# How to solve linear mapping problem?

Joined
Apr 14, 2019
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Anyone can help?

#### HallsofIvy

##### Elite Member
Joined
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That's pretty straightforward, isn't it? I presume, since you are asked to do this problem, that you know what "eigenvectors" and "eigenvalues" are!

$$\displaystyle \vec{v}$$ is an "eigenvector" of f with "eigenvalue" $$\displaystyle \lambda$$ if and only if $$\displaystyle f(\vec{v})= \lambda\vec{v}$$.

So $$\displaystyle \vec{w}= (1, 1, 2)$$ is an eigenvector of f with eigenvalue $$\displaystyle \lambda$$ if and only if $$\displaystyle f(\vec{w})= f((1, 1, 2))= [(1, 1, 2).(1, 1, 2)](1, 1, 2)+ (1, 1, 2)= 6(1, 1, 2)+ (1, 1, 2)= (6, 6, 12)+ (1, 1, 2)= (7, 7, 14)= \lambda(1, 1, 2)= (\lambda, \lambda, 2\lambda)$$.

Does there exist such a number, $$\displaystyle \lambda$$, and, if so, what is it?

#### acemi123

##### New member
Joined
Apr 14, 2019
Messages
30
The explanation is satisfying. 7.(1,1,2)
I am in trouble with this notation.
How should I understand from this notation that eigenvector and eigenvalues are asked?
And How did you understand w and v are eigenvectors?
and endomorphism? is it a function which is injective?
Thank you

#### HallsofIvy

##### Elite Member
Joined
Jan 27, 2012
Messages
6,605
The explanation is satisfying. 7.(1,1,2)
I am in trouble with this notation.
How should I understand from this notation that eigenvector and eigenvalues are asked?
The fact that choices (a) and (b) were
(a) $$\displaystyle \vec{w}$$ is an eigenvector of eigenvalue 7.
(b) $$\displaystyle \vec{w}$$ is an eigenvector of eigenvalue 6.

And How did you understand w and v are eigenvectors?
I thought I'd answered that. f is give by $$\displaystyle f(\vec{v})= (\vec{v}\cdot\vec{w})\vec{w}+\vec{v}$$ where $$\displaystyle \vec{w}= (1, 1, 2)$$'

and then the problem asks whether $$\displaystyle \vec{w}= (1, 1, 2)$$ is an eigenvector with eigenvalue 6 or 7. Didn't that make you think "Oh 'eigenvector'! I should check the definition of 'eigenvector'!"? Replacing both $$\displaystyle \vec{v}$$ and $$\displaystyle \vec{w}$$ with (1, 1, 2) in the definition of f above (I did this calculation before) we get $$\displaystyle f((1, 1, 2))= ((1, 1, 2)\cdot(1, 1, 2))(1, 1, 2)+ (1, 1, 2)= (1+ 1+ 4)(1, 1, 2)+ (1, 1, 2)= 6(1, 1, 2)+ (1, 1, 2)= 7(1, 1, 2)$$.

That is, $$\displaystyle f(\vec{w})$$ is 7 times $$\displaystyle \vec{w}$$. That is precisely the definition of "eigenvector with eigenvalue 7"!

and endomorphism? is it a function which is injective?
Thank you
Again, it is a matter of knowing the definitions! (Learning definitions is about the simplest thing you can do in mathematics but so many students don't!)

A function is an "endomorphism" is simply a function from a mathematical object (here the vector space $$\displaystyle R^3$$) to itself that "preserves the operations". That is $$\displaystyle f(\vec{u}+\vec{v})= f(\vec{u})+ f(\vec{v})$$ and $$\displaystyle f(a\vec{v})=af(\vec{v})$$.

The definition of f is $$\displaystyle f(\vec{v})= [\vec{v}\cdot(1, 1, 2)](1, 1, 2)+ \vec{v}$$. If we write $$\displaystyle \vec{v}$$ as $$\displaystyle (v_x, v_y, v_z)$$ then $$\displaystyle f((v_x, v_y, v_z)= (v_x+ v_y+ 2v_z)(1, 1, 2)+ (v_x, v_y, v_z)= (2v_x+ v_y+ 2v_z, v_x+ 2v_y+ 2v_z, v_x+ v_y+ 3v_z)$$. And if $$\displaystyle \vec{u}$$ is $$\displaystyle (u_x, u_y, u_z)$$ then $$\displaystyle f((u_x, u_y, u_z)= (u_x+ u_y+ 2u_z)(1, 1, 2)+ (u_x, u_y, u_z)= (2u_x+ u_y+ 2u_z, u_x+ 2u_y+ 2u_z, u_x+ u_y+ 3u_z)$$ and $$\displaystyle f(\vec{u}+ \vec{v})=f((u_x+ v_x, u_y+ v_y, u_z+ v_z))= [(u_x+ v_x, u_y+ v_y, u_z+ v_z)\cdot(1, 1, 2)](1, 1 2)+ (u_x+ v_x, u_y+ v_y, u_z+ v_z)= ((u_x+ v_x)1+ (u_x+ v_x), (u_y+ v_y)1+ (u_x+ v_x), (u_z+ v_z)2+ (u_z+ v_z))= (2u_z+2v_x, 2u_x+ 2v_x, 3u_z+ 3v_z)$$.

Now, is that the same as $$\displaystyle f(\vec{u})+ f(\vec{v})= ([\vec{u}\cdot(1,1, 2)](1, 1, 2)+ \vec{u})+ ([\vec{v}\cdot(1, 1, 2)](1, 1, 2)+ \vec{v})= ([u_x+ u_y+ 2u_z](1, 1, 2)+ (u_x, u_y, u_z)+ [v_x+ v_y+ 2v_z](1, 1, 2)+ (v_x, v_y, v_z)= (2u_x+ u_y+ 2u_z, u_x+ 2u_y+ 3u_z)+ (2v_x+ v_y+ 2v_z, v_x+ 2v_y+ 2v_z, v_x+ v_y+ 3v_z)$$?

A straight forward, though tedious, calculation.

Similarly for $$\displaystyle f(a\vec{v})= af(\vec{v})$$.
$$\displaystyle a\vec{v}= (av_x, av_y, av_z)$$ so $$\displaystyle f(a\vec{v})= [(av_x, av_y, av_z)\cdot(1, 1, 2)](1, 1, 2)+ (av_x, av_y, av_z)= (av_x+ av_y+ 2av_z)(1, 1, 2)+ (av_x, av_y, av_z)= (2av_x+ av_y+ 2av_z, av_x+ 2av_z+ 2av_z, 2av_x+ 2av_u+ 5av_z)\(\displaystyle . Is that the same as \(\displaystyle af(\vec{v})= a([(v_x, v_y, v_z)\cdot(1, 1, 2)](1, 1, 2)+ (v_x, v_y, v_z)= a[(v_x+ v_y+ 2v_z)(1, 1, 2)+ (v_x, v_y, v_z)]= a[(2v_x+ v_y+2v_z, v_x+ 2v_y+ 2v_z, v_x+ v_y+ 5v_z)]$$?

And what about "injective". Did you look up the definition of "injective"? A function from one vector space to another is "injective" (also called "one to one") if, whenever $$\displaystyle f(\vec{u})= f(\vec{v})$$ then $$\displaystyle \vec{u}= \vec{v}$$.

Again, let $$\displaystyle \vec{u}= (u_x, u_y, u_z)$$ and $$\displaystyle \vec{v}= (v_x, v_y, v_z)$$.

Then $$\displaystyle f(\vec{u})= [(u_x, u_y, u_z)\cdot(1, 1, 2)](1, 1, 2)+ (u_x, u_y, u_z)= (2u_x+ u_y+2u_z, u_x+ 2u_y+2u_z, u_x+u_y+ 5u_z)$$ and $$\displaystyle f(\vec{v})= [(v_x, v_y, v_z)\cdot(1, 1, 2)](1, 1, 2)+ (v_x, v_y, v_z)= (2v_x+ v_y+2v_z, v_x+ 2v_y+2v_z, v_x+v_y+ 5v_z)$$.

So $$\displaystyle f(\vec{u})= f(\vec{v})$$ means that $$\displaystyle (2u_x+ u_y+2u_z, u_x+ 2u_y+2u_z, u_x+u_y+ 5u_z)= (2v_x+ v_y+2v_z, v_x+ 2v_y+2v_z, v_x+v_y+ 5v_z)$$ so that $$\displaystyle 2u_x+ u_y+ 2u_z= 2v_x+ v_y+ 2v_z$$, $$\displaystyle u_x+ 2u_y+ 2u_z= v_x+ 2v_y+ 2v_z$$, and $$\displaystyle u_x+ u_y+ 5u_z= v_x+ v_y+ 5v_z$$.

You can solve those three equations for $$\displaystyle u_x$$, $$\displaystyle u_Y$$, and $$\displaystyle u_z$$ in terms of $$\displaystyle v_x$$, $$\displaystyle v_y$$, and $$\displaystyle v_z$$ and so show that [vex]u_x= v_x\), $$\displaystyle u_y= v_y$$, and $$\displaystyle u_z= v_z$$ so that $$\displaystyle \vec{u}= \vec{v}$$.\)
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