[SOLVED] Onto linear transfomation

**dwsmith** · May 9th 2010, 11:43 AM

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.

$\begin{bmatrix}<br /> a & b & c\\ <br /> d & e & f<br /> \end{bmatrix}$

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

**Bruno J.** · May 9th 2010, 12:31 PM

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

**dwsmith** · May 9th 2010, 12:36 PM

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?

**Defunkt** · May 9th 2010, 01:24 PM

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: $dim ~ V = dim ~ Im(T) + dim ~ Ker(T) \Rightarrow 6 = 4 + dim ~ Ker(T)$ $\Rightarrow dim ~ Ker(T) = 6-4 = 2$

**dwsmith** · May 9th 2010, 01:33 PM

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

**Defunkt** · May 9th 2010, 02:29 PM

6 = dim T(v) = 4

:P

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