# Thread: Null spaces, Ranges, Nullities, and Ranks

1. ## Null spaces, Ranges, Nullities, and Ranks

T:M(2x3) ---> M(2x2)

T $\left(\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{2 1}&a_{22}&a_{23}\end{array}\right)$ = $\left(\begin{array}{cc}2a_{11}-a_{12}&a_{13}+2a_{12}\\0&0\end{array}\right)$

Can anyone find the null space of this transformation, and explain why nullity(T) = 4?

R(T) = M(2x2), correct? So then shouldn't rank(T) be 4?

I'm just confused on all of these! Looking for anyone to shed some light on all of this for me.

2. The domain space has dimension 6, and the range space has dimension 2 notice those bottom two spots of the range space are always 0, so you don't get any dimension from those entries. A basis for the range space can just be the two matrices with 1 in the top left and then the matrix with 1 in the top right and all the rest 0. See why they span and why they are linearly independent? It should be obvious from definitions. Now apply the rank nullity theorem to see why Nullity(T)=6-2=4.

As far as finding the nullspace, it is just $\{A\in M_{2\times 3}| T(A)=0 \}=\{A\in M_{2 \times 3}| 2a_{11}=a_{12}=-\frac{a_{13}}{2} \}$. So you see once you pick one of the $a_{11},a_{12},a_{13}$ the others are set, so you get no freedom, but you are still free to choose anything for the three entries on the bottom row. This gives the 4 dimensions that we calculated above.