Question

linear algebra: orthogonal complements?

Let V be an inner product space. Show that the orthogonal complement of V is the zero subspace and the orthogonal complement of the zero subspace is V itself.

It says in the solutions that "if v is in V-perp, then (v, v)=0". I don't get this statement. How is the dot product of v and v become a 0 if they are both in V-perp? Shouldn't it be that only orthogonal vectors have a 0 from the dot product?

Answer 1

are they both supposed to be \(v\)

Answer 2

I see now...we are not dealing with a subspace of \(V\)...\[V^\perp=\{v\in V| <v,w>=0\,\,\, \forall w\in V\}\]

Answer 3

only the zero vector has the prop... \[<\vec{0},\vec{0}>=0\]

Answer 4

\[\{0\}^\perp=\{v\in V| <v,w>=0\,\,\, \forall w\in \{0\}\}\]

\[=\{v\in V| <v,0>=0\}=V\]

Answer 5

I still don't get how v with itself can become 0...:/

Answer 6

\[<v,v>=0\,\,\text{ iff }\,\,v=0\]

Answer 7

so in the statement "if v is in V-perp, then (v, v)=0", it implies that v=0?

Answer 8

if \(v\in V^\perp\) then \(<v,v>=0 \Rightarrow v=0\)

Answer 9

yes

Answer 10

let me say it another way


if \(v\in V^\perp\) then for all \(w\in V\) we have that \(<w,v>=0\)

since \(w\) could be \(v\) we also have \(<v,v>=0\), but one of the properties on an inner product is that \(<v,v>=0 \Leftrightarrow v=0\)

Answer 11

Oh, so in the textbook, I guess they assumed that w could be v and hence (v, v)=0.. Thanks so much Zarkon!

Answer 12

np