1. ## Orthogonal Subspaces

Let U and W be orthogonal subspaces.

a) Prove that $U\cap W=\{\underline{0} \}$

b) If $\{u_1,u_2,...,u_m\}$ is a linearly independent subset of U and $\{w_1,w_2,...,w_n\}$ is a linearly independent subset of W, prove that $\{u_1,u_2,...,u_m,w_1,w_2,...,w_n\}$ is a linearly independent set in V.

I find this bit a little hard to understand, and I have to multitask, so I'll have difficulty if I have to do this on my own. Any help would be greatly appreciated.

2. Originally Posted by Runty
Let U and W be orthogonal subspaces.

a) Prove that $U\cap W=\{\underline{0} \}$

If $x\in U\cap W$ then as $x\in U\,\,\,and\,\,\,x\in W$ we get $=0\Longleftrightarrow x=0$ , with <,> the inner product in the v.s.

b) If $\{u_1,u_2,...,u_m\}$ is a linearly independent subset of U and $\{w_1,w_2,...,w_n\}$ is a linearly independent subset of W, prove that $\{u_1,u_2,...,u_m,w_1,w_2,...,w_n\}$ is a linearly independent set in V.

Suppose $\sum^m_{i=1}a_1u_1+\sum^n_{j=1}b_jw_j=0\,,\,\,a_i\ ,,\,b_j \in\mathbb{F}=$ the definition field, then $\sum^m_{i=1}a_1u_1=-\sum^n_{j=1}b_jw_j\Longrightarrow \sum^m_{i=1}a_1u_1\,,\,\sum^n_{j=1}b_jw_j \in U\cap W$ $\Longrightarrow \sum^m_{i=1}a_1u_1=0=\sum^n_{j=1}b_jw_j\Longrighta rrow$ . Deduce from here and (a) above that $a_i=0=b_j\,\,\forall\,i,j$ .

Tonio

I find this bit a little hard to understand, and I have to multitask, so I'll have difficulty if I have to do this on my own. Any help would be greatly appreciated.
.

3. Those are good answers, Tonio, but our material does not make use of summation notation. Still, I guess I can use it and it'll provide a correct answer, or I could just go long-hand.

Two questions, however.

First, for part a), when you say inner product, you mean $x\bullet x$, right? I just want to make sure.

Second, for part b), when you say definition field, would you mean something like real numbers (i.e. for scalars)?

4. Originally Posted by Runty
Those are good answers, Tonio, but our material does not make use of summation notation. Still, I guess I can use it and it'll provide a correct answer, or I could just go long-hand.

Two questions, however.

First, for part a), when you say inner product, you mean $x\bullet x$, right? I just want to make sure.

Yes, I meant that: inner, scalar or dot product.

Second, for part b), when you say definition field, would you mean something like real numbers (i.e. for scalars)?

I meant the field over which the vector space containing U, W is defined.

Tonio

5. Originally Posted by Runty
Let U and W be orthogonal subspaces.

a) Prove that $U\cap W=\{\underline{0} \}$
For every $u\in U$ and every $w \in W$, the inner product of u and w is 0. Suppose v is in both subspaces. What must be true of the inner product of v with itself? Now think about the definition of "inner product".

b) If $\{u_1,u_2,...,u_m\}$ is a linearly independent subset of U and $\{w_1,w_2,...,w_n\}$ is a linearly independent subset of W, prove that $\{u_1,u_2,...,u_m,w_1,w_2,...,w_n\}$ is a linearly independent set in V.[/quote]
Remember the definition of "linearly independent". That set is linearly independent if and only if $a_1u_1+ a_2u_2+ \cdot\cdot\cdot\+ a_mu_m+ b_1w_1+ b_1w_2+ \cdot\cdot\cdot+ a_nw_n= 0$ implies that $a_1= a_2= \cdot\cdot\cdot\= a_m= b_1= b_2= \cdot\cdot\cdot= w_n= 0$. What do you get if you take the dot product of $u_1$ with math]a_1u_1+ a_2u_2+ \cdot\cdot\cdot\+ a_mu_m+ b_1w_1+ b_1w_2+ \cdot\cdot\cdot+ a_nw_n= 0[/tex]? What about the dot product with $u_2$, $w_1$, etc.

I find this bit a little hard to understand, and I have to multitask, so I'll have difficulty if I have to do this on my own. Any help would be greatly appreciated.

6. Originally Posted by HallsofIvy
For every $u\in U$ and every $w \in W$, the inner product of u and w is 0. Suppose v is in both subspaces. What must be true of the inner product of v with itself? Now think about the definition of "inner product".

b) If $\{u_1,u_2,...,u_m\}$ is a linearly independent subset of U and $\{w_1,w_2,...,w_n\}$ is a linearly independent subset of W, prove that $\{u_1,u_2,...,u_m,w_1,w_2,...,w_n\}$ is a linearly independent set in V.

Remember the definition of "linearly independent". That set is linearly independent if and only if $a_1u_1+ a_2u_2+ \cdot \cdot \cdot + a_mu_m+ b_1w_1+ b_1w_2+ \cdot \cdot \cdot + a_nw_n= 0$ implies that $a_1= a_2= \cdot \cdot \cdot = a_m= b_1= b_2= \cdot \cdot \cdot = b_n= 0$. What do you get if you take the dot product of $u_1$ with $a_1u_1+ a_2u_2+ \cdot \cdot \cdot + a_mu_m+ b_1w_1+ b_1w_2+ \cdot \cdot \cdot + a_nw_n= 0$? What about the dot product with $u_2$, $w_1$, etc.
Corrected the mistakes in Latex. This certainly makes things complicated, however. I'll see if I can figure it out.