1. ## Orthogonal projection

I'm unable to do the following problem : Let $\displaystyle V$ be a vector space over $\displaystyle \mathbb{K}$ whose dimension is $\displaystyle n$ and with an inner product $\displaystyle <,>$. Let $\displaystyle W$ be a subspace of $\displaystyle V$ with dimension $\displaystyle m$ and $\displaystyle \{ w_1,...,w_m \}$ be an orthonormal basis of $\displaystyle W$.
We define the orthogonal projection over $\displaystyle W$ as $\displaystyle P_W(v)=\sum_{i=1}^{m} <v,w_i>w_i$.
Demonstrate that $\displaystyle ||v-P_W(v) || \leq || v-w||$, $\displaystyle \forall w \in W$.
My attempt : I suppose it's true and want to see that it's really true ( don't want to fall over any contradiction).
So $\displaystyle ||v-P_W(v) || \leq || v-w||$ $\displaystyle \Leftrightarrow ||v-P_W(v) ||^2 \leq || v-w||^2$ $\displaystyle \Leftrightarrow <v,-P_W(v)> \leq <v,-w>$. And I didn't reach anything.

2. Originally Posted by arbolis
I'm unable to do the following problem : Let $\displaystyle V$ be a vector space over $\displaystyle \mathbb{K}$ whose dimension is $\displaystyle n$ and with an inner product $\displaystyle <,>$. Let $\displaystyle W$ be a subspace of $\displaystyle V$ with dimension $\displaystyle m$ and $\displaystyle \{ w_1,...,w_m \}$ be an orthonormal basis of $\displaystyle W$.
We define the orthogonal projection over $\displaystyle W$ as $\displaystyle P_W(v)=\sum_{i=1}^{m} <v,w_i>w_i$.
Demonstrate that $\displaystyle ||v-P_W(v) || \leq || v-w||$, $\displaystyle \forall w \in W$.
My attempt : I suppose it's true and want to see that it's really true ( don't want to fall over any contradiction).
So $\displaystyle ||v-P_W(v) || \leq || v-w||$ $\displaystyle \Leftrightarrow ||v-P_W(v) ||^2 \leq || v-w||^2$ $\displaystyle \Leftrightarrow <v,-P_W(v)> \leq <v,-w>$. And I didn't reach anything.
Okay, here's how I would do it:

Let $\displaystyle \{w_1,...,w_m \} \subset \{v_1,...,v_n \}$ where $\displaystyle w_i=v_i$ for $\displaystyle 1 \leq i \leq m$ and $\displaystyle \{v_i \}$ is an orthonormal basis for $\displaystyle V$, and so we have $\displaystyle v= \sum_{i=1}^ n \ {<v,v_i>v_i}$ and so if $\displaystyle u=P_W (v)= \sum_{i=1}^ m \ {<v,v_i>v_i}$ :

$\displaystyle \Vert v-u \Vert ^2 = \Vert \sum_{i=1}^ n \ {<v,v_i>v_i} - \sum_{i=1}^m \ {<v,v_i>v_i} \Vert ^2 = \Vert \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2$

Now, pick $\displaystyle w \in W$ then $\displaystyle w= \sum_{i=1}^ m \ {<w,v_i>v_i}$ and so:

$\displaystyle \Vert v-w \Vert ^2 = \Vert \sum_{i=1}^ m \ {<v-w,v_i>v_i} + \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2$ $\displaystyle = \Vert \sum_{i=1}^ m \ {<v-w,v_i>v_i} \Vert ^2 + \Vert \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2 \geq \Vert \sum_{i=m+1}^ n \ {<v,v_i>v_i} \Vert ^2$ $\displaystyle = \Vert v-u \Vert ^2$

Another thing, in your post you have $\displaystyle \Vert a-b \Vert ^2 = <a,-b>$ but it's actually $\displaystyle <a,a> - <a,b> - <b,a> + <b,b>$.

3. Thank you very much Jose. I'll think about it tomorrow. To tell you the truth I was so tired when I posted my thread that I actually didn't think a lot but rather used my short term memory to post what I did today.
I need a rest. My exam is on Monday so I'm pushing hard!
See you!