- Thread starter acemi123
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\(\displaystyle \vec{v}\) 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?

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

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The fact that choices (a) and (b) wereThe 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?

(a) \(\displaystyle \vec{w}\) is an

(b) \(\displaystyle \vec{w}\) is an

were pretty much giveaways that they were asking about 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 How did you understand w and v are eigenvectors?

and then the problem asks whether \(\displaystyle \vec{w}= (1, 1, 2)\) is an eigenvector with eigenvalue 6 or 7. Didn't

That is, \(\displaystyle f(\vec{w})\) is 7

Again, it is a matter of knowing theand endomorphism? is it a function which is injective?

Thank you

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 \(\displaystyle \vec{u}= \vec{v}\).\)