11 Orthogonal complements

11. Orthogonal complements

Orthogonal complements

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V perp is equal, to the set of all x's, all the vectors x that are a member of our $R^n$, such that x dot V is equal to zero for every vector V that is a member of our subspace.

Then what does this imply?

What is a fact that a and b are members of V prep?

√ That means a dot V, where this V is any member of our original subspace is V, is equal to 0 for any V that is a member of our subspace V. And it also means that b dot any member of our subspace is also going to be 0.

If we add them, they are also gonna be zero. So a + b is definitely a member of our orthogonal complement.

√ Now, is ca a member of V perp?

If we take ca and dot it with any member of our original subspace this is the same thing as c times a dot V. By definition this is equal to 0 because a was a member of our orthogonal complement.

So we know that V perp, or the orthogonal complement of V, is a subspace.

In the last video, I said that the null space of A is equal to the orthogonal complement of the row space of A.

And we know that row space of A is the same thing as the column space of A transpose.

So, the null space of A is the orthogonal complement of the row space.

Let's apply this generally.

The null space of A is all of the vectors x that satisfy the equation that this is going to be equal to the zero vector.

Ax = 0

Another way to write down this equation is equal to that 0.

The nullspace of A is equal to all of the x's that are members of $R^n$, such that Ax is equal to 0.

And if you have any vector that's a linear combination of these row vectors, if you dot it with any member of your null space, you're gonna get 0.

Every member of our null space is definitely orthogonal to every member of our row space.

dim(v) + dim(orthogonal complement of v) = n

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Representing vectors in $R^n$ using subspace members

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Only zero vector satisfies this.

The only place that V and orthogonal complement V overlap is zero vector.

Orthogonal complemnet of the orthogonal complent

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We know that x can be represented as the sum of two vectors. One that's in v and on that's in the orthogonal complement of v.

Then the magnitude of w squared, or the length of w squared, has got to be equal to 0. So this tells w is zero vector.

This tells us that x is a member of our subspace v.

√ What happens if we dot v and w?

That's going to give us zero.

Orthogonal complement of the nullspace

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Unique rowspace solution to Ax = b

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Any vector in $R^n$ can be represented as a sum of some vector in our null space and some vector in our rowspace.

Let's say x is a solution to Ax = b which also means x is a member of $R^n$.

And let's say $\vec {x} = \vec {r_0} + \vec {n_0}$ where $\vec {r_0} \in C(A^T) \space and \space \vec {n_0} \in N(A)$

Then we can say like this below.

So $r_0$ is equal to Ax is equal to b.

And the next question is that **is $r_0$ the only guy in our row space that is a solution to Ax is equal to b?

Let's assume there's another guy $r_1$.

These are equal to b, so there will be zero.

And that means that ($r_1 - r_0$) is in N(A)$

($r_1 - r_0$) is in a subspace and it's also in the orthogonal complement of the subspace. Then the only possible vector that can be is the zero vector.

So we get that r1 - r0 must be equal to the 0!!

This implies r1 must be equal to r0.

Conclusion: There exists a unique member of $C(A^T)$, $\vec {r_0} (\vec {r_0} \in C(A^T))$ such that r0 is a solution to Ax = b.

Any solution x to Ax=b can be written as a combination of (r0 + n0).