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### Course: Linear algebra > Unit 3

Lesson 4: Orthonormal bases and the Gram-Schmidt process- Introduction to orthonormal bases
- Coordinates with respect to orthonormal bases
- Projections onto subspaces with orthonormal bases
- Finding projection onto subspace with orthonormal basis example
- Example using orthogonal change-of-basis matrix to find transformation matrix
- Orthogonal matrices preserve angles and lengths
- The Gram-Schmidt process
- Gram-Schmidt process example
- Gram-Schmidt example with 3 basis vectors

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# Orthogonal matrices preserve angles and lengths

Showing that orthogonal matrices preserve angles and lengths. Created by Sal Khan.

## Want to join the conversation?

- Is C inverse, or C transpose, also an orthonormal matrix? Thanks!(5 votes)
- Straightforward from the definition: a matrix is orthogonal iff tps(A) = inv(A). Now, tps(tps(A)) = A and tps(inv(A)) = inv(tps(A)). This proves the claim. You can also prove that orthogonal matrices are closed under multiplication (the multiplication of two orthogonal matrices is also orthogonal): tps(AB) = tps(B)tps(A)=inv(B)inv(A)=inv(AB). Hope this helps :)(3 votes)

- does adding two nxn orthogonal matrices result in an nxn orthogonal matrix(2 votes)
- Good question,

To answer a question like this you should first try some examples. The best examples are easy examples. So let's try some 1x1 matrices. There are only two orthogonal matrices given by (1) and (-1) so lets try adding (1) + (1)=(2). (2) is not orthogonal so we have found a counterexample!.

In general you will see that adding to orthogonal matrices you will never get another since if each column is a unit vector the sum of two unit vectors cannot be a unit vector.(6 votes)

- You show cos(teta) = cos (teta_t), but i think it does not prove teta = teta_t, because teta_t also can be 360-teta, i.e., teta = -teta_t. Am i wrong or it does not matter?(3 votes)
- At4:48, in order to prove that the length of a vector does not change when represented with respect to orthogonal basis, shouldn't we start by finding length of C^-1*X rather than C*X because X represented in basis B is equal to C^-1*X ?(2 votes)
- He's using C here to represent any transformation with matrices whose columns are orthonormal basis vectors. Its not important here that it can transform from some basis B to standard basis.

We know that the matrix C that transforms from an orthonormal non standard basis B to standard coordinates is orthonormal, because its column vectors are the vectors of B. But since C^-1 = C^t, we don't yet know if C^-1 is orthonormal. All we know is that its r o w vectors are an orthonormal set.(1 vote)

- wait so x.y = x(t)y? i thought this was true iff x == y?

||cx|| = cx.cx = cx(t)cx using this logic i am.(1 vote) - At4:03Sal mentions that orthonormal matrices "rotate [vectors] around." In the last video we saw an example of an orthonormal matrix reflecting vectors. Are there any other geometric transformations that an orthonormal matrix could produce?(1 vote)
- No, those are the only two (though there can be combinations of rotation and reflection). Orthonormal matrices are special in that they preserve the relative orientation and size of vectors, so they will never scale a vector.(1 vote)

- Previous videos are referred to, but I am unaware of the order of the videos. If I knew what video was prior and this video is confusing then I feel I may need to watch the preceding videos. I don't know the order.(1 vote)
- You can access all the titles in order by just looking on the left, also by clicking on the titles at the upper left.(1 vote)

- Since the angle between two vectors is preserved under reflections (when dealing with orthogonal transformation matrices, at least), how does one go about defining
**directed angles**in R^n?(1 vote)- My guess is that in R^2 use delta(arctan(x2/x1)) between the two vectors. For R^3 and greater, which direction is positive?(1 vote)

- Just because the cosine of the angle is preserved doesn't necessarily imply the angle is as well. What could be the case is that one of the vectors in question has now the opposite direction, and then, in turn, we have the angle being 180-theta, as by definition of angle between vectors we look at the angle in front of the (a-b) side. So should we instead be talking about the cosine being preserved rather than the angle??(1 vote)
- How does one extend this proof (especially about the norms) to any finite-dimension inner product space?(1 vote)

## Video transcript

In the last couple of videos,
we've seen that if we have some matrix C that is n by n. It's a square matrix, and is
columns, column form and orthonormal set. Which just means that the
columns each have been normalized. So they each have length
of 1 if you view them as column vectors. And they're all mutually
orthogonal to each other. So if you dot it with
yourself you get 1. If you dot it with any of the
other columns, you get 0. We've seen this multiple
times. It's orthogonal to
everything else. If you have a matrix like this--
and I actually forgot to tell you the name of
this-- this is called an orthogonal matrix. We've already seen that the
transpose of this matrix is the same thing as the inverse
of this matrix. Which makes it super, duper,
duper useful to deal with. The transpose of this matrix
is equal to the inverse. Now, this statement leads to
some other interesting things about this. So, so far we've been dealing
this mainly with the change of basis. I can kind of draw the
diagram that you're probably tired of by now. Let's say that's the
standard basis. Let's say that I have
x in coordinates with another basis. We've seen I can multiply
this guy times c. To get that up there I could
multiply that guy by c inverse to get this guy right here. And, in that world, we viewed
c as just a change of basis. Were representing the same
matrix-- we're representing the same vector. We're just changing the
coordinates of how we represent it. But we also know that any matrix
product, any matrix vector product, is also a
linear transformation. So, this change of basis
is really just a linear transformation. What I want to show you in this
video, and you could view it either as a change of basis
or as a linear transformation, is that when you multiply this
orthogonal matrix times some vector, it preserves--
let me write this down-- lengths and angles. So let's have a little
touchy-feely discussion of what that means. Let's view it as a
transformation. Let's say I have some set
of vectors in my domain. Let's say they look like this. Let's say that it
looks like this. Well, let me do it like-- I'll
draw that one like that guy, and this guy like that. And there's some angle
between them. Angles are easy to visualize
in r2, r3. Maybe a little harder once we
get the higher dimensions. But that's the angle
between them. Now, if we're saying that we're
preserving the angles and the lengths, that means if
I were to multiply these vectors times c then we could
view it as a transformation. Maybe I rotate them or I--
well, you can't really. Maybe I rotate them or do
something like that. So maybe that pink vector
will now look like this. But it's going to have
the same length. This length is going to be the
same thing as that length. And even more, when I said it
preserves lengths and angles, this yellow vector's going to
look something like this. Where the angle is going
to be the same. Where this data is going
to be that data. That's what I mean by
preserves angles. If we didn't have this case,
we could imagine a transformation that doesn't
preserve angles. Let me draw one that doesn't. If this got transformed to, I
don't know, let's say this guy got a lot longer, and let's say
this guy also got longer, and I want to show
that the angle also doesn't get preserved. Not only did it get
longer, but it got distorted a little bit. So, the angle also changed. This transformation
right there is not preserving angles. So when you have a change
of basis matrix that's orthogonal, when you have a
transformation matrix that's orthogonal, all it's essentially
doing to your to your vectors, is it kind of a
rotates them around, but it's not going to really
distort them. So I'll write that in quotes
because that's not a mathematically rigorous term. So, no distortion of vectors. So, I've kind of showed
you the intuition of what that means. Let's actually prove
it to ourselves that this is the case. So, I'm saying that if this
pink vector here is x, and that this pink vector here is
c times x, I'm claiming that the length of x is equal to
the length of c times x. Let's see if that's
actually the case. The length of cx squared is the
same thing as cx dot cx. And here it's always useful
for me to kind of remind myself that if I take
two vectors-- let me do it over here. Let's say I have y dot y. This is the same thing as y
transpose, if you view them as matrices, y transpose times y. y transpose y is just y1, y2,
all the way to yn times y1, y2, all the way to yn. And if you were to do this 1
by n times n by 1 matrix product, you're going to get a 1
by 1 matrix or just a number that's going to be y1 times y1
plus y2 times y2 all the way to yn times yn. So, this is the same
thing as y dot y. I think I did this about ten or
twenty videos ago, but it's always a good refresher. So let's use this property
right here. So these two dotted
with each other. This is the same thing is taking
one of their transpose times the other one. So turn this from a vector,
vector dot product to a matrix, matrix product. So this is the same thing
as CX transpose, CX. so you can view this as a 1 by n
matrix now, times the 1 by 1 matrix which is just the
column vector cx. These are the same thing. Now, we also know that A times
B transpose is the same thing is B transpose, A transpose. We saw that a long time ago. So this thing right here is
going to be equal to X transpose, C transpose. Just switch the order and take
the transpose of each. X transpose times C transpose. And then you have
that times CX. And now we know that C transpose
is the same thing is as C inverse. This is where we need the
orthogonality of the matrix C. This is where we need it to be
a square matrix where all of its columns are mutually
orthogonal and they're all normal. And so this thing is just
going to become the identity matrix. I can write the identity matrix
there, but that's just going to disappear. So this is going to be equal
to X transpose X. X transpose is the same thing
as X dot X which is the same thing as the length
of X squared. So the length of CX squared is
the same thing as the length of X squared. So, that tells us that the
length of X, or the length of CX, is the length of x because
both of these are going to be positive quantities. So I've shown you that
orthogonal matrices definitely preserve length. Let's see if they
preserve angles. So we actually have
to define angles. Throughout our mathematical
careers, we understood what angles mean in kind
of r2 or r3. But in linear algebra, we
like to be general. And we defined an angle
using the dot product. We use the law of cosines and we
took an analogy to kind of triangle in r2. But we defined an angle or we
said the dot product V dot W is equal to the lengths, the
products of the lengths of those two vectors times
the cosine of the angle between them. Or you could say that the cosine
of the angle between two vectors, we defined as the
dot product of those two vectors divided by the lengths
of those two vectors. This was the definition so that
we can extend the idea of an angle to an arbitrarily
high dimension to r google if we had to. So let's see if it preserves. Let's see what the angle is if
we multiply these guys by C. So, if we wanted-- let's
say our new angle. So, cosine of angle C. Once we perform our
transformation. We're going to perform the
transformation on all of these characters. It's going to be CV dot CW over
the lengths of CV times the lengths of CW. Now we already know that
lengths are preserved. We already know that the length
of CW and CV are just going to be W and V. We just proved that. Let me write that. So the cosine of theta C is
equal to CV dot CW over the lengths of V times W. Because we've already shown
that it preserves lengths. We'll see what this
top part equals. So we can just use the
general property. The dot product is equal to the
transpose of one guy as kind of a matrix times
the second guy. So this is equal to CW
transpose times CV. And all of that over
these lengths. Like the W. And this is going to be equal
to-- I'm going to write it down here to have some space. We can switch these guys and
take their transpose. So, it's W transpose times
C transpose times CV. All of that over their lengths,
the product of their lengths V and W. And this is the identity
matrix. That's the identity matrix, and
this is going to be equal to W transpose times V over the
products of their lengths. And this is the same
thing as V dot W. This is V dot W over
their lengths. Which is cosine of theta. So, you notice, by our
definition of an angle as the dot product divided by the
vector lengths, when you perform a transformation or you
can imagine a change of basis either way, with an
orthogonal matrix C the angle between the transformed vectors
does not change. It is the same as the angle
between the vectors before they were transformed. Which is a really neat
thing to know. The change of bases or
transformations with orthogonal matrices don't
distort the vectors. They might just kind of rotate
them around or shift them a little bit, but it doesn't
change the angles between them.