prove its linear?

Easy, just write down the formula you noob

>>8291045
What's the formula? I don't think I'll be covering this stuff in lecture until mid September so I don't really have any material on it, but all the stuff on differentiation for this first topic I've covered so I'm moving ahead

My book and khan academy doesn't give me shit

>>8291047
Bump for late night help? There's really no one else I can ask and I can't sleep right now... I-I love you /sci/...

>How would I prove this is linear /sci/?
Show that [math]T(x_1,y_1,z_1) + T(x_2,y_2,z_2) = T(x_1+x_2,y_1+y_2,z_1+z_2)[/math] and [math]cT(x,y,z) = T(cx,cy,cz)[/math]

Goddamn, that's an awkward transformation.
OK, so we're taking the determinant of... that.

>>8291140 (cont.)
...which I see now will just be some constant.

And it's pretty easy to show that mapping to a constant doesn't hold for either property.
Non-linear.

Literally just write it out, look up the formula for determinant of a 3x3 matrix if you have to.


This is basically the cross product between vectors (1,2,3) and (-2,6,4) except the x,y,z are just numbers instead of unit vectors. That cross product was linear-hence why it was called a product--try and do the same for this.

>>8291162
>>8291151
>>8291140
Oh OK is see.

So the next part of the question says:

"Calculate T(-2,1,5)"

Does it want me to substitute it in? If so where do I put the numbers? Its only worth 2 marks as opposed to the 4 from the other question so it can't be that difficult...

>>8291047
>What's the formula? I don't think I'll be covering this stuff in lecture until mid September so I don't really have any material on it
So, there's two different formulas, and they're both recursive shit sprawls.

The determinant of an nxn matrix A is:
[math]det(A)=\sum_{i=0}^{n}(-1)^{i+j}det(M_{i,j})[/math]
Where [math]0 < j \leq n[/math], j is constant, and [math]M_{i,j}[/math] is then (n-1)x(n-1) matrix formed by deleting the i-th row and j-th column of A.
... [math]det(A)=\sum_{j=0}^{n}(-1)^{i+j}det(M_{i,j})[/math] with constant i works, too

The determinant of a 2x2 matrix is (upper left * lower right) - (upper right * lower left)

>>8291166
>Does it want me to substitute it in?
Your transformation is "plug these numbers into the matrix, then find the matrix's determinant"

*the

>>8291166
If you're not familiar with a determinant, it would help to learn that first. Here x, y and z are the three top numbers, so yeah to calculate T(-2,1,5) put them in the top row and work it out.

To make your life easier you could expand the determinant leaving x, y and z there, and get a formula that is easier to deal with.

So would that be the formula that goes x-x1 y-y1 z-z1 and then x2 -x1 .... z2 -z1 and then x3 - x1 ... z3 - z1?

It looks like that format but I'm not too sure and i dont want to make a stupid mistake over it

>>8291177
>you could expand the determinant leaving x, y and z there
This will work *really* nicely because x, y, and z all share a row.
If you solve the determinant by stepping through j with i=1, you'll just never touch x,y, or z and end up with a constant.
Any i or j will find the same determinant (x,y, and z will cancel to 0 no matter what), but i=1 is the easiest choice.

Ever so slightly just going to hijack this thread real quick as it is somewhat related to my question I had on my mind while I was lurking through here.

Lets say I have a linear map T: V --> W

How would I define the kernel of T and prove it is a subspace of V?

I figured this would be a decent opportunity to ask as this seems to be a vector related thread.

File: EPSON001.jpg (212KB, 2187x1687px) Image search: [Google]

212KB, 2187x1687px

>>8291183
T is a non-linear transformation that takes a vector in [math]\mathbb{R}^3[/math] and maps it to -10 in [math]\mathbb{R}[/math]

>>8291199
didn't OPs question ask to show that it is linear though?

>>8291199
No, wait. Shit. Screwed up my basic math.
That final bit should be
[math](-10)-(10)+(10)=10[/math]
so T maps everything to 10.

>>8291201
It's not.

To be linear is has to satisfy the two properties in >>8291140.
That transformation satisfies neither.

Right so I just watched this video
https://www.youtube.com/watch?v=21LWuY8i6Hw

So I only do the determinant for the top row? Never mind only just reading >>8291199 informative post.

Thanks for that.

So if it's non linear does that mean the question is wrong

It says "Prove it is linear" is this just one of those questions where you have to say its not linear?

>>8291207
>Question says "Prove it is linear"
>Not "Prove whether this is a linear transformation or non linear"

Of course it's linear it has to be according to the question.

How can you ask a student to prove something that is incorrect?

>>8291199
What the fuck?

[math]\begin{vmatrix}x & y & z \\ 1 & 2 & 3 \\ -2 & 6 & 4\end{vmatrix}=x\begin{vmatrix} 2 & 3 \\ 6 & 4\end{vmatrix}-y\begin{vmatrix} 1 & 3 \\ -2 & 4\end{vmatrix}+z\begin{vmatrix} 1 & 2 \\ -2 & 6 \end{vmatrix}[/math]

>>8291211
Oh. Right. Sssssssssshit.
I forgot that you multiply each chunk by the element indexed at i,j
I probably should have looked that formula up before ranting for so long.

>>8291190
>define the kernel of T
The kernel of T is the set of inputs that are mapped to [math]\vec{0}[/math] by a map.
The nullspace of A is the set of inputs that are mapped to [math]\vec{0}[/math] by a matrix.
Linear maps are of the form [math]T:A\vec{x}[/math], so the kernel of T is the nullspace of A.

>prove it is a subspace of V
A subspace is a subset of a vector space that satisfies three conditions:
>contains [math]\vec{0}[/math]
Which it does, [math]A\vec{0}=\vec{0}[/math]
>[math](\vec{u}+\vec{v}) is in the subspace for any \vec{u} and \vec{v} in the subspace[/math]
[math]A(\vec{u}+\vec{v})=A\vec{u}+A\vec{v}=\vec{0}+\vec{0}=\vec{0}[/math]
i.e. [math]A(\vec{u}=\vec{v})=\vec{0}[/math]
>[math]A(c\vec{u})=\vec{0} for any \vec{0} in the subspace and for any constant c[/math]
[math]A(c\vec{u})]c(A\vec{u})=c\vec{0}=\vec{0}

>>8291190

Let T : V→W be a linear map

Kernel: K = {u∈V | T(u) = 0} where 0 denotes the zero vector of W
In plain English:
the kernel of T is any vector in V that maps to 0 vector in W.

Clearly, K is a subset of V, since it consists only of vectors in V. To show that any subset of a vector space is a subspace, we must show that this subset is closed under vector addition and scalar multiplication.

If u1, u2 ∈ K, then T(u1) = 0 and T(u2) = 0
Since T is linear, then T(a+b) = T(a) + T(b) for ANY vectors a, b in V (not just those in K). But in particular, for u1, u2 ∈ K (which must therefore be in V) we get:
T(u1+u2) = T(u1) + T(u2) = 0 + 0 = 0
So if u1, u2 ∈ K, then u1+u2 ∈ K
Therefore, K is closed under vector addition

If u ∈ K, then T(u) = 0
Since T is linear, then T(kv) = k T(v), where k is a constant, and v is ANY vector in V. But in particular, for u ∈ K (which must therefore be in V) we get:
T(ku) = k T(u) = k * 0 = 0
So if u ∈ K, then ku ∈ K

>>8291211
I don't see an issue, it was amended here >>8291202

It's literally just going to be -10x -10y + 10z

What's the issue?

What >>8291140 said. Need to show that your transformation satisfies these properties.

Use property 3a and 3b from here.

http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/least-squares-determinants-and-eigenvalues/properties-of-determinants/MIT18_06SCF11_Ses2.5sum.pdf

>>8291190
Don't worry about it, funnily enough after calculating T(-2,1,5) I have to write down the basis for the kernel of T and state its dimension.

File: whoo.png (20KB, 948x550px) Image search: [Google]

20KB, 948x550px

>>8291216
Not even gonna try fixing that latex.

>>8291222
When I was explaining determinants I left that part out >>8291172
It's an easy fix in the problem, but all my explanations got messed up, too.

>>8291233
*for each u in the subset

I'm badly in need of sleep. G'night.

>>8291228
>>8291140
Am I correct in saying that T(x1,y1,z1) = T(-10,-10,10)? And then for T(x2,y2,z2) I just have to do what >>8291199 did?

My only concern is say for example x2, its going to be (y*4)-(6*z) and I'm not sure how to tackle that?

>>8291166
You said it's a 4 mark question, there's no way you need all of what's going on in this thread for it.

File: Determinant.png (14KB, 771x502px) Image search: [Google]

14KB, 771x502px

Using the properties from >>8291228 or as stated by >>8291140

Don't need to evaluate the determinant. Just use its properties.

>Malfunctioning keyboard stopping Latex usage.

>>8291243> I just have to do what >>8291199 did?
They did stuff wrong.

There are three things getting added together.
Each of which should get multiplied by [math]a_{i,j}[/math] for their relevant i and j.
([math]a_{i,j}[/math] is the element of the matrix A that is index at row i, column j)

That's [math]a_{1,1}=x[/math], [math]a_{1,2}=y[/math], and [math]a_{1,3}=z[/math]

T(x1,y1,z1) ≠ T(-10,-10,10)
What they were saying is wrong, but that's not what they were saying.
They said >>8291199 >>8291202 T(x1,y1,z1) = 10
The correct solution is T(x1,y1,z1) = -10x1 -10y1 + 10z1

>x2, its going to be (y*4)-(6*z)
You... just... plug... (y*4)-(6*z)... in... for... x
How did you pass highschool algebra?

>>8291257
>x2, its going to be (y*4)-(6*z)
>You... just... plug... (y*4)-(6*z)... in... for... x

Isn't that literally what he just said?

"So for this, I just replace it with this"
"Hurr you just plug it in"

Yeah because that's totally not the same thing.

>The correct solution is T(x1,y1,z1) = -10x1 -10y1 + 10z1
Which is exactly what I explained to him here >>8291222

>>8291280
>My only concern is say for example x2, its going to be (y*4)-(6*z) and I'm not sure how to tackle that?
>and I'm not sure how to tackle that?

>>8291312
Fine.

>>8291256
So this is literally the answer?

Right, I'm going to read through this and hope I can memorize the theory for future use.

>>8291043
That is only a matter of expanding the determinant. In expanded form the three variables will appear in separate terms, and will each have degree one. That makes it a linear combination of x, y, and z. I get this:

T(x, y, z) = -10x - 10y + 10z

This is the way its done, everything else is wrong.

>>8291335

Yes.

** When proving the linearity of a map, it's important to mention the conditions necessary for linearity. So all you need to show is that the first two conditions are satisfied from here.

https://en.wikipedia.org/wiki/Linear_map

Which when applied to your problem will be of the form mentioned by >>8291140

Use properties 3a and 3b as given here.

http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/least-squares-determinants-and-eigenvalues/properties-of-determinants/MIT18_06SCF11_Ses2.5sum.pdf

I've combined both to give you the solution in >>8291256

Good luck !

>>8291373
Thanks!!!

>>8291348
No, expanding the determinant is an ugly way of doing it and it doesn't generalize at all. Objectively a bad proof.

>>8291211
Or
T(x,y,z)=x|2,3;6,4|+y|3,1;4,-2|+z|1,2;-2,6|
That always seemed much more natural to me than changing the sign, although it only works in R^3 I think.

>>8291222
What are you talking about? That wasn't amended at >>8291202
because that guy said everything is mapped to 10 and x,y,z are irrelevant lol.