Finding Orthogonal Matrices: 2 Solutions and Help

[Poetria]

Homework Statement

Find four distinct orthogonal 2x2 matrices, each of which has top-left entry equal to $-\frac {1} {\sqrt{2}}$

Relevant Equations

Definition of an orthogonal matrix:
$M^T=M^{-1}$

I have found two such matrices:

$\begin{pmatrix} -cos( \frac {\pi} {4}) & sin(\frac {\pi} {4})\\ sin(\frac {\pi} {4}) & cos(\frac {\pi} {4})\end{pmatrix}$$\begin{pmatrix} -cos( \frac {\pi} {4}) & -sin(\frac {\pi} {4})\\ -sin(\frac {\pi} {4}) & cos(\frac {\pi} {4})\end{pmatrix}$

Any hint how to find the other two? I have tried several solutions but all of them were wrong.

 

[BvU]

I have found two such matrices

How ?

Any hint how to find the other two?

Do the explicit matrix multiplication and solve the equations ...

$\$

 

[Poetria]

Oh, ok, I will try. Many thanks. :)

The first matrix is a rotational one: it's easy if you know its properties. The determinant is -1.
Then I have changed the signs of both sines and verified if the second matrix is orthogonal.

How ?

Do the explicit matrix multiplication and solve the equations ...

$\$

 

[BvU]

The first matrix is a rotational one

You sure ? I thought those have determinant 1 ...

$\$

 

[Poetria]

You sure ? I thought those have determinant 1 ...

$\$

Of course, I am wrong. I confused it with this one:

$\begin{pmatrix} cos(\pi/4) & -sin(\pi/4)\\ sin(\pi/4) & cos(\pi/4)\\\end{pmatrix}$

Anyway a rotational matrix was my point of departure. But I needed a negative cos(pi/4) in the top left corner.

 

[PeroK]

Any hint how to find the other two? I have tried several solutions but all of them were wrong.

I think you need to reflect a little more.

 

[Poetria]

I think you need to reflect a little more.

I think so. I will get to the bottom of it sooner or later. :)

 

[Poetria]

Well, I have studied the definition of the orthogonal matrix:
https://mathworld.wolfram.com/OrthogonalMatrix.html
"The rows of an orthogonal matrix are an orthonormal basis. That is, each row has length one, and are mutually perpendicular."

There is a constraint: -1/sqrt(2) in the left top corner of the matrix.
So I have
$\sqrt {(-\frac {1} {\sqrt(2)})^{2}+x^{2}} =1$
$x= \frac {1} {\sqrt(2)}$ v $x= -\frac {1} {\sqrt(2)}$

There are two possibilities for the first row:
$\begin{pmatrix} -\frac {1} {\sqrt(2)} & \frac {1} {\sqrt(2)}\end{pmatrix}$
$\begin{pmatrix} -\frac {1} {\sqrt(2)} & -\frac {1} {\sqrt(2)}\end{pmatrix}$

The dot product of the first and second row has to be equal to zero.

vector (-1/sqrt(2), 1/sqrt(2)) dot vector (y, z) = 0

$\frac {z} {\sqrt(2)} - \frac {y} {\sqrt(2)}=0$

Therefore y=z.
But the length of the second row must be 1.
Therefore $y=\frac {1} {\sqrt(2)}$ v $y=-\frac {1} {\sqrt(2)}$

The only thing I could do to obtain other matrices is change $\theta$, e.g. instead of $\sin(\frac{\pi} {4})$ give $\sin(\frac{9*\pi} {4})$ but it's the same numerically.

 

[PeroK]

Not all orthogonal matrices are rotations. In fact, only two are rotations. The others involve rotations and a reflection.

Or, you can simply work through the options for $\pm \frac 1 {\sqrt 2}$ in each entry. That gives 8 possibilities with minus in the first entry to check.

 

[Poetria]

Wow. :) I got it. :) Wonderful. Many thanks :)

 

[PeroK]

Wow. :) I got it. :) Wonderful. Many thanks :)

A bit of quiet reflection was all it needed!