# Orthogonal matrix

## Homework Statement

My book used the term "orthogonal 3 by 3 matrix" and I couldn't find where that was defined. Does that mean that the column vectors are orthogonal or that the row vectors are orthogonal? Or are those two things equivalent?

## The Attempt at a Solution

It just means it's a real unitary matrix.

If A is a real matrix, then if A is orthogonal, the rows of A form an orthonormal set as WELL as the columns.

How do you prove that the rows (of a real n by n matrix) are orthogonal iff the columns are orthogonal?

What's the consequence of a real n by n matrix being orthogonal?

(A^T)A = I

What does this say about the columns?

What's the consequence of a real n by n matrix being orthogonal?

(A^T)A = I

What does this say about the columns?

Thats orthonormal not orthogonal.

Thats orthonormal not orthogonal.

Since it's an n x n matrix, it is a square matrix. It is easy to see that any square matrix with orthonormal columns is an orthogonal matrix. By definition of orthogonal matrix, it is a square "blank" matrix U such that $$U^{T}U=I$$. What is "blank" and why? Then, show that the rows of U form an orthonormal basis of $$R^n$$

Since it's an n x n matrix, it is a square matrix. It is easy to see that any square matrix with orthonormal columns is an orthogonal matrix. By definition of orthogonal matrix, it is a square "blank" matrix U such that $$U^{T}U=I$$. What is "blank" and why? Then, show that the rows of U form an orthonormal basis of $$R^n$$

But not every orthogonal matrix is orthonormal. Its not true that if U is orthogonal then U U^T = I. All we know is that U U^T is diagonal.

True, but U is a square matrix. Your original question was "orthogonal 3x3 matrix".

A square matrix is a orthogonal if and only if the columns form an orthonormal basis in $$R^n$$, thus it must also be true that the rows of U form an orthonormal basis of $$R^n$$ (isn't this what you're trying to prove?)

Theorem: U be a n x n matrix
1. U is orthogonal
2. The columns of Q are an orthonormal set of vectors in Rn
3. The rows of Q are an orthonormal set of vectors in Rn

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No, I think he's trying to prove that any square matrix with orthogonal columns must have orthogonal rows. (I'm not sure if this is true or not, but I haven't put too much thought into it.)

Theorem: U be a n x n matrix
1. U is orthogonal
2. The columns of Q are an orthonormal set of vectors in Rn
3. The rows of Q are an orthonormal set of vectors in Rn

That theorem is false.

Let n = 2 and U =

$$\left(\begin{array}{cc} 1 & 0 \\ 0 & 2 \\ \end{array}\right)$$
The second column and the second row are both not normalized.

No, I think he's trying to prove that any square matrix with orthogonal columns must have orthogonal rows. (I'm not sure if this is true or not, but I haven't put too much thought into it.)

Yes, that is what I am trying to prove.

Your matrix isn't orthogonal since it has a determinant of 2.

If A is orthogonal then AA^T = I

Lets have a real orthogonal 3x3 (square) matrix then with rows u1 = (a1,a2,a3), u2 = (b1,b2,b3), u3 = (c1,c2,c3)

solve for AA^T = I since it must be invertible.

You should end up finding that a1^+a2^2+a3^2 =1, b1^2+b2^2+b3^2 = 1 and c1^2+c2^2+c3^2 = 1.

All the other equations are equal to zero.

This shows that u1 dot u1 = 1, u2 dot u2 = 1, and u3 dot u3 = 1.

Thus AA^T = I shows that the rows of A form an orthonormal set of vectors and you just have to solve for A^TA = I since it is true for the columns.

http://en.wikipedia.org/wiki/Orthogonal_matrix

I am an idiot. In fact my book does define the term orthogonal and it is just like the Wikipedia page. I thought the definition was just that either the column vectors or the row vectors were pairwise orthogonal and I was trying to figure out which it was. Sorry.

Whoops, for the theorem I wrote, what I meant to say that "if one of the following is true, then the rest is true" for (1), (2), and (3)