# Questions tagged [orthonormal]

For questions related to orthonormality. A set of vectors in an inner product space is called orthonormal if each vector is a unit vector, and vectors are pairwise orthogonal.

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### Best approximation of a function among closed linear manifolds

Let $H$ be an infinite-dimensional Hilbert space and consider a $n-dimensional$ closed linear manifold generated by a subset of orthonormal basis, say, $M = span(\{u_1,u_2,\cdots,u_n\})$. Of course, ...
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### number of elements of a basis of a subspace of R4

If I've got $U$ a subspace of $\mathbb{R}^4$ $$U = < (1,-1,0,0),(0,1,1,1),(2,1,0,1) >$$ And I want to find an orthonormal base for the subspace $U$ My doubt is: Can I make an orthonormal ...
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### Adjoint of right shift operator on orthonormal basis $(e_{n})_{n\in\mathbb{N}}$ of $\ell^{2}(\mathbb{N})$

I'm sorry if this question is a duplicate. Suppose $(e_{n})_{n\in\mathbb{N}}$ is the usual orthonormal basis of $\ell^{2}(\mathbb{N})$. We can define an operator $v\colon H\to H$ by $ve_{n}:=e_{n+1}$....
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### Prove: $\|x\|^2=\sum_{i=1}^k|\langle x,e_i\rangle|^2\iff x\in\operatorname{span}\{e_1,\ldots,e_k\}$

Let $\{e_1,\ldots,e_k\}$ be an orthonormal set in a unitary space $V$. Prove: $$\|x\|^2=\sum\limits_{i=1}^k|\langle x,e_i\rangle|^2\iff x\in\operatorname{span}\{e_1,\ldots,e_k\}$$ My attempt: My ...
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### Orthonormal sequence and Gram schmidt

Let $V$ be a Inner product space, $(\epsilon_1....\epsilon_k)$ orthonormal sequence, let $\vec{v_1},\vec{v_2}\in V$. $\quad$ $GS(u_1...u_k)$ is the set of the vectors that recieved after doing GS ...
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### Orthonormal basis with parameters

I have the set : $$W=[(x_1,x_2,x_3,x_4)\in{\mathbb{R}^4}\quad| \quad x_1-x_2-x_3-x_4=0]$$ I need to find orthonormal basis, So I found just one vector and I have only parameters in my answer and I am ...
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### Use the Gram-Schmidt process to find the orthonormal basis for the row space of the matrix $A$.

Use the Gram-Schmidt process to find the orthonormal basis for the row space of the matrix $A$. The matrix $A$ is as follows: \begin{bmatrix}1&1&0&0\\-1&3&0&1\\-3&1&-...
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