So, I had a polynomial given as follows: $f=x_1x_2-x_2x_3$

( The task is to transform the quadratic form into canonical form, but I have to do it by using main theorem, I cannot do it by simply finding eigenvalues and then do a diagonal matrix with the corresponding values.

I constructed a corresponding matrix:

$$\begin{pmatrix}0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 0\end{pmatrix}$$

Also, how can I determine the rank of the form, if there are three different eigenvalues? And one of them is zero?

Then I found eigenvalues: $\lambda_1=0$ and $\lambda_{2,3}=\pm\frac{1}{\sqrt{2}}$

Then I kind of struggle with eigenvectors.

First, with the $\lambda_1=0$
$\frac{1}{2}y_1-\frac{1}{2}y_3=0 $ Suppose that $y_1=1$ then $y_3=1$ So the eigenvalue for $\lambda_1=0$ is (1,0,1)

Then with the $\lambda_2=+\frac{1}{\sqrt{2}}$ the matrix looks like

$$\begin{pmatrix}-\frac{1}{\sqrt{2}} & \frac{1}{2} & 0 \\\frac{1}{2} &-\frac{1}{\sqrt{2}} & -\frac{1}{2} \\0 & -\frac{1}{2} & -\frac{1}{\sqrt{2}}\end{pmatrix}$$

So, supposing that $y_1=\frac{1}{\sqrt{2}}$, $y_2=1$ and $y_3= \frac{1}{\sqrt{2}}$.

So the eigenvector when $\lambda_2=+\frac{1}{\sqrt{2}}$ is $(\frac{1}{\sqrt{2}}, 1, \frac{1}{\sqrt{2}}$) After some source it seems as the values for this eigenvector are incorrect.

For the third eigenvalue $\lambda_3=-\frac{1}{\sqrt{2}}$ the matrix looks like:

$$\begin{pmatrix}\frac{1}{\sqrt{2}} & \frac{1}{2} & 0 \\\frac{1}{2} &\frac{1}{\sqrt{2}} & -\frac{1}{2} \\0 & -\frac{1}{2} & \frac{1}{\sqrt{2}}\end{pmatrix}$$

If we suppose that $y_1=-\frac{1}{\sqrt{2}}$ then $y_2=1$ and $y_3=\frac{\sqrt{2}}{2}$

So that the eigenvector for $\lambda_3=-\frac{1}{\sqrt{2}}$ equals to $(-\frac{1}{\sqrt{2}},1,\frac{\sqrt{2}}{2})$

I suppose that computed values are incorrect, if so, can you give me the correct way to find them? Secondly how from there can I construct the canonic form? ( My main task was to transform the given polynomial from quadratic form to canonical form, but it was to be done by using the theorem).

  • $\begingroup$ It would be useful if you could give more context. What is the main theorem ? What exactly are you supposed to do ? What is the ground field ? $\endgroup$ – Captain Lama May 30 at 21:00
  • $\begingroup$ Very similar question to the one you have asked four hours ago.... math.stackexchange.com/q/3245417 $\endgroup$ – Jean Marie May 30 at 21:06
  • $\begingroup$ @CaptainLama 1) Have to find eigenvalues. 2)I need to write a canonical form for a given quadratic form in the orthonormal eigenvector basis of matrix A ( the one constructed from given quadratic form). 3) Quadratic form in this base is diagonal matrix. I think that either I haven't computed the eigenvectors correctly or they are not orthogonal in this case. $\endgroup$ – Ieva Brakmane May 30 at 22:13

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