5.5 Orthogonal Matrices

There are two criteria for a matrix to be orthogonal (see copy below).

I get the first point. But I don’t understand why the second point is required.

Why do we need the magnitude to be 1 when it’s a matrix?

In ex 5.10, Q7a we have an example where the third column is not unit magnitude. In this question, my mental image right now is that the ‘third column vector’ of the matrix stretches to [0, 0, 2] but maintains its orthogonality…

For example if we had two vectors:
i = [1, 0]
j = [0, 2]
dot(i, j) = 0
These are orthogonal even though the 2nd dimension stretches beyond the unit magnitude.

Many thanks in advance

A matrix is called orthogonal if it satisfies the following two criteria:

  1. All of its columns are pairwise orthogonal. That means that the dot product between any two columns is exactly 0.

  2. Each column i has ||Qi|| = 1, in other words, each column is unit magnitude. Remember that the magnitude of a vector (in this case, the column of a matrix) is computed as the dot product of that vector with itself.

(Page 124 & 125).

Hi Matthew. Good question. The confusion here comes from the label “orthogonal matrix.” You would think that orthogonal columns are the only necessary condition. It would be better if we called it an “orthonormal matrix” or “orthounit matrix” to imply that the columns are normalized to unit length. I don’t know the origins of the label “orthogonal matrix” but that’s become the standard terminology.

Anyway, the point is that we want Q’Q=I, in other words, the matrix times its transpose gives the identity matrix. That can only happen if the columns are (1) orthogonal and (2) unit-length. Without the second condition, we’d have Q’Q=D, in other words, a diagonal matrix. Which is also fine, of course, but Q’Q=I has some remarkable properties that are relevant for eigendecomposition, SVD, and geometric transformations. You’ll learn more about these topics later in the book, but if you like, you can skip forwards to skim through, e.g., chapter 15.8.

Thank you for the quick response Mike!

I see I just need to accept it as a definition and understand that there are some cool properties that will be explored in later chapters.

It’s clarified that my mental image was correct which is super helpful!