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

[rad0786]

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

 

[Hurkyl]

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?

 

[rad0786]

in honesty, i have no idea what that last = sign means, the book just puts it their on other examples. I am 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 :(

 

[Hurkyl]

I meant guess a basis!

 

[rad0786]

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

 

[Hurkyl]

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?

 

[rad0786]

Those questions are making this harder :(

They are independint, but NO, they are not an image. because 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] ?

 

[Hurkyl]

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?

 

[rad0786]

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

 

[Hurkyl]

Can you find any vectors in I am T?

 

[rad0786]

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

 

[Hurkyl]

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.

 

[rad0786]

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?

 

[Hurkyl]

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.

 

[Hurkyl]

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

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

 

[rad0786]

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

YES! what you said above is what I am 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 doesn't 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

 

[Hurkyl]

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

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 I am 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 I am T until you get a basis)

 

[rad0786]

Yes, I am 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. let's 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? I am 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?

 

[Hurkyl]

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

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

 

[rad0786]

okay listen... say we have a space with M22 [a a; b b]/a b are real numbers.

then basis is { [1 1; 0 0] [0 0 1 1] }

Im trying to apply that to this case.

And since you showed that {matrix [3 1;2 4] is i an element of the image of T. (e.g. take a = 2, b = 1, c = 0, d = 2) } I see that {1 1; 0 1} fits as part of the basis?

hmm.. i thought the same thing for [1 1; 1 0]

 

[Hurkyl]

okay listen... say we have a space with M22 [a a; b b]/a b are real numbers.

then basis is { [1 1; 0 0] [0 0 1 1] }

Yes, that is a basis. A reason it's a basis is because you can prove:

(1) Those two vectors are linearly independent.
(2) Those two vectors span your space.

I see that {1 1; 0 1} fits as part of the basis?

I can't figure out how you're coming up with that matrix. It certainly cannot be an element of I am T (and thus cannot be part of a basis for I am T) because the equation [1 1; 0 1] = [a+b b+c; c+d d+a] does not have a solution with [a b; c d] in M22.


Aside: you do realize that {<43, 16>, <3.892, -167>} is a basis for R², right? (I'm just guessing where you're going wrong, and hoping this hits the mark)

 

[Hurkyl]

By the way, I'm very tired, so I'm off to bed. Sorry I couldn't stay up until you got through this!

 

[rad0786]

im getting [1 1; 0 1] from (a=1 b=1 c=0 d=1) Perhaps i went competely wrong?

and yes the two vectors <2, 1> and <4.7, 93.856> form a basis for R² because a) they are linearly independet (not multipes of each other also) and b) they both span R2

 

[rad0786]

Its okay hurkly! thanks so much and I really really appreciate how much you did for me so far!

Thank you!

 

[Hurkyl]

im getting [1 1; 0 1] from (a=1 b=1 c=0 d=1) Perhaps i went competely wrong?

Let's go back to the set builder notation:

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

If you choose a=1, b=1, c=0, and d=1, we can verify that [1 1; 0 1] is indeed in M22, so this tells us that [1+1 1+0; 0+1 1+1] = [2 1; 1 2] is an element of I am T.

(More intuitively, you specified an element A = [1 1; 0 1] of M22, therefore TA is in im(T))

 

[rad0786]

Hey! okay, i just read over the entire posts from the beigging to try to make it out :)

so if we have:

[3 1;2 4] Thats in the image of T since (a = 2, b = 1, c = 0, d = 2)
[0 1;2 1] Thats in the image of T since (a = 0, b = 0, c = 1, d = 1)
[0 0;1 1] Thats in the image of T since (a = 0, b = 0, c = 0, d = 1)

[3 1;2 4] [0 1;2 1] [0 0;1 1] are linearly independent vectors and are in the image of T

so a possible basis is {[3 1; 2 4], [0 1; 2 1]; [0 0; 1 1]}

 

[Hurkyl]

That looks good. Assuming you have a proof that they're linearly independent, then adding in the fact you proved that dim I am T = 3, you know you have a basis!

 

[rad0786]

Finally!

Hurkyl, thanks so much for putting up with me, i really appreciate it! Linear Algebra is not an easily learned subject for me (and a lot of other people!) its so confusing :(

Thanks again for all that! :)