PROFESSOR: Hi, guys.
My name is Nikola, and in this video,
we're going to work out an example
of an orthogonal projection matrix.
Specifically, we are gonna compute the projection matrix
onto the plane given by the equation x plus y minus z
equals 0.
So before we start, let me just recall
what a projection matrix is.
So you've seen this sketch here a million times already.
A projection matrix takes any vector in three-space-- well,
just in this case, we are dealing with a three-space--
and projects it down onto the plane,
a two-dimensional subspace of R^3.
So I'll give you a few moments to consider
the problem for yourselves.
And then you'll see my take on it.
Hi again.
So in lecture, Professor Strang derived, in meticulous detail,
the formula for the projection matrix.
So it was given by the following slightly complicated
expression.
It's A times A transpose A inverse A
transpose where A is a matrix that
somehow encodes the subspace we're projecting on.
In particular, A has, as its columns, a_1, a_2,
I'm going to denote them-- a basis for the plane we're
projecting on.
So essentially, what we need to do
is find two such vectors that span the plane and start
computing with a matrix.
This is fairly straightforward.
One choice that works, for example, is [1, -1, 0]
for the first column, and [1, 0, 1] for the second column.
And let me write out the matrix A. So in the formula,
the slightly more complicated combination
is A transpose A inverse, so let me compute that first for you.
So A transpose A is a 2 by 2 matrix.
And it's not so hard to figure out that it's 2 and 1; 1, 2.
Now we shall invert it using the familiar formula.
1 over the determinant.
2 times 2 minus 1 is 3.
And so we switch the diagonal entries,
and we flip the signs of the off-diagonal ones.
Right.
And therefore, projection matrix is given by the following
product of matrices: ...1/3, [2, -1; -1,
2] and then transpose of A, which is [1, -1, 0; 1, 0, 1].
I'm gonna carry out this multiplication
in inhumanly fast fashion.
So I'm just gonna write down the answer, which is 1/3
[2, -1, 1; -1, 2, 2; 1, 1, 1].
So what you can do now is-- well,
you can check whether this answer actually makes sense.
One thing you can do is just-- well,
a projection matrix is supposed to take the normal
to the plane down to 0.
So you can just multiply P and the normal vector [1, 1, -1].
And if you get 0, maybe you've done a good job.
Another curious thing that I would like to point out here
is: so you see we had lots of freedom choosing
the matrix A. We could have chosen any two columns that
span the subspace, that spans the plane.
The beautiful thing about it is that in the end,
we'll get the same answer.
So I hope there will be many of you
who would say, hey, there is an easier way to do the problem.
And I'll agree with these people.
So let's see what would be an easier approach.
Well, let's go back to the sketch here.
And let's make the following observation,
that any vector is a sum of two components.
The first component is its projection onto the plane.
And the other component is its projection
onto the orthogonal complement of the plane, in this case,
onto the normal vector through the plane.
So in the language of linear algebra,
this is just b equals to its projection
onto the plane plus its projection-- I'm
gonna call it P_N-- onto the orthogonal complement
of the plane.
I'm gonna suggestively write here the identity
matrix so that you can immediately
read off a matrix equality.
Associated with this equality here, it's
the identity equals P plus P_N.
And therefore, the projection matrix
is just the identity minus the projection matrix
onto the normal vector.
Now, this object here, P_N, is much easier to compute, well,
for two reasons.
First one is that projecting onto a one-dimensional subspace
is infinitely easier than projecting
onto a higher-dimensional subspace.
And second, we already have-- well, immediately we
can read off from the equation of the plane what
the normal vector is.
So we don't have derive these guys.
We don't have to do what we did here.
So essentially, P_N will be N N transpose N inverse N
transpose.
And that's equal to [1, 1, -1]-- N transpose N inverse,
this is just a number.
It's 1 over the magnitude of the normal vector,
so that's-- then, the magnitude squared, so that's 3.
And 1, 1, -1.
I'm gonna write the answer here.
It's 1/3, 1, 1, -1; 1, 1, -1; and -1, 1, 1.
And in order to get the projection matrix-- yes?
AUDIENCE: [INAUDIBLE].
PROFESSOR: Oh.
Thank you.
And in order to get the projection matrix,
we just subtract from the identity this expression.
And you can confirm that it's-- we get the same answer as here.
I think we're done here.