Multiplicity of Eigenvalues

Example §

Orthogonally diagonalize the matrix

$$A=\left[\begin{array}{ccc}3& -2& 4\\ -2& 6& 2\\ 4& 2& 3\end{array}\right]$$

whose characteristic equation is

$$-{\lambda }^3+12{\lambda }^2-21\lambda -98=-(\lambda -7{)}^2(\lambda +2)$$

note that this characteristic equation features multiplicity within the eigenvalues, in that $\lambda =7$ appears twice.

$$\lambda =7:\left[\begin{array}{ccc}3-\lambda & -2& 4\\ -2& 6-\lambda & 2\\ 4& 2& 3-\lambda \end{array}\right]=\left[\begin{array}{ccc}-4& -2& 4\\ -2& -1& 2\\ 4& 2& -4\end{array}\right]$$

When solving the problem as usual you will notice something very important regarding the matrix reduction:

$$\left[\begin{array}{ccc}-4& -2& 4\\ 0& 0& 0\\ 4& 2& -4\end{array}\right]\to \left[\begin{array}{ccc}-4& -2& 4\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]\to \left[\begin{array}{ccc}1& \frac{1}{2}& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

Notably, $x_2$ and $x_3$ are free. When this occurs it is usually ‘best-practice’ to let one of the free variables be 1, and the other 0.

$$\text{Let }{x}_2=1\to x_1=-\frac{1}{2}$$ $$\to v_1=\left[\begin{array}{c}-\frac{1}{2}\\ 1\\ 0\end{array}\right]$$ $$\text{Let }{x}_3=1\to x_1=1$$ $$\to v_2=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$$

Note that we produced two eigenvectors for $\lambda =7$,

In general, for each distinct eigenvalue, you seek to find as many linearly independent eigenvectors as its algebraic multiplicity. However, the actual number of independent eigenvectors you can find (geometric multiplicity) may be less than the algebraic multiplicity.

Tip

• Geometric Multiplicity: This refers to the number of linearly independent eigenvectors associated with an eigenvalue. The geometric multiplicity of an eigenvalue is always less than or equal to its algebraic multiplicity.

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