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!