Basis, kernal, image, transformation, etc (not a hard one!)

1. Nov 3, 2005

I hope everyone could understand my matrix notation!

T: M22 -----> M22; T[a b; c d] = [a+b b+c; c+d d+a]

I found the the basis for ker(T) to be [1 -1; 1 -1]

im having problems finding a basis for im(T)? I know that it has to be dim3!

so far i have:

im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} = M22

How on earth do i get a basis out of that?

Anybody have any ideas?

Thanks

2. Nov 3, 2005

Hurkyl

Staff Emeritus
If you just tried picking vectors in im T at random, do you think you'll have a pretty good chance of finding three linearly independent ones? Can you think of any way to improve your odds?

Random guessing is a very underrated technique.

P.S. in

im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} = M22

I assume that last '=' was supposed to be a subset symbol?

3. Nov 3, 2005

in honesty, i have no idea what that last = sign means, the book just puts it their on other examples. Im doing at even question, so it has no solution in the solution manual

Im not gussing about the dim of the image. I KNOW that its 3 since dim(V) = dim(Ker) + dim(Im)

and i found that the dimension of the kernal is 1, and therefore, dimension of the image has to be 3.

I just need to find the basis for im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} and its got be really STUCK :(

4. Nov 3, 2005

Hurkyl

Staff Emeritus
I meant guess a basis!

5. Nov 3, 2005

does [1 0; 0 0] [0 1; 0 0] [0 0; 1 0] work?

6. Nov 3, 2005

Hurkyl

Staff Emeritus
They're certainly linearly independent. Are they in the image of T? If so, then is there any reason why they wouldn't suffice for a basis? Can you prove that they are a basis?

7. Nov 3, 2005

Those questions are making this harder :(

They are independint, but NO, they are not an image. becuase it dose not satisfy m(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22}

hmmmm...

another GUESS is [1 0; 1 0] [0 1; 1 0] [1 0; 0 1] ?????

8. Nov 3, 2005

Hurkyl

Staff Emeritus
They're questions you'd have to answer anyway, though, aren't they? After all, what is the definition of the image, and what is the definition of a basis for that space? :tongue2:

9. Nov 3, 2005

oh goodness.. this is getting so confusing now :S

Listen..

the question says " in each case, (i) find a basis of kerT and (ii) find a basis of imT. " e) T: M22 -----> M22; T[a b; c d] = [a+b b+c; c+d d+a]

i found ker(T) and my answer was [1 -1; 1 -1]

image T is im(T) = {[a+b b+c; c+d d+a]/ [a b; c d] in M22} (By definition)

so how on earth to i find the basis for [a+b b+c; c+d d+a] such that [a b; c d] are in M22?

now.... a basis is i) linearly independent, and ii) a spanning set of that space.

Now i just have no idea how to extract a basis out of that image :S

10. Nov 3, 2005

Hurkyl

Staff Emeritus
Can you find any vectors in im T?

11. Nov 3, 2005

I don't understand what you mean by that? Perhpas [1 0; 0 0] ?? is that what your trying to ask?

12. Nov 3, 2005

Hurkyl

Staff Emeritus
I want you to give me a vector that is in the image of T, and a proof that it is in the image of T.

13. Nov 3, 2005

well... [1 1; 1 1] is in imageT. and explanation is since [1 1; 1 1] satisfies [a b; c d] being in M22. how does that sound?

see.. imT = {T(v)/v in V}

So wouldn't any vector in M22 satisfy that?

14. Nov 3, 2005

Hurkyl

Staff Emeritus
Ah, I see your problem now: you're confused by the set-builder notation.

The notation

$$\mbox{im} T = \{ T(v) | v \in V \}$$

means the following:

For any w, w is in im T if and only if the equation w = T(v) has a solution with v in V.

15. Nov 3, 2005

Hurkyl

Staff Emeritus
So, the fact that

$$\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \in M_{2,2}$$

tells us that

$$T\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \in \mbox{im}\ T$$

or equivalently,

$$\left( \begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array} \right) \in \mbox{im}\ T$$

since

$$T\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) = \left( \begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array} \right)$$

16. Nov 3, 2005

Well.. sadly, that notation, I would use in theory... but when doing calculations.. i hardly pay attention to it (and sadly, thats with a lot of other equations too!)

YES!!! what you said above is what im saying!!! and so would be [3 1; 2 4] etc.....

I just have no idea what the bais of [a+b b+c; c+d d+a] is.

i made it so [a b; c d] + [b c; d a] and finding the basis of those two is just easy.

its [1 0; 0 0] [0 1; 0 0] [0 0; 1 0] [0 0; 0 1] etc... but that dosnt help in finding the basis of the ortininal one equation.

So what is the basis of [a+b b+c; c+d d+a] ??????? and if i can find that, i found the basis of imT

17. Nov 3, 2005

Hurkyl

Staff Emeritus
I don't know if you're being sloppy or confused, so I'll correct this:

It doesn't make sense to ask if [a+b b+c; c+d d+a] has a basis: it is not a vector space, it is simply a matrix.

Now, the set of all matrices that can be written in that form, with a, b, c, d all being real numbers, is a vector space, and it does make sense to ask if it is a vector space.

Now, back to the question at hand...

I agree, the matrix [3 1;2 4] is indeed an element of the image of T. (e.g. take a = 2, b = 1, c = 0, d = 2)

Therefore, the set {[3 1;2 4]} is a linearly independent subset of the image of T with one vector in it.

Since you know a basis for im T must have three vectors in it, we know that the set {[3 1;2 4]} is one-third of a basis for T, right?

(This is what I meant by random guessing: just keep picking vectors in im T until you get a basis)

18. Nov 3, 2005

Yes, im following.

any M22 matrix is in the image if it sasisfies {[a+b b+c; c+d d+a]/ [a b; c d] is in M22} That is a FACT!

OKAY... I see how its working out now. lets say that we have [5 3 2 4], then (a=2 b=3 c=0 d=2)

so far, we have a bais { [1 1; 1 0] [ ] [ ] }

I suppose the other to parts of the basis is [0 1; 1 1] and [1 0; 1 1]?

Basis is { [1 1; 1 0] [0 1; 1 1] 1 0; 1 1] }

Am i getting warmer? Im thinking right now... what about [1 1; 0 1]? why won't that fit it? (other than the fact that i know dim imT = 3 .... ?? It won't fit in because we can only have a max or 3 vectors since exactly 1 of a,b,c,d has to = 0?

Last edited: Nov 3, 2005
19. Nov 3, 2005

Hurkyl

Staff Emeritus
I can't figure out where you're going wrong...

[1 1; 1 0] is not an element of im T. Why do you think that it is?

20. Nov 3, 2005