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Let $r\leq n$ be a natural number and let $\{v_1,v_2,\dots ,v_r\}$ and $\{w_1,w_2,\dots ,w_r\}$ be two linearly independent subsets of $\mathbb{R}^n$ such that $\langle v_i,v_j \rangle = \langle w_i,w_j \rangle \ \forall \ 1\leq i,j \leq r$, where $\langle ,\rangle$ denotes the standard inner product on $\mathbb{R}^n$. Prove that there exists an orthogonal operator $T$ on $\mathbb{R}^n$ such that $T(v_i)=w_i$ for all $1\leq i \leq r$.

An orthogonal mapping preserves inner products, and it is also known that $T$ is orthogonal on $\mathbb{R}^n$ if $\langle T(\alpha),T(\beta)\rangle = \langle \alpha ,\beta \rangle$ for all $\alpha,\beta \in \mathbb{R}^n$. But the question here demands to be proved that the existence of such an orthogonal map. So do I need to find an example of such map or is there another way to prove that in general? Any help will be appreciated.

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You define $Tv_j=w_j$ for all $j=1,\ldots,r$, and since both sets are linearly independent, $T$ gets defined by linearity on their span.

But now you have the problem that maybe $r<n$. Expanding both sets to bases does not work right away, because you have no way to preserve the inner product relations. So what you do is take $\{v_{r+1},\ldots,v_n\}^\perp$ to be an orthonormal basis of $\{v_1,\ldots,v_r\}^\perp$, and $\{w_{r+1},\ldots,w_n\}$ an orthonormal basis of $\{w_1,\ldots,w_r\}^\perp$. Then define $Tv_j=w_j$ for all $j=1,\ldots, n$. The inner product relations are preserved because $\{v_j,v_k\rangle=\delta_{kj}=\langle w_j,w_k\rangle$ for $j,k>r$, and when one index is below $r$ and the other above, the inner product is zero by the orthogonality.

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Let the subspace

$V = \text{span}(\{ v_1, v_2, \ldots, v_r \}) = \text{span}(\{ w_1, w_2, \ldots, w_r \}); \tag 1$

we define $T$ on $V$ by setting

$Tv_k = w_k, \; 1 \le k \le r, \tag 2$

and extending $T$ by linearity to all of $V$, viz:

$T\left ( \displaystyle \sum_1^r \alpha_i v_i \right ) = \displaystyle\sum_1^r \alpha_i Tv_i = \sum_1^r \alpha_i w_i; \tag 3$

that such an extension is possible follows from the linear independence of the $v_i$ and the $w_i$ as is well-known.

If

$\alpha, \beta \in V \tag 4$

we may write

$\alpha = \displaystyle \sum_1^r \alpha_i v_i, \; \beta = \sum_1^r \beta_j v_j, \tag 5$

whence

$\left \langle T\alpha, T\beta \right \rangle = \left \langle T \left (\displaystyle \sum_1^r \alpha_i v_i \right), T \left (\displaystyle \sum_1^r \beta_j v_j \right) \right \rangle = \left \langle \displaystyle \sum_1^r \alpha_i Tv_i, \displaystyle \sum_1^r \beta_j Tv_j \right \rangle = \displaystyle \sum_{i, j = 1}^r \alpha_i \beta_j \langle Tv_i, Tv_j \rangle = \sum_{i, j = 1}^r \alpha_i \beta_j \langle w_i, w_j \rangle = \sum_{i, j = 1}^r \alpha_i \beta_j \langle v_i, v_j \rangle$ $= \left \langle \displaystyle \sum_1^r \alpha_i v_i, \displaystyle \sum_1^r \beta_j v_j \right \rangle = \langle \alpha, \beta \rangle, \tag 6$

which proves that $T:V \to V$ is orthogonal.

In the event that $r < n$, we extend $T$ from $V$ to all of $\Bbb R^n$ as follows: let $W \subset \Bbb R^n$ be the orthogonal compliment of $V$, $W = V^\bot$; then we have

$\Bbb R^n = V \oplus W; \tag 7$

we observe that any $x \in \Bbb R^n$ may be represented uniquely as

$x = v + w, \; v \in V, w \in W; \tag8$

existence of such a decomposition follows directly from (7), i.e. from the fact that $W = V^\bot$; to see uniqueness, suppose

$x = v_1 + w_1 = v_2 + w_2, \; v_1, v_2 \in V, w_1, w_2 \in W; \tag 9$

then

$v_1 - v_2 = w_2 - w_1, \tag{10}$

which forces

$v_1 - v_2 = 0 = w_2 - w_1, \tag{11}$

since

$v_1 - v_2 \in V, \; w_1 - w_2 \in W = V^\bot, \tag{12}$

and hence

$\Vert v_1 - v_2 \Vert^2 = \langle v_1 - v_2, v_1 - v_2 \rangle = \langle v_1 - v_2, w_2 - w_1 \rangle = 0, \tag{13}$

implying $v_1 = v_2$, and thus $w_1 = w_2$ via (11); so the decomposition (8) is unique; this uniqueness means we may unabiguously define an extension $T_E$ of $T$ from $V$ to $V \oplus W = \Bbb R^n$ by

$T_E(v + w) = Tv + w, \; v \in V, w \in W; \tag{14}$

that is,

$T_E = T \oplus I: V \oplus W \to V \oplus W; \tag{15}$

we can formally show $T_E$ is orthogonal on all of $\Bbb R^n$ if we observe that, for $v_1, v_2 \in V$, $w_1, w_2 \in W$,

$\langle v_1 + w_1, v_2 + w_2 \rangle$ $= \langle v_1, v_2 \rangle + \langle v_1, w_2 \rangle + \langle w_1, v_2 \rangle + \langle w_1, w_2 \rangle = \langle v_1, v_2 \rangle + \langle w_1, w_2 \rangle; \tag{16}$

thus,

$\langle T_E(v_1 + w_1), T_E(v_2 + w_2) \rangle = \langle (T \oplus I)(v_1 + w_1), (T \oplus I)(v_2 + w_2) \rangle$ $= \langle Tv_1 + w_1, Tv_2 + w_2 \rangle = \langle Tv_1, Tv_2 \rangle + \langle w_1, w_2 \rangle$ $= \langle v_1, v_2 \rangle + \langle w_1, w_2 \rangle = \langle v_1 + w_1, v_2 + w_2 \rangle, \tag{17}$

which shows that $T_E = T \oplus I$ is orthogonal on $\Bbb R^n$.

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