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

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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].


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].


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.

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