- #1

icystrike

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U⊥ = {w ∈ Rn : w is orthogonal to U} .

Prove that

(i) U⊥ is a subspace of Rn,

(ii) dimU + dimU⊥ = n.

__Attempt.__

i)

U. ( U⊥)T=0

If U⊥ does not passes the origin , the above equation cannot be satisfied.

Therefore U⊥ passes the origin.

U.( U⊥+ U⊥)T=U. ( U⊥)T+U.( U⊥)T=0+0=0

U.(k U⊥)T=k[U. (U⊥)T]=k.0=0

Therefore U⊥ is a subspace in Rn

ii) let U have rank r.

U⊥ is the nullspace of the U transpose.

and if U is a matrix of mxn , UT is nxm.

Therefore , the sum of dim of the two matrices is exactly n.