Why do these eigenvalues give the zero vector for 3 eigenvectors?

Question

Given the quadratic form $q(x_{1}, x_{2}, x_{3}, x_{4}) = x_{1}^2 + x_{2}^2 - x_{3}^2 - x_{4}^2 + 4x_{1}x_{2} - 2x_{1}x_{4} + 2x_{3}x_{4}$, the eigenvalues I am getting are $\lambda = -1, -2.27, .14, 3.13$. However, all of these eigenvalues except $\lambda = -1$ are resulting in the zero-vector for the their respective eigenvectors, which is not possible by definition. What is wrong here?

Edit: The characteristic equation I am getting is $\lambda^4 - 8\lambda^2-6\lambda + 1 = 0$.

Answer 1

Your calculated eigenvalues look correct. When I compute the eigenvectors, I get nonzero results— so perhaps only your eigenvector calculations are off.

In matrix form, the quadratic form looks like

$$\mathbf{A} \equiv \begin{bmatrix}1 & 2 & 0 & -1 \\ 2 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ -1 & 0 & 1 & -1\\ \end{bmatrix}$$

The characteristic equation is:

$$\chi(\lambda) = \lambda^4-8\lambda^2-6\lambda + 1 = 0$$

As I find it, the eigensystem is: \begin{align*} \lambda_1 \approx 3.13264 & \qquad v_1=(-3.89066, -3.64869, 0.241976, 1)\\ \lambda_2 \approx -2.27307 & \qquad v_2=(0.487572, -0.297929, -0.785501, 1)\\ \lambda_3 \approx -1 & \qquad v_3 = (1, -1, 1, 0)\\ \lambda_4 \approx 0.140435 & \qquad v_4=(-0.263577, 0.613281, 0.876858, 1) \end{align*}

And the eigenvectors behave as expected.

Answer 2

You only get eigenvectors for the exact eigenvalue.
If you used $-2.27$ for the eigenvalue, then an entry that should have been zero during row operations becomes non-zero, and hence an extra pivot.

Answer 3

As far as the eigenvalues, the other answers are fine.

A milder thing to do is find out how to make your symmetric matrix "congruent" to a diagonal matrix, then invoke Sylvester's Law of Inertia. Below, $PQ=QP=I.$

Two eigenvalues are positive and two negative.

$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 2 & 0 & - 1 \\ 2 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 1 \\ - 1 & 0 & 1 & - 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - 2 & 0 & - \frac{ 1 }{ 3 } \\ 0 & 1 & 0 & \frac{ 2 }{ 3 } \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & \frac{ 1 }{ 3 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - 1 & - \frac{ 2 }{ 3 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & \frac{ 1 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 2 & 0 & - 1 \\ 0 & 1 & 0 & - \frac{ 2 }{ 3 } \\ 0 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 2 & 0 & - 1 \\ 2 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 1 \\ - 1 & 0 & 1 & - 1 \\ \end{array} \right) $$