Eigenvalues and Eigenvectors

6. Eigenvalues and Eigenvectors

6.1 What Are Eigenvalues and Eigenvectors?

In linear algebra, eigenvalues and eigenvectors are fundamental concepts used to analyze the behavior of linear transformations. For a square matrix $A$, an eigenvector $\mathbf{v}$ is a non-zero vector that, when multiplied by $A$, results in a scalar multiple of $\mathbf{v}$. The scalar $\lambda$ is called the eigenvalue corresponding to $\mathbf{v}$.

Mathematically, the relationship is expressed as:
[
A \mathbf{v} = \lambda \mathbf{v}
]
Where:
- $A$ is a square matrix.
- $\mathbf{v}$ is the eigenvector.
- $\lambda$ is the eigenvalue.

6.2 Finding Eigenvalues

To find the eigenvalues of a matrix $A$, we solve the characteristic equation:
[
\text{det}(A - \lambda I) = 0
]
Where:
- $\text{det}(A-\lambda I)$ is the determinant of $A-\lambda I$.
- $I$ is the identity matrix of the same size as $A$.
- $\lambda$ is the eigenvalue.

This equation is a polynomial in $\lambda$, and the solutions to the equation are the eigenvalues of the matrix $A$.

6.3 Finding Eigenvectors

Once the eigenvalues $\lambda$ are found, the corresponding eigenvectors can be found by solving the system:
[
(A - \lambda I) \mathbf{v} = 0
]
This equation gives a system of linear equations whose solution gives the eigenvectors.

---

Example 1: Finding Eigenvalues and Eigenvectors of a 2x2 Matrix

Given the matrix:
[
A = $$\left[\begin{array}{ccc}4& 12& 3\end{array}\right]$$
]

Step 1: Find the eigenvalues by solving the characteristic equation.

The characteristic equation is:
[
\text{det}(A - \lambda I) = 0
]
[
A - \lambda I = $$\left[\begin{array}{ccc}4& 12& 3\end{array}\right]$$ - \lambda $$\left[\begin{array}{ccc}1& 00& 1\end{array}\right]$$ = $$\left[\begin{array}{ccc}4-\lambda & 12& 3-\lambda \end{array}\right]$$
]

Now, calculate the determinant:
[
\text{det}(A - \lambda I) = (4 - \lambda)(3 - \lambda) - (2)(1) = 0
]
[
(4 - \lambda)(3 - \lambda) - 2 = 0
]
[
12 - 7\lambda + \lambda^2 - 2 = 0
]
[
\lambda^2 - 7\lambda + 10 = 0
]

Solve the quadratic equation:
[
\lambda^2 - 7\lambda + 10 = 0
]
Using the quadratic formula:
[
\lambda = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)}
]
[
\lambda = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm \sqrt{9}}{2}
]
[
\lambda = \frac{7 \pm 3}{2}
]
Thus, the eigenvalues are:
[
\lambda_1 = \frac{7 + 3}{2} = 5, \quad \lambda_2 = \frac{7 - 3}{2} = 2
]

Step 2: Find the eigenvectors corresponding to each eigenvalue.

For ${\lambda }_1=5$, we solve:
[
(A - 5I) \mathbf{v} = 0
]
[
$$\left[\begin{array}{ccc}4-5& 12& 3-5\end{array}\right]$$ $$\left[\begin{array}{c}{v}_1{v}_2\end{array}\right]$$ = $$\left[\begin{array}{c}00\end{array}\right]$$
]
[
$$\left[\begin{array}{ccc}-1& 12& -2\end{array}\right]$$ $$\left[\begin{array}{c}{v}_1{v}_2\end{array}\right]$$ = $$\left[\begin{array}{c}00\end{array}\right]$$
]

This simplifies to the system:
[
-v_1 + v_2 = 0 \quad \Rightarrow \quad v_1 = v_2
]
Thus, the eigenvector corresponding to ${\lambda }_1=5$ is:
[
\mathbf{v_1} = $$\left[\begin{array}{c}11\end{array}\right]$$
]

For ${\lambda }_2=2$, we solve:
[
(A - 2I) \mathbf{v} = 0
]
[
$$\left[\begin{array}{ccc}4-2& 12& 3-2\end{array}\right]$$ $$\left[\begin{array}{c}{v}_1{v}_2\end{array}\right]$$ = $$\left[\begin{array}{c}00\end{array}\right]$$
]
[
$$\left[\begin{array}{ccc}2& 12& 1\end{array}\right]$$ $$\left[\begin{array}{c}{v}_1{v}_2\end{array}\right]$$ = $$\left[\begin{array}{c}00\end{array}\right]$$
]

This simplifies to the system:
[
2v_1 + v_2 = 0 \quad \Rightarrow \quad v_2 = -2v_1
]
Thus, the eigenvector corresponding to ${\lambda }_2=2$ is:
[
\mathbf{v_2} = $$\left[\begin{array}{c}1-2\end{array}\right]$$
]

---

Practice Problems

1. Find the eigenvalues and eigenvectors of the matrix:
   [
   A = $$\left[\begin{array}{ccc}6& 22& 4\end{array}\right]$$
   ]

2. Find the eigenvalues and eigenvectors of the matrix:
   [
   B = $$\left[\begin{array}{ccc}1& 43& 1\end{array}\right]$$
   ]

3. Given the matrix:
   [
   C = $$\left[\begin{array}{ccccccc}5& 4& 24& 5& 22& 2& 5\end{array}\right]$$
   ]
   Find the eigenvalues and eigenvectors.

Would you like solutions to these problems, or should we move on to Diagonalization of Matrices?

---

7. Diagonalization of Matrices

7.1 What is Diagonalization?

A matrix $A$ is said to be diagonalizable if it can be written as:
[
A = P D P^{-1}
]
Where:
- $P$ is an invertible matrix whose columns are the eigenvectors of $A$.
- $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $A$.
- $P^{-1}$ is the inverse of $P$.

In other words, diagonalization transforms a matrix into a diagonal matrix $D$ using a similarity transformation with $P$. This process simplifies matrix operations, such as computing powers of $A$ and solving systems of linear equations.

7.2 Conditions for Diagonalization

For a matrix $A$ to be diagonalizable, the following conditions must hold:
1. The matrix must be square (i.e., the number of rows and columns must be equal).
2. There must be enough linearly independent eigenvectors to form the matrix $P$. This means that $A$ must have $n$ linearly independent eigenvectors for an $n\times n$ matrix.

If a matrix does not have enough linearly independent eigenvectors, it is not diagonalizable.

7.3 Steps to Diagonalize a Matrix

To diagonalize a matrix $A$, follow these steps:
1. Find the eigenvalues of $A$ by solving the characteristic equation:
[
\text{det}(A - \lambda I) = 0
]
2. Find the eigenvectors corresponding to each eigenvalue.
3. Form the matrix $P$ by placing the eigenvectors as columns.
4. Form the diagonal matrix $D$ with the eigenvalues on the diagonal.
5. Verify the diagonalization by checking if $A=PD{P}^{-1}$.

---

Example 1: Diagonalizing a 2x2 Matrix

Consider the matrix:
[
A = $$\left[\begin{array}{ccc}4& 12& 3\end{array}\right]$$
]

Step 1: Find the eigenvalues.

Solve the characteristic equation:
[
\text{det}(A - \lambda I) = 0
]
[
$$\left[\begin{array}{ccc}4-\lambda & 12& 3-\lambda \end{array}\right]$$
]
The determinant is:
[
(4 - \lambda)(3 - \lambda) - 2 = 0
]
[
12 - 7\lambda + \lambda^2 - 2 = 0
]
[
\lambda^2 - 7\lambda + 10 = 0
]
Solving the quadratic equation:
[
\lambda = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(10)}}{2(1)}
]
[
\lambda = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2}
]
Thus, the eigenvalues are:
[
\lambda_1 = 5, \quad \lambda_2 = 2
]

Step 2: Find the eigenvectors.

For ${\lambda }_1=5$:
[
(A - 5I) \mathbf{v} = 0
]
[
$$\left[\begin{array}{ccc}-1& 12& -2\end{array}\right]$$ $$\left[\begin{array}{c}{v}_1{v}_2\end{array}\right]$$ = $$\left[\begin{array}{c}00\end{array}\right]$$
]
This simplifies to:
[
v_1 = v_2
]
So, the eigenvector corresponding to ${\lambda }_1=5$ is:
[
\mathbf{v_1} = $$\left[\begin{array}{c}11\end{array}\right]$$
]

For ${\lambda }_2=2$:
[
(A - 2I) \mathbf{v} = 0
]
[
$$\left[\begin{array}{ccc}2& 12& 1\end{array}\right]$$ $$\left[\begin{array}{c}{v}_1{v}_2\end{array}\right]$$ = $$\left[\begin{array}{c}00\end{array}\right]$$
]
This simplifies to:
[
v_2 = -2v_1
]
So, the eigenvector corresponding to ${\lambda }_2=2$ is:
[
\mathbf{v_2} = $$\left[\begin{array}{c}1-2\end{array}\right]$$
]

Step 3: Form the matrix $P$.

The matrix $P$ is formed by placing the eigenvectors as columns:
[
P = $$\left[\begin{array}{ccc}1& 11& -2\end{array}\right]$$
]

Step 4: Form the diagonal matrix $D$.

The diagonal matrix $D$ is formed by placing the eigenvalues on the diagonal:
[
D = $$\left[\begin{array}{ccc}5& 00& 2\end{array}\right]$$
]

Step 5: Verify the diagonalization.

Check if $A=PD{P}^{-1}$:
[
A = $$\left[\begin{array}{ccc}1& 11& -2\end{array}\right]$$ $$\left[\begin{array}{ccc}5& 00& 2\end{array}\right]$$ $$\left[\begin{array}{ccc}1& 11& -2\end{array}\right]$$^{-1}
]
You can verify this by performing matrix multiplication, and you should find that it equals $A$.

---

Practice Problems

1. Diagonalize the matrix:
   [
   A = $$\left[\begin{array}{ccc}5& 42& 1\end{array}\right]$$
   ]

2. Diagonalize the matrix:
   [
   B = $$\left[\begin{array}{ccc}6& 22& 3\end{array}\right]$$
   ]

3. Given the matrix:
   [
   C = $$\left[\begin{array}{ccccccc}7& 1& 11& 7& 11& 1& 7\end{array}\right]$$
   Find its diagonalization.

Would you like solutions to these problems, or should we proceed with Multivariate Calculus?