# Practice English Speaking&Listening with: Orthogonal Vectors and Subspaces | MIT 18.06SC Linear Algebra, Fall 2011

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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.

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?

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?

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.

The Description of Orthogonal Vectors and Subspaces | MIT 18.06SC Linear Algebra, Fall 2011