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Can somebody tell me where I made a mistake? How would you approach such an exercise? Was my way too complicated?

$ B:\mathbb{R}^3\times \mathbb{R}^3\to \mathbb{R},\quad B(x,y)=\sum \limits_{i\neq j}^{}x_{i}y_{j}+2\sum \limits_{i=1}^{3}x_{i}y_{i}=x_{1}\cdot (2y_{1}+y_{2}+y_{3})+x_{2}\cdot (y_{1}+2y_{2}+y_{3}) +x_{3}\cdot (y_{1}+y_{2}+2y_{3}) $

The exercise was to first determine an orthonormal basis for $(1,1,1)^{\perp }$ ( the orthogonal complement of (1,1,1)) regarding $B$.

So I guessed $v:=(-1,1,0)^T$ as one orthogonal vector of $(1,1,1)$, which is true because $B(\begin{pmatrix} 1\\1\\1 \end{pmatrix}, \begin{pmatrix} -1\\1\\0 \end{pmatrix}) =0 $

Then I did the following:

(i) $B(\begin{pmatrix} 1\\1\\1 \end{pmatrix}, \begin{pmatrix} y_{1}\\y_{2}\\y_{3} \end{pmatrix}) = 4y_{1}+4y_{2}+4y_{3} =0 $

(ii) $B(\begin{pmatrix} -1\\1\\0 \end{pmatrix}, \begin{pmatrix} y_{1}\\y_{2}\\y_{3} \end{pmatrix})= -y_{1}+y_{2} =0 \leftrightarrow y_{1}=y_{2}$

So I defined $y_{1}=y_{2}=\mu $ with $\mu\in \mathbb{R}$. If you put that value in (i) you get $y_{3}=-2\mu $, $w:= \mu \begin{pmatrix} 1\\1\\-2 \end{pmatrix})$

Now I calculated $B(\begin{pmatrix} 1\\1\\1 \end{pmatrix}, \begin{pmatrix} 1\\1\\1 \end{pmatrix}) =12, B(\begin{pmatrix} -1\\1\\0\end{pmatrix}, \begin{pmatrix} -1\\1\\0 \end{pmatrix}) =2, B(\begin{pmatrix} 1\\1\\-2 \end{pmatrix}, \begin{pmatrix} 1\\1\\-2 \end{pmatrix}) =6\\ \rightarrow\Vert (1,1,1)^T \| =\sqrt{12}, \quad \Vert (-1,1,0)^T \| =\sqrt{2},\quad \Vert (1,1,-2)^T \| =\sqrt{6}$

So $C=(\frac{(-1,1,0)^T}{\sqrt{2}}, \frac{(1,1,-2)^T}{\sqrt{6}})$ is an orthonormal basis for $(1,1,1)^{\perp }$ regarding B.

The next exercise was to determine a orthonormal basis of $\mathbb{R}^3$ regarding $B$.

Since $D=(\frac{(1,1,1)^T}{\sqrt{12}},\frac{(-1,1,0)^T}{\sqrt{2}}, \frac{(1,1,-2)^T}{\sqrt{6}}) $are linear independent, D should be a orthonormal basis of $\mathbb{R}^3$ regarding $B$.

But then $Q$ with the vectors of $D$ as rows should be an orthogonal matrix, so that $Q\cdot Q^{T}=E_{3}$, with $E_{3}$ being the 3x3 identity matrix.

$\begin{pmatrix} \frac{\sqrt{12}}{12} & \frac{-\sqrt{2}}{2} & \frac{\sqrt{6}}{6} \\ \frac{\sqrt{12}}{12} &\frac{\sqrt{2}}{2} &\frac{\sqrt{6}}{6} \\\frac{\sqrt{12}}{12}&0&\frac{-\sqrt{6}}{3} \end{pmatrix}\cdot \begin{pmatrix} \frac{\sqrt{12}}{12} &\frac{\sqrt{12}}{12} &\frac{\sqrt{12}}{12} \\ \frac{-\sqrt{2}}{2} &\frac{\sqrt{2}}{2} &0 \\\ \frac{\sqrt{6}}{6}& \frac{\sqrt{6}}{6}&\frac{-\sqrt{6}}{3} \end{pmatrix} \neq E_{3}$

That is the reason why I think that I made a mistake.

PS: I'm not used to writing about math in English; please ask if something doesn't make sense to you.

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You're so close. When you write $$ Q^t Q = I, $$ that means that the dot-product of the $i$th column of q with the $j$th column of $Q$ is $\delta_{ij}$, the $ij$-element of the identity. But the "dot product" here is the ordinary dot product. You want to make this claim using YOUR inner product. For that, you aim to get $$ Q^t B Q = I $$ where $B$ is the matrix representing your inner product. If you need to do it with rows instead of columnss, then you're hoping for $$ Q B Q^t = I. $$

Post-comment additions You've got the matrix $Q$ transposed. You said that you wanted the basis vectors as rows of $Q$, but your last equation has them as columns. This isn't very important, however.

The matrix $Q$ (with your vectors as rows) has the property that $$ Q B Q^t = I $$ You can see that here numerically, from a matlab program I wrote. (In matlab, multiplying means "matrix multiply", and A' means "the transpose of matrix A". A semicolon at the end of a line suppresses output.)

B = [2 1 1; 1 2 1; 1 1 2]
d = [1 1 1]
v = [-1 1 0]
v * B * d'
w = [1 1 -2]
w * B * d'
v * B * v'
w * B * w'
d * B * d'
v = v/sqrt(2); 
w = w/sqrt(6);
d = d/sqrt(12);
Q = [d;v;w]
Q * B * Q'

The output of this program is just what you'd expect (except that I've added comments after //, and deleted some blanks lines):

B =    
     2     1     1
     1     2     1
     1     1     2    
d =
     1     1     1
v =    
    -1     1     0
ans =    
     0  // so inner prod (using B) of d and v is zero
w =    
     1     1    -2
ans =    
     0  // same for inner prod of d and w
ans =    
     2  // v has squared-length 2 (using the B inner product!)
ans =    
     6  // w has squared-length 6  
ans =    
    12  // d has squared-length 12
// we divide each by its (B-based) length to get rows of Q
// Notice that the length of these rows, with usual dot-prod, is not 1!
Q =    
    0.2887    0.2887    0.2887
   -0.7071    0.7071         0
    0.4082    0.4082   -0.8165    
ans =    // so Q B Q^t is the identity (up to small numerical differences)
    1.0000         0   -0.0000
         0    1.0000         0
   -0.0000   -0.0000    1.0000

You've raised the concern that "det Q" is not 1, which it should be if $Q$ is orthogonal...and that's correct, if it's orthogonal with respect to the usual inner product. But it's actually orthogonal with respect to the inner product defined by $B$, which gives an altogether different result, as you discovered.

In short: aside from using columns instead of rows when you built $Q$, everything you did is just fine. Your matrix $Q$ really is orthogonal (in the $B$-metric).

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  • $\begingroup$ But the representing Matrix $B$ is the identity matrix, because D is an orthonormal basis, and then there would not be a difference to $QQ^t = I$ since $Q^t B Q =Q^t I Q =Q^t Q = I$, or am I wrong? $\endgroup$ – CherryBlossom1878 Mar 11 at 19:26
  • $\begingroup$ Yes, you are wrong. In the standard basis, the matrix for your inner product is $\pmatrix{2 & 1 & 1\\ 1 & 2 & 1 \\ 1 & 1 & 2}$; you can read the $ij$ coefficient in this matrix as the coefficient of $x_i y_j$ in your very first displayed equation. $\endgroup$ – John Hughes Mar 11 at 19:29
  • $\begingroup$ BTW, your question was written really nicely, and your mathematical english was entirely understandable. $\endgroup$ – John Hughes Mar 11 at 19:29
  • $\begingroup$ One way to check this is the right matrix is to compute $\langle\pmatrix{1&0&0}, \pmatrix{0 & 1 & 0} \rangle$, which I claim is $\pmatrix{1&0&0}\pmatrix{2 & 1 & 1\\ 1 & 2 & 1 \\ 1 & 1 & 2}\pmatrix{0\\1\\0} = 1$, which is the correct value, while the ordinary dot product of these two vectors is $0$. (Of course, to really check, you need to do inner products of all pairs of basis vectors, not just one example!) $\endgroup$ – John Hughes Mar 11 at 19:32
  • $\begingroup$ Thank you for the positive feedback and your answer :) So M:=$\pmatrix{2 & 1 & 1\\ 1 & 2 & 1 \\ 1 & 1 & 2}$ is a representing Matrix of $B$ in relation to the standard basis of $\mathbb{R}^3$ and I actually computed $Q^tMQ=I$, so my basis D seems to be a orthonormal basis of $\mathbb{R}^3. $ What I don't understand is why you choose M in relation to the standard basis? I thought that I need to choose M in relation to my basis D, so that in this case the gramian matrix M is the identity matrix. $\endgroup$ – CherryBlossom1878 Mar 11 at 21:00

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