Let T: $M_{2x2}(R) \to R^2$ be the linear transformation defined by

$$ T\left( \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \right) = \begin{bmatrix} a - b \\ c - d \\ \end{bmatrix} $$

What I think I know: (a) Find Ker(T), a basis for Ker(T) and nullity(T)
solving the homogenous system yields $a=b, c=d$, then $Ker(T)=span \{ \begin{bmatrix} 1 & 1 \\ 0 & 0 \\ \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 1 \\ \end{bmatrix} \}$
It follows that the Basis for Ker(T) is the two matrices above and nullity(T) = 2

What I don't know: (b) Find a basis for Range(T) and Rank(T)

I understand that if you have a Matrix A representing the linear transformation the Range(T) is just the Col(A). How do I find a basis for the Range when I am missing that Matrix?

It also seems that a simple observation would be that $Rank(T)=2$ because

$dim(v)=nullity(T)+rank(T)$ and thus $Rank(T) = 4 - 2 = 2$

but I am unaware of how this observation helps me solve the problem at hand.

Thank you!

  • $\begingroup$ Hint: Rank($T$) is $2$ means the dimension of the Range space of $T$ is $2$ and this a subspace of $\mathbb {R}^2$, so actually the range of T is the whole $\mathbb {R}^2$, so you can choose any basis of $\mathbb {R}^2$, that will be the basis for Range of $T$ $\endgroup$ – R. Singh May 11 '17 at 4:26
  • $\begingroup$ @R.Singh thank you very much! $\endgroup$ – Josh Bornstein May 11 '17 at 4:28

The rank of $T$ is $2$ and the dimension of the space of column vectors is $2$. Therefore the image is the whole space of column vectors: $\Bbb R^2$.


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