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 R^4 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 in 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 basis
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 x_1, x_2, x_3 and x_4,
we see here that this matrix has a rank of 2.
Now, we're looking at vectors which live in R^4,
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 x_4 equal some constant,
and x_3 equal another constant.
So specifically, I'm going to let x_4 equal b, and x_3
equal a.
Now what we do is we take a look at these two equations,
and this bottom equation will say
that x_2 is equal to negative x_3 plus
x_4, which is going to equal negative a-- x_4-- plus b.
And then the top equation says that x_1 is equal to negative
2*x_2 minus 2*x_3 minus 3*x_4.
And if I substitute in, x_2 is -a plus b.
x_3 is a.
And x_4 is b.
So when the dust settles, the a's cancel
and I'm left with minus 5b.
So we can combine everything together
and we end up obtaining [x_1, x_2, x_3, x_4] equals -5b,
x_2 is minus a plus b, x_3 is a, and x_4 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 R^4 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 c_1
times the vector [1, 2, 2, 3] plus c_2 times the vector [1,
3, 3, 2] plus the vector c_3 [0, -1, 1, 0] plus c4
[-5, 1, 0, 1].
OK?
So c_1 and c_2 are multiplying the vectors in S,
and c_3 and c_4 are multiplying the vectors in S perp.
So the question is, given any v, can I
find constants c_1, c_2, c_3, c_4, 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, c_1, c_2, c_3, c_4.
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,
c_1, c_2, c_3, c_4.
This then gives us a unique decomposition for v
in terms of a piece which is in S
and a 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.