$ defined by:

$T(A) = AB$ where B = $\begin{bmatrix} 1 & -1 \\ -1 & 1 \\ \end{bmatrix}$

So AB = $\begin{bmatrix} a-b & b-a \\ c-d & d-c \\ \end{bmatrix}$

after multiplying arbitrary matrix A by matrix B. My question is, what would be the basis for the range of this transformation? I have found the kernel and $dim(ker(T)) = nullity(T) = 2$, so I know the $rank(T) = 2$ but i'm really struggling to find the basis for the range here.

If someone could help me out here I would really appreciate it. Thanks!

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  • $\begingroup$ From the rank-nullity theorem, it seems like I should have 2 elements in the basis for the range. Although that is helpful as well thanks. $\endgroup$ – FuegoJohnson Aug 1 '17 at 20:34
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To find a basis for the range, just apply the transformation to basis elements of your domain.

In this case, using the standard basis, we find that $$T\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}\right) = \begin{bmatrix} 1 & -1 \\ 0 & 0 \\ \end{bmatrix}$$ and $$T\left(\begin{bmatrix} 0 & 0 \\ 1 & 0 \\ \end{bmatrix}\right) = \begin{bmatrix} 0 & 0 \\ 1 & -1 \end{bmatrix}$$ Which give a basis (note that the other basis vectors give us multiples of these, so we are finished).

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  • $\begingroup$ @MichaelLee I'm not sure exactly what you mean - $T(A)$ is defined to be $AB$, so finding $T$ is the same as right-multiplying by $B$. $\endgroup$ – platty Aug 1 '17 at 20:32
  • $\begingroup$ Oh, whoops, I misread the question. $\endgroup$ – Michael Lee Aug 1 '17 at 20:32
  • $\begingroup$ ok I was really confused, but this makes sense. thanks a bunch! $\endgroup$ – FuegoJohnson Aug 1 '17 at 20:35
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Note that for $A=\left[\begin{array}{rr} a & b \\ c & d \end{array}\right]$ we have \begin{align*} T(A) &= BA \\ &= \left[\begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{rr} a & b \\ c & d \end{array}\right] \\ &= \left[\begin{array}{rr} a - b & -a + b \\ c - d & -c + d \end{array}\right] \\ &= a\cdot\left[\begin{array}{rr} 1 & -1 \\ 0 & 0 \end{array}\right] +b\cdot\left[\begin{array}{rr} -1 & 1 \\ 0 & 0 \end{array}\right] +c\cdot\left[\begin{array}{rr} 0 & 0 \\ 1 & -1 \end{array}\right] +d\cdot \left[\begin{array}{rr} 0 & 0 \\ -1 & 1 \end{array}\right] \\ &= (a-b)\cdot\left[\begin{array}{rr} 1 & -1 \\ 0 & 0 \end{array}\right]+(c-d)\cdot\left[\begin{array}{rr} 0 & 0 \\ 1 & -1 \end{array}\right] \end{align*} This shows that $$ \DeclareMathOperator{image}{image}\image(T)= \DeclareMathOperator{Span}{Span}\Span\left\{\left[\begin{array}{rr} 1 & -1 \\ 0 & 0 \end{array}\right],\left[\begin{array}{rr} 0 & 0 \\ 1 & -1 \end{array}\right]\right\} $$ Showing that $\left\{\left[\begin{array}{rr} 1 & -1 \\ 0 & 0 \end{array}\right],\left[\begin{array}{rr} 0 & 0 \\ 1 & -1 \end{array}\right]\right\}$ is linearly independent is simple.

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