Let V be a finite-dimensional vector space and let \mathcal{B}=\left\{\alpha_{1}, \dots, \alpha_{n}\right\}B={α 1…,α n } be a basis for V. Let (\quad | \quad)(∣) be an inner product on V. If c_{1}, \dots, c_{n}c 1,…,c nare any n scalars, show that there is exactly one vector \alphaα in V such that \left(\boldsymbol{\alpha} | \boldsymbol{\alpha}_{i}\right)=c_{i}, j=1, \ldots, n(α∣α )=c ij=1,…,n.

Let V be a finite-dimensional vector space and let B ={α 1…,α n } be a basis for V . Let ( | ) be the inner product of V. If c 1,...,c n , are n is a scalar, then show that there exist one vector α∈ V such that (α∣αj)=cj, j=1 ,…,n.

We have given that V is a finite dimensional vector space.

Consider α∈ V defined as

α = α₁β₁ + α₂β₂ + -----+ αₙβₙ where,

β₁ = α₁

β₂= α₂ - (α₁ |β₁)/β₁/β₂ |(β₁)

similarly, __

βⱼ = αⱼ - ( ∑ (αⱼ|βⱼ) / βᵢ|(βᵢ) βᵢ ) , j = 2,n bar

As defined aᵢ, i = 1, j bar

α₁ = c₁/ (β₁ )|β₁

___

α₂ = c₁/ (β₁ )|β₁ - a₁ ( α₂|β₂/β₁|β₂)

proceeding like as we get

_______

αⱼ = 1/β₁|(β₁) (cⱼ - ∑ αⱼ ( αⱼ |βⱼ)

then conider

(α|αⱼ )= ( α₁β₁ + α₂β₂ + -- + αₙβₙ |αⱼ )

= α₁β₁ + α₂β₂ + -- + αₙβₙ |βⱼ ) + ∑ α₁β₁+ α₂β₂ +---+αⱼβᵢ /βᵢβᵢ)

______

(α|αⱼ) =αⱼ(βⱼ |βⱼ) + (cⱼ - ∑ αⱼ ( αⱼ |βⱼ) βⱼ |βⱼ

__ ___

= (cⱼ - ∑ αᵢ ( αⱼ |βᵢ) + ∑αⱼ |βᵢ

(α|αⱼ) = cj, j = 1,n

Hence proved.

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