i have this matrix: \begin{eqnarray} \nonumber \Pi_O(Ti) &=& \frac{ma^2 }{3}\left(% \begin{array}{lll} \frac{1}{2}& -\frac{1}{4}& 0 \\ -\frac{1}{4}& \frac{1}{2}& 0 \\ 0& 0& 1 \end{array} \right)_{(O, \vec x, \vec y, \vec z)} \end{eqnarray}

using this equation: \begin{eqnarray} \nonumber \det(\Pi_O(Ti)-\lambda\cdot I)=|\Pi_O(Ti)-\lambda\cdot I|=0 \end{eqnarray} the eigenvalues of this matrix are obtained as follows \begin{eqnarray} \nonumber\left\{\begin{array}{lll} \lambda_1=\frac{ma^2}{6}+\frac{ma^2}{12}=\frac{ma^2}{4}\\ \lambda_2=\frac{ma^2}{6}-\frac{ma^2}{12}=\frac{ma^2}{12}\\ \lambda_3=\frac{ma^2}{3} \end{array} \right. \end{eqnarray} son, we can write the diagonal matrix of $\Pi_O(Ti)$, \begin{eqnarray} \nonumber D &=& \left( \begin{array}{ccc} \lambda_1 &0 &0 \\ 0 & \lambda_2 &0 \\ 0 &0 & \lambda_3 \end{array} \right) \\ \nonumber &=& ma^2\left( \begin{array}{ccc} \frac{1}{4} &0 &0 \\ 0 & \frac{1}{12} &0 \\ 0 &0 & \frac{1}{3} \end{array} \right) \end{eqnarray}

when I try to compute the diagonal matrix from the eigenvectors of the matrix $\Pi_O(Ti)$, by using the following equation:

\begin{eqnarray} \nonumber D=P\cdot \Pi_O(Ti) P^{-1} \end{eqnarray} I can't get this result because the determinant of matrix p is zero:

\begin{eqnarray} \nonumber P&=&\left( \begin{array}{ccc} - & - &- \\ \vec V_1 & \vec V_2 & \vec V_3 \\ - &- &- \end{array}\right) \\ \nonumber &=& \frac{1}{\sqrt{2}}\left( \begin{array}{ccc} 1 &1& 0\\ 1 & -1 & 0 \\ 0 & 0 &0 \end{array} \right) \end{eqnarray}

in the following i show you how i get the eigenvectors:

\begin{eqnarray} \nonumber \left(\Pi_O(Ti)-\lambda_1\cdot I\right)\vec V_1&=&\left( \begin{array}{ccc} \left( \frac{ma^2}{6}-\frac{ma^2}{4}\right) & -\frac{ma^2}{12}& 0 \\ -\frac{ma^2}{12}& \left(\frac{ma^2}{6}-\frac{ma^2}{4}\right)& 0 \\ 0& 0& \left(\frac{ma^2}{3}-\frac{ma^2}{4}\right) \end{array}\right)\cdot\left( \begin{array}{lll} x_1 \\ y_1 \\ z_1 \end{array}\right) \\ \nonumber &=& \left\{ \begin{array}{lll} -\frac{ma^2}{12} x_1 -\frac{ma^2}{12}y_1=0 \\ -\frac{ma^2}{12}x_1 -2\frac{ma^2}{24}y_1=0\\ z_1=0 \end{array}\right. \end{eqnarray} on obtient l'équation suivante: \begin{eqnarray} \nonumber \begin{array}{lll} x_1+y_1=0\Leftrightarrow x_1=-y_1 \end{array} \end{eqnarray} finalement le vecteur propre $\vec V_1$ associé à la valeur propre est donné par: \begin{eqnarray} \vec V_1= \frac{1}{\sqrt{2}}\left( \begin{array}{lll} 1 \\ -1 \\ 0 \end{array}\right)_{(\vec i, \vec j, \vec k)} \nonumber \end{eqnarray} on a considèré ce vecteur comme un vecteur normé, puisqu'il sert à décrire une base qui sera diangonale.

\underline{\textbf{\textcolor{blue}{Le vecteur propre $\vec V_2$ associé à la valeur propre $\lambda_2$}}} \begin{eqnarray} \nonumber \left(\Pi_O(Ti)-\lambda_2\cdot I\right)\vec V_2&=&\left( \begin{array}{ccc} \left( \frac{ma^2}{6}-\frac{ma^2}{12}\right) & -\frac{ma^2}{12}& 0 \\ -\frac{ma^2}{12}& \left(\frac{ma^2}{6}-\frac{ma^2}{12}\right)& 0 \\ 0& 0& \left(\frac{ma^2}{3}-\frac{ma^2}{12}\right) \end{array}\right)\cdot\left( \begin{array}{lll} x_2 \\ y_2 \\ z_2 \end{array}\right) \\ \nonumber &=& \left\{ \begin{array}{lll} \frac{ma^2}{12} x_2 -\frac{ma^2}{12}y_2=0 \\ -\frac{ma^2}{12}x_2 +2\frac{ma^2}{24}y_2=0\\ z_2=0 \end{array}\right. \end{eqnarray} Alors on obtient l'équation suivante: \begin{eqnarray} \nonumber x_2-y_2=0 \Rightarrow x_2=y_2 \end{eqnarray} Finalement le vecteur normalisé associé à la valeur propre est donné par: \begin{eqnarray} \vec V_2= \frac{1}{\sqrt{2}}\left( \begin{array}{lll} 1 \\ 1 \\ 0 \end{array}\right)_{(\vec i, \vec j, \vec k)} \nonumber \end{eqnarray} \underline{\textbf{\textcolor{blue}{Le vecteur propre $\vec V_3(x_3, y_3, z_3)$ associé à la valeur propre $\lambda_3$}}} \begin{eqnarray} \nonumber \left(\Pi_O(Ti)-\lambda_3\cdot I\right)\vec V_3&=&\left( \begin{array}{lll} \left(\frac{ma^2}{6}-\frac{ma^2}{3}\right) & -\frac{ma^2}{12}& 0 \\ -\frac{ma^2}{12}&\left( \frac{ma^2}{6}-\frac{ma^2}{3}\right)& 0 \\ 0& 0& \left(\frac{ma^2}{3}-\frac{ma^2}{3}\right) \end{array}\right)\cdot\left( \begin{array}{lll} x_3 \\ y_3 \\ z_3 \end{array}\right) \\ \nonumber &=& \left( \begin{array}{lll} -\frac{ma^2}{6} & -\frac{ma^2}{12}& 0 \\ -\frac{ma^2}{12}&-\frac{ma^2}{6}& 0 \\ 0& 0& 0 \end{array}\right)\cdot\left( \begin{array}{lll} x_3\\ y_3 \\ z_3 \end{array}\right) \end{eqnarray} \begin{eqnarray}\nonumber \left\{ \begin{array}{lll} -2\frac{ma^2}{12}\cdot x_3-\frac{ma^2}{12}\cdot y_3=0 \\ -\frac{ma^2}{12}\cdot x_3-2\frac{ma^2}{12}\cdot y_3=0\\ z_3 =0 \end{array}\right. \Rightarrow \left\{ \begin{array}{lll} x_3= -\frac{1}{2} y_3\\ y_3=-\frac{1}{2} x_3\\ z_3 =0 \end{array}\right. \end{eqnarray} alors: \begin{eqnarray} \vec V_3=\vec 0 \nonumber \end{eqnarray}

  • $\begingroup$ Maltrix ? I think you've been drinking too much beer. $\endgroup$ – Rene Schipperus Apr 6 '18 at 13:57
  • $\begingroup$ can you help me please, i want to calclate the P matrix correctly $\endgroup$ – moradov Apr 6 '18 at 14:05

Zero is never an eigenvector. You computed the matrix product that led to your last system of equations incorrectly: its last element is $0$, not $z$. Thus, the last equation should really have been $0=0$ instead of $z=0$, making $z$ free.

That said, you needn’t have gone through the generic computation for the last eigenvector. One can find by inspection that $(0,0,1)^T$ is an eigenvector with eigenvalue $\frac{ma^2}3$. Even if you didn’t spot that, you have a real symmetric matrix, so it’s diagonalizable and its eigenspaces are mutually orthogonal. You have three distinct eigenvalues and have found eigenvectors for two of them, so an eigenvector for the remaining eigenvalue is their cross product: $$\frac1{\sqrt2}(1,-1,0)^T\times\frac1{\sqrt2}(1,1,0)^T=(0,0,1)^T.$$ You should always look for symmetries and other special properties of your problem in order to simplify it or save yourself some work.

When you do this sort of calculation again, I’d suggest leaving the common factor, in this case $\frac{ma^2}3$, out of your eigenvector computations. It just clutters things up and makes it more likely that you’ll make an error. The eigenvectors are unchanged by this, but remember to scale the eigenvalues appropriately.


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