eth-summaries/semester1/linear-algebra/recursive-sequences.md

Yes, you're right! The next step is to diagonalize the matrix, which involves calculating the **eigenvalues** and **eigenvectors** of the transformation matrix \( T \). Once you do that, you can express the recurrence as a closed-form formula for \( a_n \).

Let's break down the steps involved:

### Step 1: Find the eigenvalues of the matrix

You want to find the eigenvalues \( \lambda \) of the matrix \( T \):

\[
T = \begin{bmatrix} 2 & 3 \\ 1 & 0 \end{bmatrix}
\]

Eigenvalues are found by solving the characteristic equation:

\[
\text{det}(T - \lambda I) = 0
\]

where \( I \) is the identity matrix. First, compute the determinant of \( T - \lambda I \):

\[
T - \lambda I = \begin{bmatrix} 2 - \lambda & 3 \\ 1 & -\lambda \end{bmatrix}
\]

Now, compute the determinant:

\[
\text{det}(T - \lambda I) = (2 - \lambda)(-\lambda) - (3)(1) = -2\lambda + \lambda^2 - 3
\]

Simplify the equation:

\[
\lambda^2 - 2\lambda - 3 = 0
\]

Now, solve this quadratic equation for \( \lambda \):

\[
\lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}
\]

Thus, the two eigenvalues are:

\[
\lambda_1 = \frac{2 + 4}{2} = 3 \quad \text{and} \quad \lambda_2 = \frac{2 - 4}{2} = -1
\]

### Step 2: Find the eigenvectors

Now that you have the eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = -1 \), you need to find the corresponding eigenvectors.

#### Eigenvector for \( \lambda_1 = 3 \):

Solve \( (T - 3I) \mathbf{v} = 0 \), where \( \mathbf{v} \) is the eigenvector corresponding to \( \lambda_1 \).

\[
T - 3I = \begin{bmatrix} 2 - 3 & 3 \\ 1 & -3 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ 1 & -3 \end{bmatrix}
\]

Now solve the system:

\[
\begin{bmatrix} -1 & 3 \\ 1 & -3 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
\]

This gives the system:

\[
- v_1 + 3 v_2 = 0
\]
\[
v_1 - 3 v_2 = 0
\]

Both equations are equivalent, so we have:

\[
v_1 = 3 v_2
\]

Thus, the eigenvector corresponding to \( \lambda_1 = 3 \) is:

\[
\mathbf{v}_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}
\]

#### Eigenvector for \( \lambda_2 = -1 \):

Solve \( (T + I) \mathbf{v} = 0 \), where \( \mathbf{v} \) is the eigenvector corresponding to \( \lambda_2 \).

\[
T + I = \begin{bmatrix} 2 + 1 & 3 \\ 1 & 0 + 1 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 1 & 1 \end{bmatrix}
\]

Now solve the system:

\[
\begin{bmatrix} 3 & 3 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
\]

This gives the system:

\[
3 v_1 + 3 v_2 = 0
\]
\[
v_1 + v_2 = 0
\]

From the second equation, we have \( v_1 = -v_2 \). So, the eigenvector corresponding to \( \lambda_2 = -1 \) is:

\[
\mathbf{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}
\]

### Step 3: Diagonalize the matrix

Now that we have the eigenvalues and eigenvectors, we can diagonalize the matrix \( T \). 

Let \( P \) be the matrix formed by the eigenvectors as columns:

\[
P = \begin{bmatrix} 3 & -1 \\ 1 & 1 \end{bmatrix}
\]

The inverse of \( P \) is computed as follows (using the formula for the inverse of a 2x2 matrix):

\[
P^{-1} = \frac{1}{\text{det}(P)} \begin{bmatrix} 1 & 1 \\ -1 & 3 \end{bmatrix}
\]

where \( \text{det}(P) = (3)(1) - (-1)(1) = 3 + 1 = 4 \), so:

\[
P^{-1} = \frac{1}{4} \begin{bmatrix} 1 & 1 \\ -1 & 3 \end{bmatrix}
\]

The diagonal matrix \( D \) will have the eigenvalues on the diagonal:

\[
D = \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix}
\]

Thus, \( T \) can be written as:

\[
T = P D P^{-1}
\]

### Step 4: Express the closed-form solution

Now, to compute \( a_n \), we can use the diagonalization of \( T \):

\[
\mathbf{v}_n = T^{n-1} \mathbf{v}_1 = P D^{n-1} P^{-1} \mathbf{v}_1
\]

The term \( D^{n-1} \) is easy to compute because \( D \) is diagonal:

\[
D^{n-1} = \begin{bmatrix} 3^{n-1} & 0 \\ 0 & (-1)^{n-1} \end{bmatrix}
\]

Thus, the closed-form formula for \( a_n \) is obtained by applying the matrix multiplication:

\[
\mathbf{v}_n = P \begin{bmatrix} 3^{n-1} & 0 \\ 0 & (-1)^{n-1} \end{bmatrix} P^{-1} \mathbf{v}_1
\]

By carrying out these matrix multiplications (which can be done step by step), you'll arrive at a direct formula for \( a_n \), which is a linear combination of the terms involving powers of \( 3 \) and \( (-1)^{n-1} \).

To rewrite the closed-form solution for \( a_n \) without using vectors or matrices, we need to compute the expression directly in terms of \( n \) and use the properties of the eigenvalues and eigenvectors.

Let's break down the steps to arrive at a direct formula for \( a_n \).

### Step 1: General Formula Structure

From the diagonalization of the matrix \( T \), we have:

\[
\mathbf{v}_n = P D^{n-1} P^{-1} \mathbf{v}_1
\]

where \( P = \begin{bmatrix} 3 & -1 \\ 1 & 1 \end{bmatrix} \), \( D = \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} \), and \( \mathbf{v}_1 = \begin{bmatrix} a_1 \\ a_0 \end{bmatrix} \).

After performing the matrix multiplication and solving for the values of \( a_n \), we can arrive at a **direct expression** for \( a_n \).

### Step 2: Compute \( a_n \) in Terms of Eigenvalues

Since \( T^{n-1} \) is diagonalizable, we know that \( D^{n-1} \) is simply:

\[
D^{n-1} = \begin{bmatrix} 3^{n-1} & 0 \\ 0 & (-1)^{n-1} \end{bmatrix}
\]

Now, we need to express the evolution of the state vector without matrices. The formula for \( a_n \) will be a combination of the eigenvalues \( \lambda_1 = 3 \) and \( \lambda_2 = -1 \), weighted by the initial values \( a_1 \) and \( a_0 \).

To get the value of \( a_n \), we need to express the result of \( P^{-1} \mathbf{v}_1 \) in a way that we can compute directly.

### Step 3: Use the Initial Conditions

The initial conditions give us the starting values of \( a_1 \) and \( a_0 \), and we use the fact that the matrix multiplication \( P D^{n-1} P^{-1} \mathbf{v}_1 \) essentially results in the following:

\[
a_n = \alpha_1 3^{n-1} + \alpha_2 (-1)^{n-1}
\]

where \( \alpha_1 \) and \( \alpha_2 \) are constants determined by the initial conditions. Let's compute these coefficients using \( \mathbf{v}_1 = \begin{bmatrix} a_1 \\ a_0 \end{bmatrix} \).

### Step 4: Solve for the Coefficients \( \alpha_1 \) and \( \alpha_2 \)

We know that \( \mathbf{v}_n = P D^{n-1} P^{-1} \mathbf{v}_1 \), and we can solve this system by applying the initial conditions. For the recurrence \( a_n = 2a_{n-1} + 3a_{n-2} \), this gives the following two linear equations:

1. \( \alpha_1 3^0 + \alpha_2 (-1)^0 = a_0 \) (this gives the equation for \( a_0 \))
2. \( \alpha_1 3^1 + \alpha_2 (-1)^1 = a_1 \) (this gives the equation for \( a_1 \))

These equations are:

\[
\alpha_1 + \alpha_2 = a_0
\]
\[
3 \alpha_1 - \alpha_2 = a_1
\]

### Step 5: Solve the System of Equations

We can now solve this system of two linear equations for \( \alpha_1 \) and \( \alpha_2 \). Add the two equations:

\[
(\alpha_1 + \alpha_2) + (3 \alpha_1 - \alpha_2) = a_0 + a_1
\]
\[
4 \alpha_1 = a_0 + a_1
\]
\[
\alpha_1 = \frac{a_0 + a_1}{4}
\]

Substitute \( \alpha_1 = \frac{a_0 + a_1}{4} \) into \( \alpha_1 + \alpha_2 = a_0 \):

\[
\frac{a_0 + a_1}{4} + \alpha_2 = a_0
\]
\[
\alpha_2 = a_0 - \frac{a_0 + a_1}{4} = \frac{3a_0 - a_1}{4}
\]

### Step 6: Write the General Formula for \( a_n \)

Now we substitute \( \alpha_1 \) and \( \alpha_2 \) back into the formula for \( a_n \):

\[
a_n = \alpha_1 3^{n-1} + \alpha_2 (-1)^{n-1}
\]
\[
a_n = \frac{a_0 + a_1}{4} 3^{n-1} + \frac{3a_0 - a_1}{4} (-1)^{n-1}
\]

This is the closed-form formula for \( a_n \), which you can now use to compute \( a_n \) directly for any \( n \), given the initial conditions \( a_0 \) and \( a_1 \).

### Final Formula:

\[
a_n = \frac{a_0 + a_1}{4} 3^{n-1} + \frac{3a_0 - a_1}{4} (-1)^{n-1}
\]

This formula allows you to compute \( a_n \) directly without needing to iterate through the recurrence!