As in the previous problem, we consider the matrix

Question

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^{	ext {th}} component of H^{100} \mathbf{x}.)

(H^{100} \mathbf{x})^{(1)} = 
 unanswered 
(H^{100} \mathbf{x})^{(2)} = 
 unanswered 
(H^{100} \mathbf{x})^{(3)} =

GPT 3.5 · Answer

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}}.