# Orthogonal and Diagonal Matrices

1. Sep 10, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
Find all 2 x 2 and 3 x 3 orthogonal matrices which are diagonal. Construct an example of a 3 x 3 orthogonal matrix which is not diagonal.

2. Relevant equations
Diagonal Matrix = All components are 0 except for the diagonal, for a 2x2 matrix, this would mean components a and d may or may not be 0, while b and c must be 0.
Orthogonal IF: A^-1=A^T

3. The attempt at a solution
So, if I have a diagonal 2x2 matrix, then it is automatically considered orthogonal for all values of a and d.

I cannot think of a possibility where this doesn't hold.

for a 3x3 matrix, this problem seems substantially more difficult as I have to calculate the inverse of the 3x3 diagonal matrix. The transpose should be easy to calculate. It is simply the same matrix.
So, if the transpose of this 3x3 matrix has to equal the inverse, we don't need to do the inverse calculation.

We can say all 3x3 matrices that are diagonal are also orthogonal.

Now to construct an example of a 3x3 orthogonal matrix which is not diagonal, is a bit more difficult. I see that it has to be over the field of real numbers, but not getting further on this yet.

2. Sep 10, 2015

### RUber

Are you certain of this definition? I have seen some definitions that simply use the idea that the dot product of rows or columns is zero.
Using this definition...Is there not some magnitude requirement so that A inverse can be equal to A transpose?
i.e.:
$\begin{pmatrix} a& 0 &0 \\0& b &0 \\0& 0 &c \end{pmatrix}^T = ?, \quad \begin{pmatrix} a& 0 &0 \\0& b &0 \\0& 0 &c \end{pmatrix}^{-1} = ?$

For the construction, you can either use a physical example, like a rotated and tiled x,y,z axis, or just pick any old vector, and construct a second and third vector to complete the orthogonal set. Gram-Schmitt can be useful for this.

3. Sep 10, 2015

### vela

Staff Emeritus
I can. For example, the transpose of the matrix
$$A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$$ is not the inverse of A.

4. Sep 11, 2015

### RJLiberator

Ah, I had made a foolish mistake in my original calculation of the inverse of a diagonal matrix. The result is actually [ 1/a, 0, 0, 1/d ]

Would this mean a*d must equal 1 for a matrix to be orthogonal and diagonal?

5. Sep 11, 2015

### RJLiberator

After taking the 3x3 inverse of a diagonal matrix, I see that it has a similar condition to a 2x2 matrix in this situation.

The transpose of a diagonal matrix is just itself.
The inverse of the matrix here, is the reciprocal of itself.

6. Sep 11, 2015

### RJLiberator

Since I have come to this conclusion,

It seems the only orthogonal and Diagonal matrixes are the inverse matrices since a=1/a b=1/b c =1/c and so on.

Is this the conclusion to the first part of this problem stating "find all possible orthogonal and diagonal 2x2 3x3 matrices?"

7. Sep 11, 2015

### RUber

Looks right to me. So, a combination of ones and negative ones.
Any luck on the second part?

8. Sep 11, 2015

### RJLiberator

It's a bit more difficult.

I see what you stated above, but I am unaware of Gram-schmitt for this problem.

A rotated X,Y,Z axis seems like a great idea, one that I should explore, but not quite sure how to start it at this moment.

9. Sep 11, 2015

### RUber

Maybe a more general approach is called for.
You have the constraints that $A^TA = I$
Which can be seen by terms as:
$\begin{pmatrix} a&b&c\\d&e&f \\g&h&i \end{pmatrix}\begin{pmatrix} a&d&g\\b&e&h \\c&f&i \end{pmatrix} = \begin{pmatrix} 1&0&0\\0&1&0 \\0&0&1 \end{pmatrix}$
Multiplying this out, you get a set of 6 constraints for 9 variables, meaning that you have 3 free variables. So...if you pick a number between zero and one for a, d, and g, you should be able to solve for the other equations by solving the system.

10. Sep 11, 2015

### RJLiberator

I spoke with my linear algebra teacher for this problem.

He suggested the possibility of matrix

\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{bmatrix}

Which has the rotation matrix in the bottom right 2x2 section, by 90 degrees.

The Transpose of this matrix does equal the inverse of the matrix which concludes that it is orthogonal but not diagonal.

So this is an example constructed, which satisfies the problem requirement. Permitting I have a good understanding of diagonal and orthogonal matrices in my first part here, I've successfully completed this assignment.

I would like to explore your route here in your last post, as it seems more general and fundamental then me winging an example up.

Last edited: Sep 11, 2015
11. Sep 11, 2015

### RUber

Granted, that is not diagonal...but it is clearly the same basic structure as the identity.

For a bit more of a challenge, you could rotate the system about the z-axis 45 degrees counter-clockwise...now your x-hat is pointing to $( \frac{\sqrt{2}}{2} \, \frac{\sqrt{2}}{2} \, 0)$ and your y-hat is pointing to $( -\frac{\sqrt{2}}{2} \, \frac{\sqrt{2}}{2} \, 0)$. Your z-axis is still the same and can be represented by (0,0,1).

Or more generally, say you have a unit vector ( a, b, c ) such that a^2 + b^2 + c^2 = 1.
You can choose ( d, e, f ) such that one of the terms is zero, say, d. Then you know that be = -cf, which means f = kb and e = -kc, where k is a scaling factor to get you to unit length. By definition, $k = \frac{1}{\sqrt{1-a^2}}.$
And so on and so forth to complete the matrix satisfying the constraints.

Last edited: Sep 11, 2015
12. Sep 11, 2015

### RJLiberator

Indeed, there is some good information for me to look at here.

I thank you for helping me through this problem and giving me some thoughts to work on as I continue this thought process.

For the most part, I do understand what you are saying here in the latest post, it is starting to click, I believe I just need more of a fundamental understanding of matrix operations and what's going on before I truly understand it.
That should come soon in my semester :).

Regards.