WEBVTT

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DAVID SHIROKOFF: Hi everyone.

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I'm Dave.

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Now today, I'd like
to tackle a problem

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in orthogonal subspaces.

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So the problem we'd like to
tackle: given a subspace S,

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and suppose S is spanned by
two vectors, [1, 2, 2, 3]

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and [1, 3, 3, 2].

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We have a question here which
is to find a basis for S perp--

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S perp is another subspace
which is orthogonal to S.

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And then secondly, can every
vector in R^4 be uniquely

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written in terms
of S and S perp.

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So I'll let you think
about this for now,

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and I'll come back in a minute.

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Hi everyone.

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Welcome back.

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OK, so why don't we
tackle this problem?

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OK, so first off,
what does it mean

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for a vector to be in S perp?

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Well, if I have a
vector x, and S perp,

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and x is in S perp,
what this means is

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x is going to be orthogonal
to every vector in S. Now

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specifically, S is spanned
by these two vectors.

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So it's sufficient that x be
perpendicular to the two basis

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vectors in S.

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So specifically, I can take
[1,  2, 2, 3] and dot it with x,

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and it's going to be 0.

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So I'm treating x as
a column vector here.

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In addition, x must also be
orthogonal to [1, 3, 2, 2].

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So any vector x
that's an S perp must

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be orthogonal to both
of these vectors.

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So what we can do
is we can write

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this as a matrix equation.

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And we do this by
combining these two vectors

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as rows of the matrix.

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So if we step back and take
a look at this equation,

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we see that what
we're really asking

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is to find all x that are in
the null space of this matrix.

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So how do we find x in the
null space of a matrix?

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Well what we can do is we
can row reduce this matrix

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and try and find a basis
for the null space.

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So I'm going to just
row reduce this matrix.

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And notice that
by row reduction,

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we don't actually change
the null space of a matrix.

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So if I'm only interested
in the null space,

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this system is going
to be equivalent

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to-- I can keep the
top row the same.

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And then just to
simplify our lives,

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we can take the second
row and subtract

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one copy of the first row.

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Now, if I do that, I
obtain 0, 1, 1, -1.

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Now, to parameterize the null
space, what I'm going to do

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is I'm going to write
x out as components.

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So if I write x with components
x_1, x_2, x_3 and x_4,

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we see here that this
matrix has a rank of 2.

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Now, we're looking at
vectors which live in R^4,

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so we know that the null space
is going to have a dimension

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which is 4 minus 2.

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So that means there should be
two vectors in the null space

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of this matrix.

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To parameterize these
two-dimensional vectors, what

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I'm going to do is I'm going
to let x_4 equal some constant,

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and x_3 equal another constant.

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So specifically, I'm going
to let x_4 equal b, and x_3

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equal a.

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Now what we do is we take a
look at these two equations,

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and this bottom
equation will say

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that x_2 is equal
to negative x_3 plus

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x_4, which is going to equal
negative a-- x_4-- plus b.

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And then the top equation says
that x_1 is equal to negative

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2*x_2 minus 2*x_3 minus 3*x_4.

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And if I substitute
in, x_2 is -a plus b.

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x_3 is a.

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And x_4 is b.

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So when the dust
settles, the a's cancel

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and I'm left with minus 5b.

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So we can combine
everything together

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and we end up obtaining [x_1,
 x_2, x_3, x_4] equals -5b,

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x_2 is minus a plus b,
x_3 is a, and x_4 is b.

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And now what we can do is
we can take this vector

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and we can decompose
it into pieces

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which are a multiplied
by a vector,

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and b multiplied by a vector.

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So you'll note that this is
actually a times [0, -1, 1, 0]

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plus b times [-5, 1, 0, 1].

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OK?

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So we have successfully
achieved a parameterization

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of the null space of this
matrix as some constant a

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times a vector
[0, -1, 1, 0] plus b

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times a vector [-5, 1, 0, 1].

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And now we claim that this
is the entire space, S perp.

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So S perp is going to be spanned
by this vector and this vector.

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Now notice how, if I were
to take either of these two

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vectors in S and dot it with
any vector in the null space,

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by construction, it
automatically vanishes.

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So this concludes part one.

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Now for part two.

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Can every vector v in R^4 be
written uniquely in terms of S

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and S perp?

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The answer is yes.

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So how do we see this?

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Well, if I have a
vector v, what I can do

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is I can try and write
it as some constant c_1

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times the vector [1, 2, 2, 3]
plus c_2 times the vector [1,

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3, 3, 2] plus the vector
c_3 [0,  -1, 1, 0] plus c4

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[-5, 1, 0, 1].

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OK?

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So c_1 and c_2 are
multiplying the vectors in S,

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and c_3 and c_4 are multiplying
the vectors in S perp.

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So the question is,
given any v, can I

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find constants c_1,
c_2, c_3, c_4, such

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that this equation holds?

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And the answer is yes.

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Just to see why it's
yes, what we can do

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is we can rewrite this
in matrix notation,

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and there's kind
of a handy trick.

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What I can do is I
can take these columns

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and write them as
columns of the matrix.

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And this whole
expression is actually

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equivalent to this
matrix multiplied

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by the constant,
c_1, c_2, c_3, c_4.

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And on the right-hand
side, we have the vector v.

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Now, by construction,
these vectors

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are linearly independent.

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And we know from linear
algebra that if we

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have a matrix with linearly
independent columns,

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the matrix is invertible.

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What this means is, for any
v on the right-hand side,

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we can invert this matrix and
obtain unique coefficients,

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c_1, c_2, c_3, c_4.

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This then gives us a
unique decomposition for v

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in terms of a
piece which is in S

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and a piece which is in S perp.

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And in general this can be
done for any vector space.

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Well I'd like to
conclude this problem now

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and I hope you had a good time.