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Let $\{\tilde w_1,...,\tilde w_n\}$ be a linearly independent set in $\mathbb{R}^n$ and let $\{\tilde v_1,...,\tilde v_n\}$ denote the orthonormal basis obtained by performing the Gram-Schmidt procedure on $\{\tilde w_1,...,\tilde w_n\}$ (which is normalized).

Prove that the matrix $T$ defined by $t_{ij} = \langle \tilde w_j, \tilde v_i \rangle \quad \forall 1 \leq i, j \leq n$ is invertible.

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  • $\begingroup$ How is $T$ defined? What inner product do you mean? $\endgroup$ – DGRasines Jun 18 '14 at 14:30
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    $\begingroup$ I edited your question, please make sure that I didn't change the meaning. @dgrasines517, I think we can assume the standard scalar product, i.e. $\langle u,v\rangle = \sum u_iv_i$ $\endgroup$ – k1next Jun 18 '14 at 14:38
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I'll drop the tilde. Gram-Schmidt creates an orthonormal basis, thus we know that the matrix \begin{align} Q = \Bigl( v_1 |...|v_n\Bigr) \end{align} is orthonormal.

The $w_j$ are linearly independent, thus the matrix \begin{align} W=\Bigl(w_1|...|w_n\Bigr) \end{align} is invertible.

$t_{ij}$ is $\langle w_i,v_j\rangle$ and thus $T=W\cdot Q$

Then $\det(T)=\det(W\cdot Q)=\det(W)\cdot \det(Q) =\det(W)\cdot 1 \not =0$ since $W$ is invertible and $Q$ orthonormal. Thus, $T$ is invertible.

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