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So my professor gave me this question:

I have to find the basis of the eigenvalues of this matrix

\begin{pmatrix} \cos(q) & \sin(q)\\ \sin(q) & -\cos(q)\\ \end{pmatrix}

so I calculate the eigenvalues and I found it is 1 and -1. so now I need to find the basis of the kernel of those matrices

\begin{pmatrix} 1-\cos(q) & -\sin(q)\\ -\sin(q) & 1+\cos(q)\\ \end{pmatrix}

\begin{pmatrix} -1-\cos(q) & -\sin(q)\\ -\sin(q) & -1+\cos(q)\\ \end{pmatrix}

so actually I need to find

\begin{pmatrix} a \\ b\\ \end{pmatrix}

that will be function as basis for the kernel of each one of those matrices.

but how do I do it ?

I know how to do it when there is no angles meaning I would just compare it to zero.

for example in the first matrix I get these two equations :

$x-xcos(q)-ysin(q)=0$

$-xsin(q)+y+ycos(q)=0$

so I get this equation:

$x(-sin(y)+((1-cos(y))/(sin(y)))+((cos(y)-cos(y)cos(y))/(sin(y))))=0$

which is true for every $x$ and then I can assign some value to $x$ for example zero and then $y$ equal zero also but after I am checking it, it is not true. could u please help me ? so how can I do it ?

All he told us is that $q\neq{0}$ and $q\neq\pi$ and $q\neq{2}\pi $

and nothing more.

Widawensen
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wantToLearn
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  • Here $q$ is fixed, because it determines the matrix. So keep the $q$ and look for non-trivial vectors $(x,y)$ such that $$\pmatrix{1-\cos q&-\sin q\cr-\sin q&1+\cos q\cr}\pmatrix{x\cr y\cr}=\pmatrix{0\cr0\cr}.$$ Same for the other matrix. But you can save yourself some trouble because the matrix is symmetric, so the eigenvectors belonging to different eigenvalues are orthogonal. – Jyrki Lahtonen May 18 '13 at 17:10
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    In other words, I don't quite understand, how $q$ disappeared from your calculations and $\sin y$, $\cos y$ appeared in place of $\sin q$ and $\cos q$. Where you expecting the same eigenvectors to work for all values of $q$? That is not the case, because this matrix corresponds to a reflection about line making an angle $q/2$ with the $x$-axis. BTW, that is also a BIG hint for finding the eigenvectors! – Jyrki Lahtonen May 18 '13 at 17:14
  • so according to what u said. I assign q to be pi/2 when I calculate the kernel I get the basis for this kernel is x=y=1. but after caluculating again the multiplication of this (x,y) with the matrix I do not get (0,0) so I do not understand yet how to do it. – wantToLearn May 18 '13 at 17:20
  • If $q=\pi/2$, then the matrix $$\pmatrix{1-\cos q&-\sin q\cr-\sin q&1+\cos q\cr}= \pmatrix{1&-1\cr-1&1\cr}.$$ Then $$\pmatrix{1&-1\cr-1&1\cr}\pmatrix{1\cr1\cr}=\pmatrix{0\cr0\cr}.$$ Something went wrong in your calculation? – Jyrki Lahtonen May 18 '13 at 17:23
  • This I know. but q can be any value greater than zero and smaller than pi. so I have to find something that is true for all q – wantToLearn May 18 '13 at 17:29
  • No. The eigenvectors depend on $q$. This is different from the case of rotation matrices, when the same (complex) eigenvectors work for all the rotations. In this case the linear transformations are reflections, and they have real eigenvectors. Obviously different reflections have different eigenvectors, because a vector along the mirror is its own mirror image (= an eigenvector for $\lambda=1$), but a vector pointing orthogonally out of the mirror is the reverse of its mirror image (=an eigenvector for $\lambda=-1$). Different mirrors - different eigenvectors. – Jyrki Lahtonen May 18 '13 at 17:35
  • @Jyrki Lahtonen so what are u exactly say ? that I can not find a bases that do not depend on the angles ? – wantToLearn May 18 '13 at 17:43
  • Yes. I am fairly sure that the question asks you to find a basis of eigenvectors for all the matrices. But a different basis for different matrices. – Jyrki Lahtonen May 18 '13 at 17:46
  • @wantToLearn do u by any chance study in the Hebrew university ? :) – vondip May 20 '13 at 10:47

1 Answers1

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Observe that $\cos(q-\pi/2)=\sin q$, and $\sin(q-\pi/2)=-\cos q$. So, the linear transformation corresponding to the matrix sends ${\bf i}\,(=\pmatrix{1\\0})$ to $\bf u$ in the unit circle at angle $q$ (the first column) and sends ${\bf j}\,(=\pmatrix{0\\1})$ to $\bf v$ in the unit circle at angle $q-\pi/2$.

If you draw it, it could be clear that the reflection through the line $e$ at angle $q/2$ also does the same to the basis vectors $\bf i$ and $\bf j$. As both are linear and coincide on a basis, we get that this is the matrix of reflection through $e$. From this, we easily get the eigenvectors: vectors parallel to $e$ will be fixed (i.e. they correspond to eigenvalue $1$) and vectors orthogonal to $e$ will be reversed (i.e. they correspond to eigenvalue $-1$).

Berci
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  • Can u be more clarify your answer please. I do not understand how this facts can help me solved that. I mean all I need to do is to find two basas for the two matrices. How do I do it ? what is the process of doing it ? – wantToLearn May 18 '13 at 17:38
  • Well, this is a special case when we could explicitly tell the geometric transformation behind the matrix, and then argue according to that. A basis of eigenvectors is $(\pmatrix{\cos(q/2)\ \sin(q/2)},\ \pmatrix{-\sin(q/2)\ \cos(q/2)})$. – Berci May 18 '13 at 17:54
  • First of all I want to thank you. Second I truly saw that this is the eigenvectors of those matrices. but I tried to calculate it myself on a draft and I achieved those equation : x-xcos(q)=ysin(q) && -xsin(q)+y+ycos(q)=0. but those two equation t=does not compute anything because I get x=x((sin(q)(sin(q))+((cos(q)(cos(q))). which mean 0=0. so what am I doing wrong. I will be grateful. – wantToLearn May 19 '13 at 17:26
  • I ment to you regarding the last comment – wantToLearn May 19 '13 at 17:52