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In this section we’re going to be talking about a special
kind of matrix called an orthogonal matrix.
This is also going to be a fairly short section (at least in relation to
many of the other sections in this chapter anyway) to close out the
chapter. We’ll start with the following
definition.
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Definition 1 Let
Q be a square matrix and suppose
that
then we call Q an orthogonal matrix.
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Notice that because we need to have an inverse for Q in order for it to be orthogonal we
are implicitly assuming that Q is a
square matrix here.
Before we see any examples of some orthogonal matrices (and
we have already seen at least one orthogonal matrix) let’s get a couple of
theorems out of the way.
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Theorem 1 Suppose that Q is a square matrix then Q is orthogonal if and only if .
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Proof : This is a
really simple proof that falls directly from the definition of what it means
for a matrix to be orthogonal.
In this direction we’ll assume that Q is orthogonal and so we know that ,
but this promptly tells us that,
In this direction we’ll assume that ,
since this is exactly what is needed to show that we have an inverse we can see
that and so Q
is orthogonal.
The next theorem gives us an easier check for a matrix being
orthogonal.
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Theorem 2 Suppose that Q is an matrix, then the following are all
equivalent.
(a) Q is orthogonal.
(b) The
columns of Q are an orthonormal set
of vectors in under the standard Euclidean inner product.
(c) The
rows of Q are an orthonormal set of
vectors in under the standard Euclidean inner product.
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Proof : We’ve
actually done most of this proof already.
Normally in this kind of theorem we’d prove a loop of equivalences such
as . However, in this case if we prove and we can get the above loop of equivalences by
default and it will be much easier to prove the two equivalences as we’ll see.
The equivalence is directly given by Theorem 2 from the previous
section since that theorem is in fact a more general version of this
equivalence.
The proof of the equivalence is nearly identical to the proof of Theorem 2
from the previous section and so we’ll leave it to you to fill in the details.
Since it is much easier to verify that the columns/rows of a
matrix or orthonormal than it is to check in general this theorem will be useful for
identifying orthogonal matrices.
As noted above, in order for a matrix to be an orthogonal
matrix it must be square. So a matrix
that is not square, but does have orthonormal columns will not be
orthogonal. Also, note that we did mean
to say that the columns are orthonormal.
This may seem odd given that we call the matrix “orthogonal” when
“orthonormal” would probably be a better name for the matrix, but traditionally
this kind of matrix has been called orthogonal and so we’ll keep up with
tradition.
In the previous section we were finding QR-Decompositions and if you recall the matrix Q had columns that were a set of orthonormal vectors and so if Q is a square matrix then it will also
be an orthogonal matrix, while if it isn’t square then it won’t be an
orthogonal matrix.
At this point we should probably do an example or two.
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Example 1 Here
are the QR-Decompositions that we
performed in the previous section.
From Example 1
From Example 2
In the first
case the matrix Q is,
and by
construction this matrix has orthonormal columns and since it is a square
matrix it is an orthogonal matrix.
In the second
case the matrix Q is,
Again, by construction this matrix has orthonormal
columns. However, since it is not a
square matrix it is NOT an orthogonal matrix.
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Example 2 Find
value(s) for a, b, and c for which the following matrix will be orthogonal.
Solution
So, the columns
of Q are,
We will leave
it to you to verify that ,
and and so all we need to do if find a, b,
and c for which we will have ,
and .
Let’s start
with the two dot products and see what we get.
From the first
dot product we can see that, . Plugging this into the second dot product
gives us, . Using the fact that we now know what c is in terms of a and plugging this into we can see that .
Now, using the
above work we now know that in order for the third column to be orthogonal
(since we haven’t even touched orthonormal yet) it must be in the form,
Finally, we
need to make sure that then third column has norm of 1. In other words we need to require that ,
or we can require that since we know that the norm must be a
positive quantity here. So, let’s
compute ,
set it equal to one and see what we get.
This gives us
two possible values of a that we
can use and this in turn means that we could used either of the following two
vectors for
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A natural question is why do we care about orthogonal
matrices? The following theorem gives
some very nice properties of orthogonal matrices.
Proof : We’ll
prove this set of statements in the order :