Orthogonal Subspaces

**Runty** · March 9th 2010, 05:58 AM

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.

**tonio** · March 9th 2010, 06:40 AM

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 $<x,x>=0\Longleftrightarrow x=0$ , 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.

.

**Runty** · March 9th 2010, 06:55 AM

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)?

**tonio** · March 9th 2010, 08:17 AM

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

**HallsofIvy** · March 9th 2010, 09:04 AM

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.

**Runty** · March 9th 2010, 09:26 AM

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.

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