If \(\mathbf{b}=(v_1,\ldots,v_n)\) is an ordered basis of an \(\mathbb{R}\)-vector space \(V,\) then there is always an inner product \(\langle\cdot{,}\cdot\rangle: V\times V\to \mathbb{R}\) such that \(\mathbf{b}\) is an ordered orthonormal basis of \(V\) with respect to \(\langle\cdot{,}\cdot\rangle.\)

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The functions \(f_1(x) = x^2+1\) and \(f_2(x)=x\) are orthogonal with respect to the inner product \(\langle\cdot{,}\cdot\rangle:\mathsf C([-1,1])\times \mathsf C([-1,1])\to \mathbb{R}\) defined by \[\langle f,g\rangle = \int_{-1}^{1}f(x)g(x)\mathrm dx\]

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• False

If \(\mathbf{A}\in M_{n,n}(\mathbb{R})\) is symmetric, it admits a decomposition \(\mathbf{A}=\mathbf{C}^T\mathbf{C},\) where \(\mathbf{C}\in M_{n,n}(\mathbb{R})\) is upper triangular.

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• False

Any \(n\)-dimensional Euclidean space admits an orthonormal set with \(n\) elements.

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• False

An orthonormal set \(\mathcal S\subset V\) in a Euclidean space \((V, \langle\cdot{,}\cdot\rangle)\) is a basis for the subspace it generates.

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• False

Given a unit vector \(v\) in a Euclidean space \((V,\langle\cdot{,}\cdot\rangle)\) one can always generate an orthonormal basis of \(V\) which contains \(v.\)

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• False

In a Euclidean space \((V,\langle\cdot{,}\cdot\rangle),\) two vectors \(u\) and \(v\) are linearly independent if \(\langle u,v\rangle =0.\)

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• False

In a Euclidean space \((V,\langle\cdot{,}\cdot\rangle),\) two non-zero vectors \(u\) and \(v\) are linearly independent if \(\langle u,v\rangle =0.\)

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• False

Let \(U\) be a subspace of a Euclidean space \((V,\langle\cdot{,}\cdot\rangle).\) If \(\mathbf{b}=(v_1,\ldots,v_k)\) is an ordered orthonormal basis of \(U\) and \(\mathbf{c}= (v_{k+1},\ldots, v_n)\) is an ordered basis of \(U^\perp,\) then \((v_1,\ldots,v_n)\) is an ordered orthonormal basis of \(V.\)

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• False