1
$\begingroup$

Let $T\colon M_{23} \to M_{33}$ be the linear transformation defined by $T(A)=\begin{pmatrix} 2 & -1 \\ 1 & 2 \\ 3&1 \end{pmatrix}\,A$, for $A\in M_{23}$. Find a basis for the kernel and range of $T$.

I don't know how to exactly approach this question. All I know is that the kernel of $T$ would be the nullspace. I row reduced $\begin{pmatrix} 2 & -1 \\ 1 & 2 \\ 3&1 \end{pmatrix}$ and got $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0&0 \end{pmatrix}$ but I don't know what to do from here.

  • 0
    Well, the rank of that matrix defining $\,T\,$ is two, if your reduction of the matrix is correct, so its rank is $\,2\,$. Now apply the rank-nullity theorem.2012-06-26
  • 0
    @DonAntonio: How would the rank-nullity theorem help? Because we're supposed to find the basis for the kernel. I know the rank is 2, the nullity is 0; so there are 2 columns?2012-06-26
  • 0
    @Deborah Have you learned in the past how to write equations of lines/planes in parametric form? This may be very useful here....2012-06-26
  • 0
    @Deborah May be helpful: http://math.stackexchange.com/questions/161033/the-rank-of-a-multiplication-map-l-colon-m-2-times-3-to-m-3-times-3/161036#1610362012-06-26
  • 0
    @Deborah: if you first know the dimension of the kernel then you go on to find its basis...On the other hand, this is not too hard: just take the space of all solutions of the homogeneous system defined by your matrix.2012-06-26
  • 1
    @Don, have you noticed that the domain is not ${\bf R}^2$?2012-06-27
  • 0
    @Gerry, yes but it doesn't matter: by checking the rank and nullity of the matrix definining the map is possible to deduce stuff...2012-06-27

2 Answers 2