# calculate kernels of matrices with angles

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 those two equation :

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!=0 and q!=pi and q!=2pi

and nothing more.

-
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
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

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 $-1$).
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