Fix linear algebra folder structure

semester1/linear-algebra/recursive-sequences.md

@@ -1,269 +0,0 @@
-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!