How do you calculate eigenvectors on this matrix? \begin{pmatrix}4&1&-3\\ 1&2&-1\\ -3&-1&3\end{pmatrix} I found the eigenvalues: Lamba1 = 6.94338, Lambda2 = 1.60909, Lambda3 = 0.447525, but every time I tried to get the eigenvectors, I get a zero vector for each lambda. I used row reduction and Cramer's rule. Now I am lost. I don't know what to do anymore.
Also on a website that calculates eigenvectors for you, I obtained v1 = [-1.16825, -0.438618, 1], v2= [-2.64178, 9.31626, 1], v3=[0.810038, 0.122361, 1]. I understand that if I fixed the last value to 1 for each vector, I get the other two by solving my equations. But what is the logic behind it? How do you go from a zero vector to those one?
Any help will be really appreciated!!! thank you.
The eigenvalues and eigenvectors are really unpleasant for this one. On the other hand, maybe you were told to "diagonalize" the matrix. In that case, the reasonable thing to do is congruence diagonalization, for your symmetric matrix $H,$ find nonsingular matrix $P$ such that $P^T HP = D$ is diagonal. We can actually arrange that $\det P = \pm 1.$
$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 4 } & 1 & 0 \\ \frac{ 5 }{ 7 } & \frac{ 1 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 1 & - 3 \\ 1 & 2 & - 1 \\ - 3 & - 1 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 5 }{ 7 } \\ 0 & 1 & \frac{ 1 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 5 }{ 7 } \\ \end{array} \right) $$
$$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 4 & 1 & - 3 \\ 1 & 2 & - 1 \\ - 3 & - 1 & 3 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 4 & 0 & - 3 \\ 0 & \frac{ 7 }{ 4 } & - \frac{ 1 }{ 4 } \\ - 3 & - \frac{ 1 }{ 4 } & 3 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & \frac{ 3 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 3 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & - \frac{ 3 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & - \frac{ 1 }{ 4 } \\ 0 & - \frac{ 1 }{ 4 } & \frac{ 3 }{ 4 } \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 5 }{ 7 } \\ 0 & 1 & \frac{ 1 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & - \frac{ 3 }{ 4 } \\ 0 & 1 & - \frac{ 1 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 5 }{ 7 } \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 4 } & 1 & 0 \\ \frac{ 5 }{ 7 } & \frac{ 1 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 1 & - 3 \\ 1 & 2 & - 1 \\ - 3 & - 1 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 4 } & \frac{ 5 }{ 7 } \\ 0 & 1 & \frac{ 1 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 5 }{ 7 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 4 } & 1 & 0 \\ - \frac{ 3 }{ 4 } & - \frac{ 1 }{ 7 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & \frac{ 7 }{ 4 } & 0 \\ 0 & 0 & \frac{ 5 }{ 7 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 4 } & - \frac{ 3 }{ 4 } \\ 0 & 1 & - \frac{ 1 }{ 7 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 1 & - 3 \\ 1 & 2 & - 1 \\ - 3 & - 1 & 3 \\ \end{array} \right) $$