How would I prove this is linear Veeky Forums?
Prove its linear?
Justin Sanchez
Lucas Thompson
Easy, just write down the formula you noob
Elijah Brown
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
Lucas Price
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 Veeky Forums...
Easton Smith
>How would I prove this is linear Veeky Forums?
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.
Chase Ward
(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.
Colton Allen
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.
Nathan Reed
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...
Kevin Hughes
>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)
>Does it want me to substitute it in?
Your transformation is "plug these numbers into the matrix, then find the matrix's determinant"
Leo Flores
*the
Julian Miller
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.
Cameron Reyes
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
Connor Adams
>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.
Brody Bailey
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.
Camden Martinez
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]
Christopher Allen
didn't OPs question ask to show that it is linear though?
Jeremiah Morris
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.
Brayden Powell
It's not.
To be linear is has to satisfy the two properties in .
That transformation satisfies neither.
Leo Cook
Right so I just watched this video
youtube.com
So I only do the determinant for the top row? Never mind only just reading 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?
Alexander Bailey
>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?
Parker Hill
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]
Alexander Carter
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.
>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}
Nathaniel Long
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
I don't see an issue, it was amended here It's literally just going to be -10x -10y + 10z
What's the issue?
Matthew Rodriguez
What said. Need to show that your transformation satisfies these properties.
Use property 3a and 3b from here.
Kayden Taylor
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.
Michael Perry
Not even gonna try fixing that latex.
When I was explaining determinants I left that part out It's an easy fix in the problem, but all my explanations got messed up, too.
Kevin Carter
*for each u in the subset
I'm badly in need of sleep. G'night.
Austin Kelly
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 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?
Nolan Wright
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.
Benjamin Morgan
Using the properties from or as stated by Don't need to evaluate the determinant. Just use its properties.
>Malfunctioning keyboard stopping Latex usage.
Aiden Moore
> I just have to do what 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 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?
Luke Cook
>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
Christopher Davis
>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?
Connor James
Fine.
Logan Peterson
So this is literally the answer?
Right, I'm going to read through this and hope I can memorize the theory for future use.
Carter Jackson
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.
Brayden Peterson
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.
en.wikipedia.org
Which when applied to your problem will be of the form mentioned by Use properties 3a and 3b as given here.
I've combined both to give you the solution in Good luck !
Michael Bailey
Thanks!!!
Nolan Clark
No, expanding the determinant is an ugly way of doing it and it doesn't generalize at all. Objectively a bad proof.
Dylan King
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.
Isaiah Davis
What are you talking about? That wasn't amended at because that guy said everything is mapped to 10 and x,y,z are irrelevant lol.