# Transformation, kernel, image, etc

• June 20th 2009, 01:19 PM
arbolis
Transformation, kernel, image, etc
This is an exercise I had on the Linear Algebra final exam I failed. I could only do a part and I'm still stuck for most of it. So I'll write it entirely and I'll ask help only for part 1) until we finish it, then ask for further help.

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Let $\bold B = \{ \alpha_1, \alpha_2, \alpha_3 \}$ be the ordered basis of $\mathbb{R}^3$ where $\alpha_1=(1,0,1)$, $\alpha_2=(0,1,1)$, $\alpha_3=(0,-1,1)$.
Let $T : \mathbb{R}^3 \to \mathbb{R}^4$ be defined by $T(\alpha_1)=(1,-2,1,0)$, $T(\alpha_2)=(2,1,0,-1)$, $T(\alpha_3)=(1,0,1,0)$.
1)Give an explicit description of $\ker T$, calculate its dimension and give a basis.
2)Give an explicit description of $\Im T$, calculate its dimension and give a basis.
3)Calculate $[T]_{\bold B}^{\bold B'}$, where $\bold B' =\{ (1,1,0,0), (0,0,1,1), (1,0,0,4), (0,0,0,1) \}$.
4)If $\alpha=(-3,-1,3)$, find the coordinates of $T(\alpha)$ with respect to $\bold B'$.
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1) My attempt : only thoughts. I must find all the vectors $v \in \mathbb{R}^3$ such that $T(v)=0$. I'm at a loss. I don't even know if the transformation is linear. I've absolutely no idea about how to find the kernel of it.
• June 20th 2009, 01:43 PM
CaptainBlack
Quote:

Originally Posted by arbolis
This is an exercise I had on the Linear Algebra final exam I failed. I could only do a part and I'm still stuck for most of it. So I'll write it entirely and I'll ask help only for part 1) until we finish it, then ask for further help.

-------------------------------------------------------------
Let $\bold B = \{ \alpha_1, \alpha_2, \alpha_3 \}$ be the ordered basis of $\mathbb{R}^3$ where $\alpha_1=(1,0,1)$, $\alpha_2=(0,1,1)$, $\alpha_3=(0,-1,1)$.
Let $T : \mathbb{R}^3 \to \mathbb{R}^4$ be defined by $T(\alpha_1)=(1,-2,1,0)$, $T(\alpha_2)=(2,1,0,-1)$, $T(\alpha_3)=(1,0,1,0)$.
1)Give an explicit description of $\ker T$, calculate its dimension and give a basis.
2)Give an explicit description of $\Im T$, calculate its dimension and give a basis.
3)Calculate $[T]_{\bold B}^{\bold B'}$, where $\bold B' =\{ (1,1,0,0), (0,0,1,1), (1,0,0,4), (0,0,0,1) \}$.
4)If $\alpha=(-3,-1,3)$, find the coordinates of $T(\alpha)$ with respect to $\bold B'$.
-------------------------------------------------------------

1) My attempt : only thoughts. I must find all the vectors $v \in \mathbb{R}^3$ such that $T(v)=0$. I'm at a loss. I don't even know if the transformation is linear. I've absolutely no idea about how to find the kernel of it.

The question should have stated that $T$ is linear as otherwise the definition does not define its behaviour.

The kernel or null space of $T$ is the subspace of $\mathbb{R}^3$ whose image is the zero vector in $\mathbb{R}^4$. Now let $x=u_1\alpha_1+u_2 \alpha_2+u_3\alpha_3$ be in the null space of $T$, then:

$
T(x)=u_1 T(\alpha_1)+u_2 T(\alpha_2)+u_3 T(\alpha_3)={\bf{0}}_{4}
$

Write this out in full and you will find that the only solution has $u_1=u_2=u_3=0$, so the zero vector in $\mathbb{R}^3$ is the only element in the Kernel of $T$.

CB
• June 20th 2009, 02:32 PM
arbolis
Quote:

Originally Posted by CaptainBlack
The question should have stated that $T$ is linear as otherwise the definition does not define its behaviour.

The kernel or null space of $T$ is the subspace of $\mathbb{R}^3$ whose image is the zero vector in $\mathbb{R}^4$. Now let $x=u_1\alpha_1+u_2 \alpha_2+u_3\alpha_3$ be in the null space of $T$, then:

$
T(x)=u_1 T(\alpha_1)+u_2 T(\alpha_2)+u_3 T(\alpha_3)={\bf{0}}_{4}
$

Write this out in full and you will find that the only solution has $u_1=u_2=u_3=0$, so the zero vector in $\mathbb{R}^3$ is the only element in the Kernel of $T$.

CB

Thanks so much! It worked. I agree, the professors didn't even specify verbally that the transformation had to be linear.
So $\dim \ker T =0$ and it has no basis.

Part 2) I did it entirely. ( I can write all what I did, but I think it's a waste of time since I did it well)

Part 3) I've done all wrong. I've written the vectors of the basis $\bold B$ as a linear combination of the vectors of the basis of $\bold B'$, but I can't do that since they do not have the same dimension.
As of today I still do not know how to proceed.
• June 21st 2009, 01:10 PM
arbolis
Using the method I said in my previous, I got that $[T]_{\bold B}^{\bold B'}=\begin{bmatrix} 0&1&1 \\ 1&1&2 \\ 1&-1&-1 \\ -4&3&3 \end{bmatrix}$. However I got it all wrong. I don't know how to find what they ask me...