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.
were pretty much giveaways that they were asking about eigenvectors!
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 \(\displaystyle \vec{u}= \vec{v}\).\)