Finding image of linear transformation (difficult)

[andrey21]

1. Find the image of the linear transformation whose matrix is given by:

1 2 5 2
4 -3 1 0
10 -13 -7 -4

Homework Equations

3. Tried numerous times but struggle to get anywhere

 

[micromass]

Take a basis of the domain, preferably (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1). Then the image of this basis will span the image of the linear transformation (but it won't be a basis in general). You can use this to find the image of this transformation...

 

[andrey21]

Thanks micromass I sort of understand what u are saying could u possibly start me off with the calcuation.

 

[micromass]

Take the basis element (1,0,0,0) and calculate it's image.
The image of a vector x under a matrix A is Ax. So it's just a matrix multiplication...

 

[andrey21]

So the taking the basis element (1,0,0,0):

T(1,0,0,0) = (1,4,10,0) ?

 

[micromass]

No, T(1,0,0,0) = (1,4,10). I'm not sure how you found that zero in the end...

 

[andrey21]

Yep sorry I don't know why i put the extra zero in so:

T(1,0,0,0) = (1,4,10)
T(0,1,0,0) = (2,-3,-13)
T(0,0,1,0) = (5,1,-7)
T(0,0,0,1) = (2,0,-4)
Correct?

 

[micromass]

Yes, correct

Now the set spanned by those four vectors will be your image.

 

[andrey21]

Ok sorry if seems dumb but the set spanned, what does this mean exactly?

 

[micromass]

The set spanned by some vectors is the set of all linear combinations of these vectors.

 

[andrey21]

Im really confused how would this calcultion start?

 

[andrey21]

Setting Ax = 0 and solving to find echelon form:

1 2 5 2
0 11 19 8
0 0 0 0

 

[micromass]

There are no calculations anymore. You already found the answer...

 

[andrey21]

So that is the image of the linear transform?

 

[micromass]

The answer is just that the image is the set of all linear combinations of (1,4,10), (2,-3,-13), (5,1,-7) and (2,0,-4). Explicitly:

Image=\{\alpha(1,4,10)+\beta(2,-3,-13)+\gamma(5,1,-7)+\delta(2,0,-4)~\vert~\alpha,\beta,\gamma,\delta\in \mathbb{R}\}

You can span the image with less then 4 vectors however, but this doesn't seem to be the question...

 

[andrey21]

Ah i see well the next part of the question goes on to say:

Give three vectprs which are in the image?

 

[micromass]

Thats not to difficult is it? You've found four vectors which is in the image, so you've got three vectors to...

 

[andrey21]

Yes I understand now so I can just take 3 of the 4 vectors used. The final part of the question says show explicity that image forms a subspace. So i should just use vectors and make sure satisfy the axioms?

 

[micromass]

Yeah, there are many ways to do this. But what you suggest seems good...

 

[andrey21]

Great Thanks for all ur help micromass :)

 

[andrey21]

Sorry micromass quick qustion as i have stated above i could use vectors to prove the image forms a subspace of R^3. How would I go about doing this, sub in vectors to the image?

 

[micromass]

You could prove this in general. Let T:V--> W be a linear transformation, then im(T) is a subspace of W.
You'll only need to prove that
1) 0 in im(T)
2) v+w in im(T) whenever v,w in im(T)
3) av in im(T) whenever a in R and w in im(T).

 

[andrey21]

Ah I see so:

1) 0 in image(T) would that just be (0,0,0) ?

 

[micromass]

Yes, I am sorry, with 0 I just meant (0,0,0)...

 

[andrey21]

Rite so showing (0,0,0 is in im(T) would just be a case of subbing that into the image? Say swappping (1,4,10) for (0,0,0) ?

 

[micromass]

No. (0,0,0) is in the image because T(0,0,0,0)=(0,0,0).

 

[andrey21]

Yep I understand now. With the addition axiom shall I just add (1,0,0,0) and (0,1,0,0) for example. Therefore giving me a vector (3,1,-3) ?

And finally with the scaler multiplication axiom:

2.(1,0,0,0) = 2.(1,4,10) = (2,8,20) ?

 

[andrey21]

Is this correct micromass?

 

[micromass]

Yes, I can see what you mean and it is correct.
However, I must also tell you to watch your notation: writing that 2.(1,0,0,0) = 2.(1,4,10) is not correct. You have to write 2.f(1,0,0,0) = 2.(1,4,10)...

 

[andrey21]

Oh why is the f there just out of curiosity?

 

[micromass]

Oh, I am sorry, that f had to be a T :shy:

 

[andrey21]

Haha I see now, well again thanks for all ur help:)