Eigenvalues

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eigenvalues

The eigenvalues of a matrix, $M$, are the values $\lambda$ satisfying the equation $$ M{\bf x} = \lambda{\bf x} $$ where ${\bf x}$ is an eigenvector of a matrix

Effectively an eigenvalue is a scale factor that describes how far a matrix stretches a vector parallel to an eigenvector

In practice, the eigenvalues can be found as the solutions of the equation $$ \det(M - \lambda I) = 0 $$

Eigenvalues 1

Determine the eigenvalues of the matrix $$ M = \left(\begin{array}{cc}3 & 4 \\ 5 & 11 \end{array}\right) $$
solution - press button to display

The eigenvalues are solutions of the equation $\det(M-\lambda I) = 0$. This gives $$ \left|\begin{array}{cc}3-\lambda & 4 \\ 5 & 11-\lambda \end{array}\right| = 0 $$

Calculating this determinant gives $$ (3-\lambda)(11-\lambda) - 4\times 5 =0 \Rightarrow \lambda^2 - 14\lambda + 13 =0 $$

The solution of which are $\lambda = 13$ or $\lambda = 1$

Eigenvalues 2

Determine the eigenvalues of the following matrix: $$ M = \left(\begin{array}{ccc}-71 & 23 & 1 \\ -246 & 83 & -6 \\ -47 & 16 & 3\end{array} \right)$$

solution - press button to display

Let us first find the characteristic equation, $P(\lambda)$ of the matrix;

$$ \begin{align}P(\lambda) &= \det(M - \lambda) = \left|\begin{array}{ccc}-71-\lambda & 23 & 1 \\ -246 & 83-\lambda & -6 \\ -47 & 16 & 3-\lambda\end{array} \right| \\ &= (-71-\lambda)\left|\begin{array}{cc}83-\lambda & - 6\\ 16 & 3-\lambda\end{array}\right| -23\left|\begin{array}{cc}-246 & -6 \\ -47 & 3-\lambda \end{array}\right| + 1\left|\begin{array}{cc}-246 & 83-\lambda \\ -47 & 16 \end{array}\right|\end{align}$$

Expansion and simplification gives $$ P(\lambda) = -\lambda^3 +15\lambda^2 +150\lambda - 1000 $$

The eigenvalues are the solutions of the equation $P(\lambda) =0$. We now have a cubic to solve. In practice, a cubic can be solved by either using a graphical calculator or using the polynomial solve feature on Casio calculators! If however, such technology is not readily available, the first step would be to check to see if any of the factors of 1000 are solutions.

In fact, this equation factorises to $P(\lambda) = -(\lambda +10)(\lambda - 5)(\lambda-20)$, giving eigenvalues of $\lambda = -10,\;5,\;20$