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Consider the inner product in $\mathbb{R}^3$ with the following orthonormal basis:

$\{(1, 0,-1),(0,1,0),(-1,0,2)\}$

Show that the inner product is defined by the application $<,>:\mathbb{R}^3 . \mathbb{R}^3\rightarrow\mathbb{R}$ is given by:

$<(x_1,x_2,x_3),(y_1,y_2, y_3)>=5x_1y_1+3x_1y_3+x_2y_2+3x_3y_1+2x_3y_3$

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  • $\begingroup$ I'm stuck in the beginning. :/ I don't know what to do and my searches are not helping me. The resolution of the professor just gives the answer straight away without steps. $\endgroup$ – Henrique Dias Dec 6 '17 at 15:48
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Inner product is defined by the matrix A such that: $$<x,x>=x^TAx$$

Let v be a vector orthonormal with respect to the inner product, then:

$$<v,v>=v^TAv=v^Tv$$

Let $Q$ the matrix with an orthonormal basis as column vectors, then with respect to this basis:

$$x^TQ^TAQx=x^Tx$$

Thus:

$$Q^TAQ=I$$

Finally:

$$A=(Q^T)^{-1}Q^{-1}=Q^{-1}Q^{-1}$$

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    $\begingroup$ How did you go from $(Q^T)^{-1}Q^{-1}$ to $Q^{-1}Q^{-1}$? Seems like the final expression should be $(QQ^T)^{-1}$ instead. $\endgroup$ – amd Dec 6 '17 at 19:49
  • $\begingroup$ I think it's because of Q being orthonormal. Then, it is symmetrical. $\endgroup$ – Henrique Dias Dec 6 '17 at 20:46
  • $\begingroup$ yes exactly! On real it’s all symmetric. $\endgroup$ – gimusi Dec 6 '17 at 21:03
  • $\begingroup$ @gimusi Do You say that all matrices $Q$ with an orthonormal basis ( wrt a (symmetric) inner product) as column vectors are symmetric in the sense that $Q^T=Q$? In $\mathbb{R}^2$ take $A=I$ ( the standard inner product) and e.g. $Q=\begin{pmatrix}\sqrt{1/2}&-\sqrt{1/2}\\\sqrt{1/2}&\sqrt{1/2}\end{pmatrix}$ isn´t symmetric. Did I get You wrong? $\endgroup$ – Peter Melech Dec 8 '17 at 12:45
  • $\begingroup$ In this particulare case it's true because Q is symmetric! Maybe I wasn't so clear in my previous quote. Thanks! $\endgroup$ – gimusi Dec 8 '17 at 12:59
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$\textbf{Hint}$: The product can be represented by a matrix $A=(a_{i,j})_{i,j=1,...,3}$ and the condition that the given set is an orthonormal set gives You the conditions $$(1,0,-1)A\begin{pmatrix}1\\0\\-1\end{pmatrix}=1$$ and $$(0,1,0)A\begin{pmatrix}0\\1\\0\end{pmatrix}=1$$ and $$(-1,0,2)A\begin{pmatrix}-1\\0\\2\end{pmatrix}=1$$ and $$(1,0,-1)A\begin{pmatrix}0\\1\\0\end{pmatrix}=0$$ and so on which yields a system of equations in the matrix entries $a_{11},...,a_{33}$ and use the fact that by the symmetry of the inner product the matrix $A$ must be symmetric

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HINT: Let your basis be $\textit{B} = \{\vec{v_1}, \vec{v_2}, \vec{v_3}\}$ and you have the transformation matrix $T = [\vec{v_1}, \vec{v_2}, \vec{v_3}]$ such that $T.\vec{v} = [\vec{v}]_B$

Then $\vec{v} = T^{-1}[\vec{v}]_B$, so $\vec{v}.\vec{u} = (T^{-1}[\vec{v}]_B).(T^{-1}[\vec{u}]_B)$ where $[\vec{v}]_B,[\vec{u}]_B$ correspond to the coordinates of vectors $\vec{v}$ and $\vec{u}$ in the basis $B$, i.e. they are $x_1, x_2, x_3$ and $y_1, y_2, y_3$ in your question.

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