Question

There exists a subspace Vof${\mathbf{ℝ}}^{\mathbf{5}}$ such that$\mathit{d}\mathit{i}\mathit{m}\mathbf{\left(}\mathit{V}\mathbf{\right)}\mathbf{=}\mathit{d}\mathit{i}\mathit{m}\mathbf{\left(}{\mathit{V}}^{\mathbf{\perp }}\mathbf{\right)}$ where${\mathit{V}}^{\mathbf{\perp }}$ denotes the orthogonal complementof V.

Short answer

False

Step-by-step solution

Step by step solution Step 1: Orthogonal complement

In themathematicalfields oflinear algebraandfunctional analysis, the orthogonal complement of asubspaceW of avector spaceV equipped with abilinear formB is the set${\mathit{Q}}^{\mathbf{T}}\mathit{Q}\mathbf{=}\mathit{Q}{\mathit{Q}}^{\mathbf{T}}\mathbf{=}\mathit{I}$of all vectors in V that areorthogonalto every vector in W. Informally, it is called the prep, short for perpendicular complement. It is a subspace of V.

Determine whether the statement is true or false

For example,

Let V be a vector space over a field F equipped with a bilinear form B. We define u to be left-orthogonal to V, and v to be right-orthogonal to u when B(u, v)=0.

For a subset W of Vdefine the left orthogonal complement $W^{\perp }$to be:

$W^{\perp }=\{x\in V:B(x,y)=0$

Thus,

$\dim \left(V\right)+\dim \left(V^{\perp }\right)=5$

Which means that one is even and the other one is odd.

Hence, the statement is false.