How do I find the basis of $R^3$ in respect of the below inner product:

$\left \langle (x,y,z)(x',y',z') \right \rangle=3xx'+3yy'+zy'+z'y+zz'$

I tried this but I'm not sure it's correct

Let $B=${$(1,0,0),(0,1,0),(0,0,1)$} be a basis of $R^3$ then the Gram matrix of the basis in respect of the inner product I'm given is:

$\begin{pmatrix} \left \langle e_1,e_1 \right \rangle & \left \langle e_1,e_2 \right \rangle & \left \langle e_1,e_3 \right \rangle\\ \left \langle e_2,e_1 \right \rangle& \left \langle e_2,e_2 \right \rangle & \left \langle e_2,e_3 \right \rangle\\ \left \langle e_3,e_1 \right \rangle & \left \langle e_3,e_2 \right \rangle& \left \langle e_3,e_3 \right \rangle \end{pmatrix}$,

$e_1=(1,0,0)$ $e_2=(0,1,0)$ $e_3=(0,0,1)$ and $<,>$ the above inner product

I Calculated the eigenvectors of the above matrix and create a basis using them.

Is this a correct solution? Thanks in advance!

  • $\begingroup$ What do you mean by "basis with respect to an inner product"? For a scalar product, you can calculate the matrix for your given basis and then, since the matrix is positive definite, make a basis transform such that the new matrix of the scalar product with respect to the new basis becomes the identity matrix... $\endgroup$ – mol3574710n0fN074710n May 7 '18 at 20:01
  • $\begingroup$ @SlimJim Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi May 31 '18 at 20:14

With respect to a chosen basis we can express the matrix associated to the inner product by

$$B_e=\begin{pmatrix} \left \langle e_1,e_1 \right \rangle & \left \langle e_1,e_2 \right \rangle & \left \langle e_1,e_3 \right \rangle\\ \left \langle e_2,e_1 \right \rangle& \left \langle e_2,e_2 \right \rangle & \left \langle e_2,e_3 \right \rangle\\ \left \langle e_3,e_1 \right \rangle & \left \langle e_3,e_2 \right \rangle& \left \langle e_3,e_3 \right \rangle \end{pmatrix}\implies \vec x^TB_e \vec x'$$

Since the matrix is symmetric then (by spectral theorem) an orthogonal matrix $M$ exists, which has the eigenvectors as column, such that $B_v=M^TB_eM$ is diagonal with respect to that basis.

  • 1
    $\begingroup$ Could even make it hermitian, if you want the most general case, just replace the "T" by "*" or whatever you like to use for conjugate transpose. ;) $\endgroup$ – mol3574710n0fN074710n May 7 '18 at 20:07
  • $\begingroup$ @mol3574710n0fN074710n Thanks for pointing it out, I refer to real matrices since the OP was asking about that and also I'm more familiar with that. But your comment is very useful! $\endgroup$ – gimusi May 7 '18 at 20:09

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