# Thread: [SOLVED] Onto linear transfomation

1. ## [SOLVED] Onto linear transfomation

Let V be the real vector space of all real 2x3 matrices, and let W be the real vector space of all real 4x1 column vectors. If T is a linear transformation from V onto W, what is the dimension of the subspace {v $\in$ V: T(v) = 0}?

ker(v)=nullity; therefore, the range of v=6.

$\begin{bmatrix}
a & b & c\\
d & e & f
\end{bmatrix}$

I know the answer is 2 but how do I show it?

2. $6 = \mbox{dim }V = \mbox{dim } T(V) + \mbox{dim ker }V = 4 + \mbox{dim ker }V$

3. Originally Posted by Bruno J.
$6 = \mbox{dim }V = \mbox{dim } T(V) + \mbox{dim ker }V = 4 + \mbox{dim ker }V$
Then that leaves us with:

6 = dim V = dim T(v) + 0 = 4 + 0

6 = dim V = dim T(v) = 4

Will this equality always work for any transformation?

4. Originally Posted by dwsmith
Then that leaves us with:

6 = dim V = dim T(v) + 0 = 4 + 0

6 = dim V = dim T(v) = 4

Will this equality always work for any transformation?
6=4?

$Ker(T) = \{v\in V : T(v) = 0\}$. You want to find $dim ~ Ker(T)$. By a theorem (it is rather easy to prove), as Bruno had mentioned, we have that $dim ~ V = dim ~ Im(T) + dim ~ Ker(T)$. Since T is onto a space of dimension 4, we get that $dim ~ Im(T) = 4$. However, $dim ~ V = 6$ and so: $dim ~ V = dim ~ Im(T) + dim ~ Ker(T) \Rightarrow 6 = 4 + dim ~ Ker(T)$ $\Rightarrow dim ~ Ker(T) = 6-4 = 2$

5. Originally Posted by Defunkt
6=4?

$Ker(T) = \{v\in V : T(v) = 0\}$. You want to find $dim ~ Ker(T)$. By a theorem (it is rather easy to prove), as Bruno had mentioned, we have that $dim ~ V = dim ~ Im(T) + dim ~ Ker(T)$. Since T is onto a space of dimension 4, we get that $dim ~ Im(T) = 4$. However, $dim ~ V = 6$ and so: $dim ~ V = dim ~ Im(T) + dim ~ Ker(T) \Rightarrow 6 = 4 + dim ~ Ker(T)$ $\Rightarrow dim ~ Ker(T) = 6-4 = 2$

6= dim V

6 = dim T(v) = 4.... I don't know how you came up with 6=4 from that because 6-4=2 = dim T(v) = 4-4=0

6. 6 = dim T(v) = 4
:P