Let {𝐮1, 𝐮2, …, 𝐮n} be an orthonormal basis for R^n. Prove that 𝐯^2=|𝐯 ·𝐮1|^2+|𝐯 ·𝐮2|^2+⋯+|𝐯 ·𝐮n

step 1 Hi, in this question we're given an orthonormal basis for Rn and we want to prove that the norm

Let {𝐮1, 𝐮2, …, 𝐮n} be an orthonormal basis for R^n. Prove...

Key Concepts

Parseval's Equality

Parseval's equality is a result in inner product spaces which states that the total energy (squared norm) of a vector is equal to the sum of the squares of its coefficients with respect to an orthonormal basis. It encapsulates the idea that the representation of a vector in terms of an orthonormal basis fully accounts for its magnitude, with no loss of information in the transformation.

Orthonormal Basis

An orthonormal basis in a vector space is a set of vectors that are all mutually orthogonal and each of unit length. This property ensures that any vector in the space can be uniquely expressed as a linear combination of these basis vectors. Their orthogonality simplifies many calculations, including inner products, because the dot product of any two different basis vectors is zero.

Vector Decomposition

Any vector in an n-dimensional space can be written as a linear combination of the vectors in an orthonormal basis. The coefficients of this decomposition are given by the inner products of the vector with the basis vectors. This process is crucial because it translates properties of the vector, like its norm, into a sum involving these coefficients.

Inner Product

The inner product, or dot product, is a fundamental operation in vector spaces that, when applied to two vectors, provides a measure of their angle and magnitude relationship. In the context of an orthonormal basis, the inner product enables the extraction of the component (projection) of a vector along a given basis vector.

Pythagorean Theorem in Inner Product Spaces

When a vector is decomposed into orthogonal components, the Pythagorean theorem states that the square of the norm (length) of the vector is equal to the sum of the squares of the norms of the individual components. This property is central to the proof of Parseval's equality, as it allows us to equate the squared norm of the original vector with the sum of the squares of its projections onto the basis vectors.

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