As in the previous problem, we consider the matrix
H = I_ n - \frac{1}{n} \mathbf{1} \mathbf{1}^ T
and for simplicity let n = 3. Let \mathbf{x} = (2, -1, -2)^ T.
What is H^{100} \mathbf{x}?
(Enter the components of H^{100} \mathbf{x} below. Below, (H^{100} \mathbf{x})^{(i)} denotes the i^{\text {th}} component of H^{100} \mathbf{x}.)
(H^{100} \mathbf{x})^{(1)} =
unanswered
(H^{100} \mathbf{x})^{(2)} =
unanswered
(H^{100} \mathbf{x})^{(3)} =
To find H^{100} \mathbf{x}, we first need to find H^k \mathbf{x} for smaller values of k and observe the pattern.
For n = 3, the matrix H is:
H = I_3 - \frac{1}{3} \mathbf{1} \mathbf{1}^T
= \begin{pmatrix} 1-\frac{1}{3} & 0 & 0 \\ 0 & 1-\frac{1}{3} & 0 \\ 0 & 0 & 1-\frac{1}{3} \end{pmatrix}
= \begin{pmatrix} \frac{2}{3} & 0 & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & 0 & \frac{2}{3} \end{pmatrix}
Now, let's find H^k \mathbf{x} for several values of k:
H^2 \mathbf{x} = H(H \mathbf{x}) = \begin{pmatrix} \frac{2}{3} & 0 & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & 0 & \frac{2}{3} \end{pmatrix} \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} \frac{4}{3} \\ -\frac{2}{3} \\ -\frac{4}{3} \end{pmatrix}
H^3 \mathbf{x} = H(H^2 \mathbf{x}) = \begin{pmatrix} \frac{2}{3} & 0 & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & 0 & \frac{2}{3} \end{pmatrix} \begin{pmatrix} \frac{4}{3} \\ -\frac{2}{3} \\ -\frac{4}{3} \end{pmatrix} = \begin{pmatrix} \frac{8}{9} \\ -\frac{4}{9} \\ -\frac{8}{9} \end{pmatrix}
H^4 \mathbf{x} = H(H^3 \mathbf{x}) = \begin{pmatrix} \frac{2}{3} & 0 & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & 0 & \frac{2}{3} \end{pmatrix} \begin{pmatrix} \frac{8}{9} \\ -\frac{4}{9} \\ -\frac{8}{9} \end{pmatrix} = \begin{pmatrix} \frac{16}{27} \\ -\frac{8}{27} \\ -\frac{16}{27} \end{pmatrix}
From this pattern, we observe that H^k \mathbf{x} can be written as:
H^k \mathbf{x} = \left(\frac{2}{3}\right)^k \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}
Therefore, for H^{100} \mathbf{x}, we have:
(H^{100} \mathbf{x})^{(1)} = \left(\frac{2}{3}\right)^{100} \cdot 2 = \frac{2^{101}}{3^{100}}
(H^{100} \mathbf{x})^{(2)} = \left(\frac{2}{3}\right)^{100} \cdot -1 = -\frac{2^{100}}{3^{100}}
(H^{100} \mathbf{x})^{(3)} = \left(\frac{2}{3}\right)^{100} \cdot -2 = -\frac{2^{101}}{3^{100}}
Therefore, (H^{100} \mathbf{x})^{(1)} = \frac{2^{101}}{3^{100}}, (H^{100} \mathbf{x})^{(2)} = -\frac{2^{100}}{3^{100}}, (H^{100} \mathbf{x})^{(3)} = -\frac{2^{101}}{3^{100}}.
To calculate H^{100} \mathbf{x}, we first need to compute the matrix H^k for k = 1, 2, ..., 100. Then we can multiply H^{100} with the vector \mathbf{x}.
Let's start by calculating H^k for k = 1, 2, ..., 100.
Step 1: Define the matrix H
H = I_3 - \frac{1}{3} \mathbf{1} \mathbf{1}^ T
Step 2: Calculate H^k for k = 1, 2, ..., 100 using the formula:
H^k = H^{k-1} \cdot H
Step 3: Multiply H^{100} with the vector \mathbf{x}
(H^{100} \mathbf{x})^{(i)} = (H^{100})_{i1} \cdot x_1 + (H^{100})_{i2} \cdot x_2 + (H^{100})_{i3} \cdot x_3
Now let's calculate H^{100} \mathbf{x}.
To find the vector H^{100} \mathbf{x}, we need to raise the matrix H to the power of 100 and then multiply it by the vector \mathbf{x}.
First, let's calculate the matrix H.
H = I_ n - \frac{1}{n} \mathbf{1} \mathbf{1}^ T
Here, I_n represents the identity matrix of size n, and \mathbf{1} is the column vector of all ones.
In this problem, n = 3, so the matrix H will be a 3x3 matrix.
H = I_3 - \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}
H = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \frac{1}{3} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}
Simplifying, we get:
H = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \end{bmatrix}
Now, let's find H^{100} by raising the matrix H to the power of 100. You can do this by multiplying H by itself 100 times, or you can use matrix exponentiation algorithms.
Once we have H^{100}, we multiply it by the vector \mathbf{x}.
Finally, we can calculate each component of (H^{100} \mathbf{x}) as follows:
(H^{100} \mathbf{x})^{(1)} = [H^{100} \mathbf{x}]_1
(H^{100} \mathbf{x})^{(2)} = [H^{100} \mathbf{x}]_2
(H^{100} \mathbf{x})^{(3)} = [H^{100} \mathbf{x}]_3
To find the numerical values of (H^{100} \mathbf{x})^{(1)}, (H^{100} \mathbf{x})^{(2)}, and (H^{100} \mathbf{x})^{(3)}, we need to actually calculate H^{100} and multiply it by the vector \mathbf{x}. Unfortunately, I am unable to perform these calculations as they involve large matrix exponentiation and multiplication.