Interesting linear transformation, kerT, imT question

[mystmyst]

Homework Statement

Prove or dispove

$$T:R^n \rightarrow R^n$$ is a linear transformation

if for every $$u \in R^n$$ and for every $$v \in kerT$$,
$$T(u) \cdot v=0$$

then $$KerT = (ImT)^{\perp}$$

The Attempt at a Solution

True.

since it was given $$T(u) \cdot v=0$$ and we are dealing with all of $$R^n$$ then either:
1. $$kerT=0, imT=R^n$$
or
2. $$kerT=R^n, imT=0$$

since how else can $$T(u) \cdot v=0$$ ?

and in both those cases $$kerT=(imT)^{\perp}$$

Does anybody see a mistake here? Is the claim really false? Or is the claim true for a different reason?

Thanks!

 

[arkajad]

First of all take $$R^2$$ and consider T given by the matrix

$$T=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$

Its kernel are vectors

$$\begin{pmatrix}0\\x\end{pmatrix}$$

Its image are vectors

$$\begin{pmatrix}x\\0\end{pmatrix}$$

Evidently its kernel is the orthogonal complement to its image.

From your equation:

$$T(u) \cdot v=0$$

It follows immediately that the image is contained in the orthogonal complement of the kernel - do you see it? So, what remains is to prove or disprove that it is exactly the orthogonal complement. For this show that T maps $$kerT^\perp$$ onto $$imT$$ in a one-to-one way (this is the key). So, $$kerT^\perp$$ and $$imT$$ must have the same dimension.

 

[mystmyst]

First of all take $$R^2$$ and consider T given by the matrix

$$T=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$

Its kernel are vectors

$$\begin{pmatrix}0\\x\end{pmatrix}$$

Its image are vectors

$$\begin{pmatrix}x\\0\end{pmatrix}$$

Evidently its kernel is the orthogonal complement to its image.

Great example. So you disproved my reasoning but not my claim.

From your equation:

$$T(u) \cdot v=0$$

It follows immediately that the image is contained in the orthogonal complement of the kernel - do you see it?

I see that now but why are we concerned with the orthogonal complement of the kernel?

hm...can we say since the image is contained in the orthogonal complement of the kernel, then the kernel must be contained (actually equal to) in the orthogonal complement of the image?

So, what remains is to prove or disprove that it is exactly the orthogonal complement. For this show that T maps $$kerT^\perp$$ onto $$imT$$ in a one-to-one way (this is the key). So, $$kerT^\perp$$ and $$imT$$ must have the same dimension.

**I sort of get it. How can you show that it maps it in a one-to-one way?

So what was your conclusion? Is the claim true or false? (I have 12 points relying on this...and since my proof was not totally correct, i'll get at least 6 points off ):, and if the answer is false, i'll get everything off :( )

 

[arkajad]

hm...can we say since the image is contained in the orthogonal complement of the kernel, then the kernel must be contained (actually equal to) in the orthogonal complement of the image?

Yes, for two subspaces $$U,V$$ we have that $$U\subset V^\perp$$ is equivalent to $$V\subset U^\perp$$. Prove it for yourself and keep in mind.

How can you show that it maps it in a one-to-one way?

Show that if $$T(u)=T(u')$$ then $$u=u'$$. How? Suppose $$u,u'\in kerT^\perp$$. We have $$T(u-u')=0$$. Therefore $$u-u'\in kerT$$. But, on the other hand $$u-u'\in kerT^\perp$$ (because $$kerT^\perp$$ is a linear subspace). Now, if a vector belong to a subspace and is orthogonal to this subspace, then it is orthogonal to itself, therefore it is zero.

and if the answer is false, i'll get everything off :( )

Do not worry. Next time you will do better, I am sure!

 

[mystmyst]

So basically, the answer is True?

 

[arkajad]

Basically you need to finish now the exercise. Otherwise you will not learn!

 

[mystmyst]

haha! I don't want to learn. I just had an entire semester of learning linear algebra (: I just want to know if I'm getting around a 94 or around 88 (:

but thanks for your help!

 

[arkajad]

Well, that's your choice, but I hope there will be others reading this thread that will learn something.