DAVID SHIROKOFF: Hi everyone.
I'm Dave.
Now today, I'd like to tackle a problem
in orthogonal subspaces.
So the problem we'd like to tackle, given a subspace S,
and suppose S is spanned by two vectors, 1, 2, 2, 3, and
1, 3, 3, 2.
We have a question here which is to find
a basis for S perp--
S perp is another subspace which is orthogonal to S. And
then secondly, can every vector in R4 be uniquely
written in terms of S and S perp.
So I'll let you think about this for now, and I'll come
back in a minute.
Hi everyone.
Welcome back.
OK, so why don't we tackle this problem?
OK, so first off, what does it mean for a
vector to be an S perp?
Well if I have a vector x, and S perp, and x is in S perp,
what this means is x is going to be orthogonal to every
vector in S. Now specifically, S is spanned
by these two vectors.
So it's sufficient that x be perpendicular to the two bases
vectors in S.
So specifically, I can take 1, 2, 2, 3, and dot it with x,
and it's going to be 0.
So I'm treating x as a column vector here.
In addition, x must also be orthogonal to 1, 3, 2, 2.
So any vector x that's an S perp must be orthogonal to
both of these vectors.
So what we can do is we can write
this as a matrix equation.
And we do this by combining these two vectors as rows of
the matrix.
So if we step back and take a look at this equation, we see
that what we're really asking is to find all x that are in
the null space of this matrix.
So how do we find x in the null space of a matrix?
Well what we can do is we can row reduce this matrix and try
and find a basis for the null space.
So I'm going to just row reduce this matrix, and notice
that by row reduction, we don't actually change the null
space of a matrix.
So if I'm only interested in the null space, this system is
going to be equivalent to, I can keep the top row the same.
And then just to simplify our lives, we can take the second
row and subtract one copy of the first row.
Now, if I do that, I obtain 0, 1, 1, -1.
Now, to parameterize the null space, what I'm going to do is
I'm going to write x out as components.
So if I write x with components x1, x2, x3 and x4,
we see here that this matrix has a rank of 2.
Now, we're looking at vectors which live in R4, so we know
that the null space is going to have a dimension
which is 4 minus 2.
So that means there should be two vectors in the null space
of this matrix.
To parameterize these two dimensional vectors, what I'm
going to do is I'm going to let x4 equals some constant,
and x3 equal another constant.
So specifically, I'm going to let x4 equal b,
and x3 equal a.
Now what we do is we take a look at these two equations,
and this bottom equation will say that x2 is equal to
negative x3 plus x4, which is going to equal -a, x4 plus b.
And then the top equation says that x1 is equal to -2, x2
minus 2x3 minus 3x4, And if I substitute in,
x2 is -a plus b.
x3 is a.
And x4 is b.
So when the dust settles, the a's cancel and
I'm left with -5b.
So we can combine everything together and we end up
obtaining x1, x2, x3, x4 equals -5b.
x2 is -a plus b.
x3 is a, and x4 is b.
And now what we can do is we can take this vector and we
can decompose it into pieces which are a multiplied by a
vector, and b multiplied by a vector.
So you'll note that this is actually a times 0, -1, 1, 0
plus b times -5, 1, 0, 1.
OK?
So we have successfully achieved a parameterization of
the null space of this matrix as some constant a times a
vector 0, -1, 1, 0 plus b times a vector -5, 1, 0, 1.
And now we claim that this is the entire space, S perp.
So S perp is going to be spanned by this vector and
this vector.
Now notice how if I were to take either of these two
vectors in S and dot it with any vector in the null space,
by construction it automatically vanishes.
So this concludes part one.
Now for part two.
Can every vector v in R4 be written uniquely in terms of S
and S perp?
The answer is yes.
So how do we see this?
Well if I have a vector v, what I can do is I can try and
write it as some constant c1 times the vector 1, 2, 2, 3
plus c2 times the vector 1, 3, 3, 2 plus the vector c3, 0,
-1, 1, 0 plus c4, -5, 1, 0, 1.
OK?
So c1 and c2 are multiplying the vectors in S, and c3 and
c4 are multiplying the vectors in S perp.
So the question is, given any v, can I find constants c1,
c2, c3, c4, such that this equation holds?
And the answer is yes.
Just to see why it's yes, what we can do is we can rewrite
this in matrix notation, and there's kind of a handy trick.
What I can do is I can take these columns and write them
as columns of the matrix.
And this whole expression is actually equivalent to this
matrix multiplied by the constant, c1, c2, c3, c4.
And on the right-hand side, we have the vector v.
Now, by construction, these vectors are linearly
independent.
And we know from linear algebra that if we have a
matrix with linearly independent columns, the
matrix is invertible.
What this means is for any v on the right-hand side, we can
invert this matrix and obtain unique
coefficients, c1, c2, c3, c4.
This then gives us a unique decomposition for v in terms
of a piece which is in S, and the piece which is in S perp.
And in general this can be done for any vector space.
Well I'd like to conclude this problem now and I hope you had
a good time.