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I need to solve the following problem,

In this problem, the eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.




Now I know how to find the eigenvalues by using the fact that $|A-\lambda I|=0$, but how would I do it by inspection? Inspection is easy for matrices that have the sum of their rows adding up to the same value, but this coefficient matrix doesn't have that property.

EDIT: Originally I didn't understand what inspection meant either. After googling it this is what I found. Imagine you have the matrix, $A = \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}$

By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

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The "inspection and factoring" refers to finding the zeros of the polynomial. You still have to calculate it. – Phira Nov 7 '11 at 19:35
"how would I do it by inspection" is an oxymoron – opt Nov 7 '11 at 19:44
Ask your teacher what they mean by "inspection and factoring" if you aren't sure. – Graphth Nov 7 '11 at 22:08

The coefficient matrix is $A = \begin{pmatrix} 2 & 1 & -1 \\ -4 & -3 & -1 \\ 4 & 4 & 2 \end{pmatrix}$.

I'm not sure there is a way to do this "by inspection".

In general, the product of the eigenvalues is the determinant of a matrix, so in this case $\lambda_1\lambda_2\lambda_3=4$. The trace (the sum of the diagonals) is the sum of the eigenvalues so $\lambda_1+\lambda_2+\lambda_3=2-3+2=1$. But without knowing what one of the eigenvalues is, we need one more equation.

If you notice (for some odd reason) that $[-1\;1\;0]^T$ is an eigenvector with eigenvalue 1, then $\lambda_1=1$ so $\lambda_2\lambda_3=4$ and $\lambda_2+\lambda_3=0$. Thus $\lambda_2,\lambda_3=\pm 2\sqrt{-1}$.

In general there is no easy way to spot eigenvalues of $3 \times 3$ matrices.

With $2 \times 2$ matrices you can use the determinant and trace. The reason this works comes directly from the fact that these are the only non-trivial coefficients of the characteristic polynomial (and thus determine the eigenvalues).

When dealing with bigger matrices you need more invariants (i.e. coefficients of the characteristic polynomial) and other coefficients of the characteristic polynomial aren't as easy to use/remember.

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You can see easily that $A-I$ has two equal rows (it also has two equal columns), thus you can guess the first eigenvalue.

Then the trace and determinant give you the other two eigenvalues..

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Here is how I would do it, but solving by inspection is subjective anyway.

Let $A$ be the matrix of coefficients, then $(A \cdot x)_k = \lambda x_k$. Add the third and the second row. This gives $x_2 + x_3 = \lambda( x_2 + x_3)$, thus $\lambda = 1$ or $x_2 + x_3 = 0$. Since the trace of $A$ equals 1, the sum of other two eigenvalues is 0.

To find other eigenvalues, let $x_3 = -x_2$, giving $2 x_1 + 2x_2 = \lambda x_1$ and $-4 x_1 - 2 x_2 = \lambda x_2$. The characteristic polynomial is easily found as $\lambda^2 + 4 = 0$.

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