proof that dim V + dim V perp = n

• Nov 6th 2010, 12:54 PM
oblixps
proof that dim V + dim V perp = n
I am trying to prove that the dimension of a subspace V of R^n + the dimension of the orthogonal complement of that subspace V is equal to n for any subspace V in R^n.

So i assumed the basis for the subspace V consisted of vectors v1, v2,..., vk. now i extend the basis so that the new basis is a basis for R^n. This new basis is v1, v2, ..., vk, w1, w2, ..., w_m. where m + k = n. So in order to prove this theorem i have to show that w1, w2, ..., w_m forms a basis for the orthogonal complement of V. What i have trouble showing is that these vectors span the orthogonal complement of V. i have that any vector in R^n let's say u can be expressed as a linear combination of v1, v2, ..., vk, w1, w2, ..., w_m since that is a basis for R^n. so u = c1v1 + ... +ckvk + d1w1 + ... + d_m(w_m). i tried taking the dot product on both sides with an arbitrary vector w in the orthogonal complement of V and i got u * w = w * (d1w1 + ...+ d_m(w_m)) since w is orthogonal to every vector in V so w * v_i = 0. now i'm stuck and i'm not sure how to continue on.

I can show that w1, ... w_m are linearly independent since v1, v2, ..., vk, w1, w2, ..., w_m form a basis which means that the only linear combination of them that forms the 0 vector is for all constants to be 0. Then since i know that v1, ... vk is also a basis, the only linear combination of those vectors that equal 0 is for all the constants to equal 0 as well. and therefore the only linear combination of w1, w2, ..., w_m that equal 0 is all the constants being 0 so those vectors are independent.

now all i need to show is that those vectors span the orthogonal complement of V and the proof will be done. is my approach to showing that the vectors span correct? if anyone could give a hint or 2 to help me move on in this problem i'd greatly appreciate it.
• Nov 6th 2010, 12:58 PM
Drexel28
Quote:

Originally Posted by oblixps
I am trying to prove that the dimension of a subspace V of R^n + the dimension of the orthogonal complement of that subspace V is equal to n for any subspace V in R^n.

So i assumed the basis for the subspace V consisted of vectors v1, v2,..., vk. now i extend the basis so that the new basis is a basis for R^n. This new basis is v1, v2, ..., vk, w1, w2, ..., w_m. where m + k = n. So in order to prove this theorem i have to show that w1, w2, ..., w_m forms a basis for the orthogonal complement of V. What i have trouble showing is that these vectors span the orthogonal complement of V. i have that any vector in R^n let's say u can be expressed as a linear combination of v1, v2, ..., vk, w1, w2, ..., w_m since that is a basis for R^n. so u = c1v1 + ... +ckvk + d1w1 + ... + d_m(w_m). i tried taking the dot product on both sides with an arbitrary vector w in the orthogonal complement of V and i got u * w = w * (d1w1 + ...+ d_m(w_m)) since w is orthogonal to every vector in V so w * v_i = 0. now i'm stuck and i'm not sure how to continue on.

I can show that w1, ... w_m are linearly independent since v1, v2, ..., vk, w1, w2, ..., w_m form a basis which means that the only linear combination of them that forms the 0 vector is for all constants to be 0. Then since i know that v1, ... vk is also a basis, the only linear combination of those vectors that equal 0 is for all the constants to equal 0 as well. and therefore the only linear combination of w1, w2, ..., w_m that equal 0 is all the constants being 0 so those vectors are independent.

now all i need to show is that those vectors span the orthogonal complement of V and the proof will be done. is my approach to showing that the vectors span correct? if anyone could give a hint or 2 to help me move on in this problem i'd greatly appreciate it.

Do you know the fact that if $V\oplus V^{\perp}=\mathbb{R}^n$? I'm guessing not.

EDIT: Upon closer inspection I see that my question was stupid, since that's what it looks like you're trying to do.
• Nov 6th 2010, 01:32 PM
Also sprach Zarathustra
How to prove the above?

Proof:

There exist an orthogonal basis $\{u_1,...,u_r\}$ to $V$, also by some theorem we can extend it to orthonormal basis of $\mathbb{R}^n$: $\{u_1,...,u_n\}$.Therefor $u_{r+1},...,u_n \in V^{\perp}$. If $v\in\mathbb{R}^n$, $v$ is from the form: $v=a_1u_1+...+a_nu_n$ when $a_1u_1+...+a_ru_r \in V$ and $a_{r+1}u_{r+1}+...+a_nu_n \in V^{\perp}$, the conclusion is $\mathbb{R}^n=V+V^{\perp}$.

Now, if $w\in V\bigcap V^{\perp}$, then $=0$, therefor w=0, hence $V\bigcap V^{\perp}=\{0\}$

$\mathbb{R}^n=V+V^{\perp}$ and $V\bigcap V^{\perp}=\{0\}$ gives us $V\oplus V^{\perp}=\mathbb{R}^n$.
• Nov 6th 2010, 03:40 PM
Roam
Actually the easiest method to prove this is to use the idea that for any $m \times n$ matrix $rank(A)+nullity (A)=n$:

If $V=\{ 0 \}$, then $V^{\perp} =R^n$, in this case $dim(V)+dim(V^{\perp})=0+n=n$. If V $\ne \{ 0 \}$, then choose a basis for $V$ and form the matrix $A$ which has these basis as its row vectors. The matrix has $n$ columns since its row vectors come from $R^n$. Moreover, the row space of $A$ is $V$ and the nullspace of $A$ is $V^{\perp}$, so you will have

$dim(V)+dim(V^{\perp})= rank(A)+nullity(A)=n$
• Nov 7th 2010, 09:23 AM
oblixps
so basically in your proof you didn't have to show that u_{r+1},...,u_n \in V^{\perp} formed a basis for V perp? and how did you know that the vectors u_r+1, ... u_n are in V perp? maybe this is supposed to be obvious and i'm just thinking too much but we start with a basis for V, extend to a basis of R^n so the left over vectors must be in V perp? also you never showed that the vectors were a basis so we don't know if the vectors span V perp so how do you know that a_{r+1}u_{r+1}+...+a_n u_n is any vector is V perp?
• Nov 11th 2010, 12:23 PM
oblixps
For Also sprach Zarathustra's proof I am confused on the part when he said that u_r+1,...u_n are in V perp. how did you know that the vectors u_r+1, ... u_n are in V perp?