# Proving that V^{\bot} is a Subspace: Exercise 3.3

• E'lir Kramer
In summary, the conversation is discussing how to prove that the orthogonal complement of a subspace V is also a subspace. The definition of a subspace is unpacked and it is shown that in order to prove that the orthogonal complement is a subspace, one must show that the sum of two vectors in the complement is also in the complement and that a scalar times a vector in the complement is also in the complement. Using the properties of inner products, it is established that the sum and scalar multiple of vectors in the orthogonal complement are also orthogonal to a vector in V, thus proving that the orthogonal complement is indeed a subspace.
E'lir Kramer
This Exercise 3.3 from Advanced Calculus of Several Variables by C.H. Edwards Jr.:

If $V$ is a subspace of $\Re^{n}$, prove that $V^{\bot}$ is also a subspace.

As usual, this is not homework. I am just a struggling hobbyist trying to better myself on my own time.

The only progress I've been able make towards a formal proof is to unpack the definition of a subspace: a set of objects which is closed over two operations: $o_{1} + o_{2}$ and $a \cdot o$, where $o, o_{1}$, and $o_{2}$ are any objects in the set, and a is a real number.

So what I want to show is that for two objects $o_{1}, o_{2} \in V^{\bot}$, their sum $o_{1} + o_{2} \in V^{\bot}$, and $a \cdot o \in V^{\bot}$.

But, I think this is the sticking point: what does it take to formally establish that some vector $o$ is in $V^{\bot}$?

E'lir Kramer said:
This Exercise 3.3 from Advanced Calculus of Several Variables by C.H. Edwards Jr.:

If $V$ is a subspace of $\Re^{n}$, prove that $V^{\bot}$ is also a subspace.

As usual, this is not homework. I am just a struggling hobbyist trying to better myself on my own time.

The only progress I've been able make towards a formal proof is to unpack the definition of a subspace: a set of objects which is closed over two operations: $o_{1} + o_{2}$ and $a \cdot o$, where $o, o_{1}$, and $o_{2}$ are any objects in the set, and a is a real number.

So what I want to show is that for two objects $o_{1}, o_{2} \in V^{\bot}$, their sum $o_{1} + o_{2} \in V^{\bot}$, and $a \cdot o \in V^{\bot}$.

But, I think this is the sticking point: what does it take to formally establish that some vector $o$ is in $V^{\bot}$?

Show that the dot product of o with any vector in V is zero.

At this point in the book, he's barely touched on the the dot product, which he's calling the "usual inner product". He's only been talking about inner products at a higher level of generality as any binary operation with the three properties of positivity, "symmetry" (commutativity), and linearity. If I proved anything using the dot product, it would reduce the generality that he's clearly taking pains to establish. I'm worried that any such proof is going to get left behind in the subsequent chapters.

E'lir Kramer said:
At this point in the book, he's barely touched on the the dot product, which he's calling the "usual inner product". He's only been talking about inner products at a higher level of generality as any binary operation with the three properties of positivity, "symmetry" (commutativity), and linearity. If I proved anything using the dot product, it would reduce the generality that he's clearly taking pains to establish. I'm worried that any such proof is going to get left behind in the subsequent chapters.

What definition were you given for the orthogonal complement? It would depend on which inner product you are using, right?

Last edited:

Given a subspace $V$ of $\Re^{n}$, denote by $V^{\bot}$ all of those vectors in $\Re^{n}$, each of which is orthogonal to every vector in $V$.

Orthogonality is defined like this:

A set of nonzero vectors $v_{1}, v_{2}, ...$ in $V$ is said to be an orthogonal set if $<v_{i}, v_{j}> = 0$ whenever i ≠ j.

But your question has lead me further down the path by leading me back to the definition of orthogonality. I have to prove that $<v, o>= 0$. It's the essentially what you said, except instead of using the dot product, I should prove it using only the properties of the generalized inner product $<>$. I'll work towards that goal now. Thank you for the hint :).

In order to show that a set of vectors is a subspace, you need only prove that the sum of two vectors in the set is also in the set and that a scalar times a vector in the set is also in the set.

If u and v are in the orthogonal complement of V, then <u, a>= 0 and <v, a>= 0 for every vector in V. So what is true of <u+ v, a>? What is true of <ku, a> for k any scalar?

Thanks Ivy. I solved this one quickly once I referred back to the definition of orthogonality. It follows directly from property 3 of inner products, < ax + by, z > = a< x, z > + b< y, z >.

Thus using your variables, < u, a > + < v , a > = < u + v, a > = 0 + 0 = 0, proving that u+v is orthogonal to a.

Also by linearity,
< ku, a > = k< u, a > = k * 0 = 0

Excellent!

## 1. What is V^{\bot}?

V^{\bot} is the orthogonal complement of a given subspace V. It consists of all vectors that are perpendicular to every vector in V.

## 2. How is V^{\bot} related to V?

V^{\bot} is a complementary subspace to V. This means that V and V^{\bot} together span the entire vector space.

## 3. How can we prove that V^{\bot} is a subspace?

To prove that V^{\bot} is a subspace, we need to show that it satisfies the three main properties of a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector.

## 4. What is the significance of proving that V^{\bot} is a subspace?

Proving that V^{\bot} is a subspace is important because it helps us understand the relationship between a subspace and its orthogonal complement. It also allows us to use properties of subspaces to make conclusions about V^{\bot} and vice versa.

## 5. Can you provide an example of proving that V^{\bot} is a subspace?

Sure, let's say we have a subspace V = {(x, y) | x + y = 0}. To prove that V^{\bot} is a subspace, we need to show that it contains the zero vector (0, 0), is closed under addition, and is closed under scalar multiplication. After verifying these properties, we can conclude that V^{\bot} is indeed a subspace.

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