Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

OK, I am trying to prove the following transformation is linear, and find the basis for $\ker(T)$ and Im$(T)$ (also denoted in our textbook by $N(T)$ and $R(T)$ ). Then we're suposed to find the nullity and rank of $T$.

$T: \Bbb{R}^3 \rightarrow \Bbb{R}^2$ defined by $T(a_1, a_2, a_3) = (a_1-a_2, 2a_3)$

We want to see that the transformation preserved addition and scalar multiplication. So I define vector $a$ as $(a_1, a_2, a_3)$ and $b$ as $(b_1, b_2, b_3)$. So the first question is whether $T((a_1+b_1, a_2+b_2, a_3+b_3)$ = $T((a_1-a_2), 2a_3) + T(b_1-b_2, 2b_3)$

and when I plug in vectors $a+b$ to the transformation I get:

$((a_1+b_1)-(a_2+b_2), 2(a_3+b_3))$ which works. So addition is preserved.

The next question is whether it preserves scalar multiplication, or if $T(ca+b) = cT(a) + T(b)$ and as it happens: $T(ca_1+b_1, ca_2+b_2, ca_3+b_3) = ((ca_1+b_1-ca_2+b_2), 2(ca_3+b_3))$

and then if we break up the vectors we find that we get $(ca_1-ca_2, 2ca_3)+(b_1-b_2, 2ba_3)$ so the transformation is linear.

To find the kernel we look for the set of vectors for which $T(a_1,a_2,a_3) = 0$. That happens whenever $a_1 = a_2$ and $a_3 = 0$

But that is where I get stuck because the definition of a kernel doesn't seem to fit. What is the basis for the kernel in this case? If a kernel is a set of vectors then this is making little or no sense to me from the get-go. Because I am not sure what the basis would be if the set of vectors are all those where $a_1 = a_2$ unless it's something like $(a_1, a_2, 0)$. And the dimension of the kernel is 2, I wold think intuitively, but I want to better understand why that is so I can get through the rest of the problem.

share|improve this question
add comment

4 Answers

up vote 0 down vote accepted

Your verification that $T$ is a linear transformation could use some work. Try expanding out the first expression to show explicitly that $T(x + y) = T(x) + T(y)$. Also, check your (linear) algebra in your proof that $T$ preserves scalar multiplication.

As for the kernel, you are correct $T(a_1, a_2, a_3) = 0$ when $a_1 = a_2$ and $a_3 = 0$. So $$N(T) = \{(a, a, 0) : a \in \mathbb{R}\}$$ Can you see why $\{(1, 1, 0)\}$ is a basis for $N(T)$?

It's pretty clear that $T$ is onto. For any $(x,y) \in \mathbb{R}^2$, $(x, 0, y/2) \mapsto (x, y)$ so $R(T) = \mathbb{R}^2$. You can also use the rank-nullity theorem, which gives us

$$3 = \dim \mathbb{R}^3 = \dim N(T) + \dim R(T) = 1 + \dim R(T)$$

share|improve this answer
    
thanks, I think I see better now what was going on. if i understand you correctly since $a_1=a_2$ we end up with a vector that can be expressed with the first two coordinates being the same, and the third zero, so multiplying 1 by anything gets you the numbers you want. The dimensionality (dim(ker(T)) is 1 because it only takes one number to express the vector (a,a,0). Yes? –  Jesse Jun 22 '13 at 12:59
add comment

"... the definition of a kernel doesn't seem to fit." What do you mean?

You're looking at vectors of the form $(\lambda,\lambda,0)=\lambda(1,1,0)$, $\lambda\in\Bbb R$. This is just $\langle (1,1,0)\rangle$. This has dimension $1$.

share|improve this answer
    
I just means that when one refers to a set of vectors (as the definition does) I was having trouble connecting it to the answer I got. –  Jesse Jun 22 '13 at 12:52
add comment

First, because of paragraph 6, paragraphs 3-5 are not necessary. plugging in $ca+b$ is enough to demonstrate linearity.

Next, the kernel consists of all vectors of the form $(t,t,0)$, $t\in \mathbb{R}$. This means that $\{(1,1,0)\}$ is a basis for the kernel of this linear transformation.

share|improve this answer
add comment

A few things:

Your proof that $T$ is linear works. However, you could save time by doing this in one step, showing that $T(\vec a+ c \vec b)=T(\vec a)+c\,T(\vec b)$; plugging in $c=1$ to this equality gives you addition, and plugging in $\vec a=\vec 0$ gives you scalar multiplication.

As for the second part of the question: the kernel is indeed the set of vectors such that $a_1=a_2$ and $a_3=0$. That is, we have a system of two equations on a 3-dimensional system which means that our solution should be one-dimensional. In particular, we can state that the vector $\vec v=(1,1,0)$ forms a basis of the kernel, since any vector satisfying these equations will be a multiple of $\vec v$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.