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I'm working on a problem that states:

Find an orthonormal Basis of $\mathbb{R}^4$ with respect to the bilinear form defined through

$$P= \begin{pmatrix}1&-2&1&-1\\-2&13&-6&4\\1&-6&3&-2\\-1&4&-2&2\end{pmatrix}.$$

I think I have to find $4$ vectors $e_i$ such that $\varphi(e_i,e_j)= e_i^tPe_j=\delta_{ij}$.

My problem is: how do I find these vectors? Do I have to put $e_i=(x,y,z,w)$ and compute with the matrix to get equations?

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  • $\begingroup$ try to diagonalize it $\endgroup$ – janmarqz Apr 7 '20 at 6:12
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    $\begingroup$ Gram-Schmidt process? $\endgroup$ – Angina Seng Apr 7 '20 at 6:14
  • $\begingroup$ nah, you ought to find, first the eigen vectors, with them, you will have a change of basis which carry to you to, maybe, a version of the same inner product with your matrix $P$ but in another frame to achieve your aim $\endgroup$ – janmarqz Apr 7 '20 at 6:22
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    $\begingroup$ The eigenvalues of this matrix aren’t very “nice.” Applying the Gram-Schmidt process to the standard basis, as suggested by @AnginaSeng is probably a better way to go. $\endgroup$ – amd Apr 7 '20 at 7:36
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    $\begingroup$ Thank you all for the methods to approach this problem. At first I didn't understand why diagonalize the Matrix but the answer below made it clear. I decided to use the Gram-Schmidt process and it worked very good! $\endgroup$ – MathematicalMoose Apr 9 '20 at 8:52
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My Try

The standard basis for $\mathbb{R}^4$ is $B=\{e_1,e_2,e_3,e_4\}$ where $e_1=(1,0,0,0), e_2=(0,1,0,0), e_3=(0,0,1,0), e_4=(0,0,0,1)$ and We want to find 4 vectors $c_i$ such that $\phi(c_i,c_j)=c_i^tPc_j=\delta_{ij}$. Let $f(u,v)=u^tAv$ and let $c_1=e_1$. Using Gram-Schmidt method, we get

$c_1=(1,0,0,0)$

$c_2=e_2-\frac{f(e_2,c_1)}{f(c_1,c_1)}c_1=(0,1,0,0)-\frac{-2}{1}(1,0,0,0)=(0,1,0,0)-(-2,0,0,0)=(2,1,0,0)$

$c_3=e_3-\frac{f(e_3,c_1)}{f(c_1,c_1)}c_1-\frac{f(e_3,c_2)}{f(c_2,c_2)}c_2=(0,0,1,0)-\frac{1}{1}(1,0,0,0)-\frac{-4}{9}(2,1,0,0)=(\frac{-1}{9},\frac{4}{9},1,0)$

$c_4=e_4-\frac{f(e_4,c_1)}{f(c_1,c_1)}c_1-\frac{f(e_4,c_2)}{f(c_2,c_2)}c_2-\frac{f(e_4,c_3)}{f(c_3,c_3)}c_3=(0,0,0,1)-\frac{-1}{1}(1,0,0,0)-\frac{2}{9}(2,1,0,0)-\frac{\frac{-1}{9}}{\frac{2}{9}}(\frac{-1}{9},\frac{4}{9},1,0)=(\frac{1}{2},0,\frac{1}{2},1)$.

Then we can normalize them as below:

$i_1=(1,0,0,0)$.

$i_2=\frac{c_2}{||c_2||}=\frac{(2,1,0,0)}{\sqrt{2^2+1^2}}=(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}},0,0)$.

$i_3=\frac{c_3}{||c_3||}=\frac{(\frac{-1}{9},\frac{4}{9},1,0)}{\sqrt{(\frac{-1}{9})^2+(\frac{4}{9})^2+1^2}}=\frac{(\frac{-1}{9},\frac{4}{9},1,0)}{\frac{7\sqrt{2}}{9}}=(\frac{-1}{7\sqrt{2}},\frac{4}{7\sqrt{2}},\frac{9}{7\sqrt{2}},0)$.

$i_4=\frac{c_4}{||c_4||}=\frac{(\frac{1}{2},0,\frac{1}{2},1)}{\sqrt{\frac{1}{4}+\frac{1}{4}+1}}=\frac{(\frac{1}{2},0,\frac{1}{2},1)}{\frac{\sqrt{3}}{2}}=(\frac{1}{\sqrt{3}},0,\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}})$.

Thus the basis of $\mathbb{R}^4$ is $\{i_1,i_2,i_3,i_4\}=\left[ {\begin{array}{*{20}c} 1 & \frac{2}{\sqrt{5}} & \frac{-1}{7\sqrt{2}} & \frac{1}{\sqrt{3}}\\ 0 & \frac{1}{\sqrt{5}} & \frac{4}{7\sqrt{2}} & 0\\ 0 & 0 & \frac{9}{7\sqrt{2}} & \frac{1}{\sqrt{3}}\\ 0 & 0 & 0 & \frac{2}{\sqrt{3}}\\ \end{array} } \right]$

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    $\begingroup$ maybe you should be normalizing the $i_s$ employing the norm defined by the bilinear $P$ instead of the euclidean norm of $\mathbb R^4$, anyway this is an heroic work. Also, you use $f$ instead of $\varphi$. $\endgroup$ – janmarqz Apr 9 '20 at 0:51
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    $\begingroup$ Thank you very much both of you. I did this and the only "change" I did was taking the billinear form as norm and not the Euclidean. $\endgroup$ – MathematicalMoose Apr 9 '20 at 8:49
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    $\begingroup$ @janmarqz You are absolutely right! $\endgroup$ – xzyao Apr 9 '20 at 10:36
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Your matrix once diagonalized via $S^{\top}PS$ approximately results in $$ \tilde{P}= \left( \begin{array}{cccc} 9. & 0 & 0 & 0 \\ 0 & 2.94468 & 0 & 0 \\ 0 & 0 & 0.825505 & 0 \\ 0 & 0 & 0 & 218.291 \\ \end{array} \right), $$ where $S$ is a matrix of eigenvectors' original matrix $P$.

From $\tilde{P}$ anyone can guess that the orthonormal base should be $$ v_1=\left( \begin{array}{c} \frac{1}{3}\\ 0\\ 0\\ 0 \end{array} \right )\quad ,\quad v_2=\left( \begin{array}{c} 0\\ \frac{1}{\sqrt{2.94468}}\\ 0\\ 0 \end{array} \right) \quad ,\quad\left( v_3=\begin{array}{c} 0\\ 0\\ \frac{1}{\sqrt{0.825505}}\\ 0 \end{array} \right) \quad ,\quad v_4=\left( \begin{array}{c} 0\\ 0\\ 0\\ \frac{1}{\sqrt{ 218.291}} \end{array} \right), $$ because they satisfy $v_i^{\top}\tilde{P}v_j=\delta_{ij}$.

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  • $\begingroup$ the matrix $S^{\top}PS$ contrast with $S^{-1}PS$ which gives a diagonalization with the eigenvalues of $P$ in its diagonal $\endgroup$ – janmarqz Apr 8 '20 at 20:08

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