Lecture Three: Eigenvalues

1. Eigenvalues Using the Determinant Method

Eigenvalues of a square matrix $A$ are values $\lambda$ for which there exists a non-zero vector $v$ (called an eigenvector) such that:

$$A\cdot v=\lambda \cdot v$$

This can be rewritten as:

$$(A-\lambda I)\cdot v=0$$

For a non-trivial solution ($v\ne 0$), the matrix $(A-\lambda I)$ must be singular, meaning its determinant is zero:

$$|A-\lambda I|=0$$

This determinant equation gives a polynomial in $\lambda$, called the characteristic polynomial. The roots of this polynomial are the eigenvalues of $A$.

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Example for a $2\times 2$ Matrix:

Let:

$$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$$$$\lambda I_2=\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]$$

Then:

$$A-\lambda I=\left[\begin{array}{cc}a-\lambda & b\\ c& d-\lambda \end{array}\right]$$

The determinant is:

$$|A-\lambda I|=(a-\lambda )(d-\lambda )-bc$$

Expanding this:

$$|A-\lambda I|={\lambda }^2-(a+d)\lambda +(ad-bc)$$

The characteristic polynomial is:

$${\lambda }^2-\text{trac}(A)\lambda +|A|=0$$

where:

• $\text{trac}(A)=a+d$ is the trace (sum of diagonal entries of $A$),
• $|A|=ad-bc$ is the determinant of $A$.

The eigenvalues are found by solving this quadratic equation.

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Example for a $3\times 3$ Matrix:

Let:

$$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$$

Then:

$$A-\lambda I=\left[\begin{array}{ccc}a-\lambda & b& c\\ d& e-\lambda & f\\ g& h& i-\lambda \end{array}\right]$$

2. Eigenvalues Using the Cayley-Hamilton Theorem

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic polynomial.

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For $2\times 2$ Matrices:

The characteristic polynomial is:

$${\lambda }^2-S_1\lambda +S_2=0$$

where:

• $S_1=\text{trac}(A)={\lambda }_1+{\lambda }_2$
• $S_2=|A|={\lambda }_1{\lambda }_2$

Eigenvalues can be computed directly from the quadratic equation.

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For $3\times 3$ Matrices:

For the matrix $A$:

$$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$$

The characteristic polynomial is:

$${\lambda }^3-S_1{\lambda }^2+S_2\lambda -S_3=0$$

where:

• $S_1=\text{trac}(A)={\lambda }_1+{\lambda }_2+{\lambda }_3$
• $S_2$ is the sum of minors of $A$
• $S_3=|A|={\lambda }_1{\lambda }_2{\lambda }_3$

Eigenvalues are obtained by solving the cubic characteristic equation.

Summary:

• The determinant method directly computes eigenvalues from the characteristic polynomial $\parallel A-\lambda I|=0$.
• The Cayley-Hamilton theorem leverages the characteristic polynomial to establish relationships between $A$ and its eigenvalues, offering additional insights into $A$'s powers and structure.

Examples:

Example 1: Determinant Method for a $2\times 2$ Matrix

Let:

$$A=\left[\begin{array}{cc}4& 2\\ 1& 3\end{array}\right]$$

Step 1: Write $A-\lambda I$:

$$A-\lambda I=\left[\begin{array}{cc}4-\lambda & 2\\ 1& 3-\lambda \end{array}\right]$$

Step 2: Compute the determinant:

$$|A-\lambda I|=\left|\begin{array}{cc}4-\lambda & 2\\ 1& 3-\lambda \end{array}\right|$$$$|A-\lambda I|=(4-\lambda )(3-\lambda )-(2\times 1)$$$$|A-\lambda I|={\lambda }^2-7\lambda +12-2={\lambda }^2-7\lambda +10$$

Step 3: Solve for $\lambda$:

$${\lambda }^2-7\lambda +10=0$$

Factoring:

$${\lambda }^2-7\lambda +10=(\lambda -2)(\lambda -5)$$

Eigenvalues are:

$${\lambda }_1=2,{\lambda }_2=5$$

Example 2: Cayley-Hamilton Method for a $2\times 2$ Matrix

Let:

$$A=\left[\begin{array}{cc}4& 2\\ 1& 3\end{array}\right]$$

Step 1: Characteristic Polynomial:

$${\lambda }^2-S_1\lambda +S_2=0$$$$S_1=\text{trac}(A)=4+3=7$$$$S_2=|A|=12-2=10$$$${\lambda }^2-7\lambda +10=0$$

Step 2: Use Polynomial to Solve for Eigenvalues:

Substituting $\lambda =2$ and $\lambda =5$ satisfies the equation.

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Example 3: Determinant Method for a $3\times 3$ Matrix

Let:

$$A=\left[\begin{array}{ccc}5& 2& 0\\ 2& 5& 0\\ -3& 4& 6\end{array}\right]$$

Step 1: Write $A-\lambda I$:

$$A-\lambda I=\left[\begin{array}{ccc}5-\lambda & 2& 0\\ 2& 5-\lambda & 0\\ -3& 4& 6-\lambda \end{array}\right]$$

Step 2: Compute the determinant:

$$|A-\lambda I|=\left|\begin{array}{ccc}5-\lambda & 2& 0\\ 2& 5-\lambda & 0\\ -3& 4& 6-\lambda \end{array}\right|$$

Expanding along the third column (Remember to apply the Sign rule):

$$|A-\lambda I|=0\left|\begin{array}{cc}2& 5-\lambda \\ -3& 4\end{array}\right|-0\left|\begin{array}{cc}5-\lambda & 2\\ -3& 4\end{array}\right|+(6-\lambda )\left|\begin{array}{cc}5-\lambda & 2\\ 2& 5-\lambda \end{array}\right|$$

Compute the subdeterminants:

1. For $0\left|\begin{array}{cc}2& 5-\lambda \\ -3& 4\end{array}\right|$: skip because it's multiplied by 0.

2. For $0\left|\begin{array}{cc}5-\lambda & 2\\ -3& 4\end{array}\right|$: skip because it's multiplied by 0.

3. For $\left|\begin{array}{cc}5-\lambda & 2\\ 2& 5-\lambda \end{array}\right|$:

$$=(5-\lambda )(5-\lambda )-2(2)$$$$=(5-\lambda {)}^2-4$$$$={\lambda }^2-10\lambda +25-4$$$$={\lambda }^2-10\lambda +21$$

Substitute back:

$$|A-\lambda I|=0-0+(6-\lambda )({\lambda }^2-10\lambda +21)$$

Simplify:

$$|A-\lambda I|=(6-\lambda )({\lambda }^2-10\lambda +21)$$$$=-{\lambda }^3+16{\lambda }^2-81\lambda +126$$

Step 3: Solve for $\lambda$:

Solve the cubic $-{\lambda }^3+16{\lambda }^2-81\lambda +126=0$ to find the eigenvalues (mode + 5 + 4 on calculator):

$${\lambda }_1=3,{\lambda }_2=7,{\lambda }_3=6$$

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Example 4: Cayley-Hamilton Method for a $3\times 3$ Matrix

Let:

$$A=\left[\begin{array}{ccc}5& 2& 0\\ 2& 5& 0\\ -3& 4& 6\end{array}\right]$$

Step 1: Characteristic Polynomial:

$${\lambda }^3-S_1{\lambda }^2+S_2\lambda -S_3=0$$$$S_1=\text{trac}(A)=5+5+6=16$$$$S_2=|A|=\left|\begin{array}{cc}5& 0\\ 4& 6\end{array}\right|+\left|\begin{array}{cc}5& 0\\ -3& 6\end{array}\right|+\left|\begin{array}{cc}5& 2\\ 2& 5\end{array}\right|=81$$$$S_3=|A|=126$$$${\lambda }^3-16{\lambda }^2+81\lambda -126=0$$

Step 2: Solve for Eigenvalues:

Solve the cubic ${\lambda }^3-16{\lambda }^2+81\lambda -126=0$ to find the eigenvalues (mode + 5 + 4 on calculator):

$${\lambda }_1=3,{\lambda }_2=7,{\lambda }_3=6$$