0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.