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Let $T: \text{Mat}_{2 \times 3} \rightarrow \text{Mat}_{2 \times 2}$ be defined by

$$T \left(\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix}\right) = \begin{pmatrix} a_{11} + a_{12} & a_{12} + a_{13} \\ a_{21} + a_{22} & a_{13} + a_{23} \end{pmatrix}.$$

Find bases for $R(T)$ and $N(T)$.

I found the basis for $N(T)$ which is $\begin{pmatrix} 1 & -1 & 1 \\ 0 & 0 & -1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 & 0 \\ 1 & -1 & 0 \end{pmatrix}$.

I don't understand how to find it for $R(T)$ though as $3$ of the entries are interdependent. I know one of the vectors for the basis will be $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, but to generate the other entries I am not sure I understand. If I put $1$ in entry $[1,1]$ how will the others react?

Thanks

Michael Joyce
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user144806
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  • Question reminiscent of http://math.stackexchange.com/questions/751337/finding-bases-for-rt-and-nt --- do you two have the same teacher? – Gerry Myerson Apr 22 '14 at 11:09

1 Answers1

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Hint: You've computed that the null space of $T$ is two dimensional. Since the domain vector space if six dimensional, what does the rank-nullity theorem tell you about the rank of $T$? In this case, once you know the rank, you can quickly come up with a basis of $R(T)$.

Michael Joyce
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  • Oh thank you, I got confused and used the dimension theorem but used rank(T) + nullity(T) = dimension of the codomain instead of the domain. So the rank of T will be 4 and they would be the 4 matrices with a 1 and 0 everywhere else for each of the 4 entries right? – user144806 Apr 22 '14 at 01:19
  • That's one (natural) choice of basis. Any four linearly independent matrices in $\text{Mat}{2 \times 2}$ will give a basis for $R(T) = \text{Mat}{2 \times 2}$. – Michael Joyce Apr 22 '14 at 01:29