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Roots With Matrices

We can find roots using matrices. To find the cube root of 2, we use the matrix  
\[A= \left( \begin{array}{ccc} y & 2 & 2x \\ x & y & 2 \\ 1 & x & y \end{array} \right) \]
.
This matrix is constructed by building diagonals. First a 1 in the bottom left corner, above this a diagonal of  
\[x\]
's (an approximation to  
\[ \sqrt[3]{2}\]
, which we guess), above this a diagonal of  
\[y\]
's (an approximation to  
\[\sqrt[3]{2^2}\]
, which we guess), above this a diagonal of 2's, and finally in the upper right, a  
\[2x\]
.
If we take a non zero vector  
\[\mathbf{v}\]
  and repeatedly multiply this vector by  
\[A\]
. The components of  
\[A^n \mathbf{v|}\]
  will tend to the ration  
\[2^{2/3} \colon 2^{1/3} \colon 1 \]
.
Let our guess for  
\[x= \sqrt[3]{2}\]
  be 1 and let our guess for  
\[y=\sqrt[3]{2^2}\]
  be 2. Then  
\[A= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right)\]

Let  
\[\mathbf{v}= \begin{pmatrix}1\\1\\1\end{pmatrix}\]
.
Then  
\[A \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}6\\5\\4\end{pmatrix}\]
.
\[A^2 \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}6\\5\\4\end{pmatrix} = \begin{pmatrix}30\\24\\19\end{pmatrix}\]
.
\[A^3 \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}30\\24\\19\end{pmatrix} = \begin{pmatrix}146\\116\\92\end{pmatrix}\]
.
\[A^4 \mathbf{v}= \left( \begin{array}{ccc} 2 & 2 & 2 \\ 1 & 2 & 2 \\ 1 & 1 & 2 \end{array} \right) \begin{pmatrix}146\\116\\92\end{pmatrix} = \begin{pmatrix}708\\562\\446\end{pmatrix}\]
.
Hence  
\[2^{2/3} \simeq \frac{708}{446}=1.587443946\]
  and  
\[2^{1/3} \simeq \frac{562}{446}=1.260089686\]
.
Compare these with the true values  
\[2^{2/3} = 1.587401052\]
  and  
\[2^{1/3} = 1.259921050\]
, both to 10 significant figures.
To find an approximation to  
\[5^{1/4}\]
  use the matrix  
\[A= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \]
.
The 1's in the second diagonal are an approximation to  
\[5^{1/4}\]
, the 2's an approximation to  
\[5^{2/4}\]
, the 4's an approximation to  
\[5^{3/4}\]
.
Let  
\[\mathbf{v}= \begin{pmatrix}1\\1\\1\\1\end{pmatrix}\]
.
Then  
\[A \mathbf{v}= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \begin{pmatrix}1\\1\\1\\1\end{pmatrix} = \begin{pmatrix}24\\16\\12\\8\end{pmatrix}\]
.
\[A^2 \mathbf{v}= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \begin{pmatrix}24\\16\\12\\8\end{pmatrix} = \begin{pmatrix}316\\212\\144\\96\end{pmatrix}\]
.
\[A^3 \mathbf{v}= \left( \begin{array}{cccc} 4 & 5 & 5 & 10 \\ 2 & 4 & 5 & 5 \\ 1 & 2 & 4 & 5 \\ 1 & 1 & 2 & 4 \end{array} \right) \begin{pmatrix}316\\212\\144\\96\end{pmatrix} = \begin{pmatrix}4004\\2680\\1796\\1200\end{pmatrix}\]
.
Hence  
\[5^{3/4} \simeq \frac{4004}{1200}=3.336666667\]
,  
\[5^{2/4} \simeq \frac{2680}{1200}=2.233333333\]
  and  
\[5^{1/4} \simeq \frac{1796}{1200}=2.1.496666667\]
.
Compare these with the true values  
\[5^{3/4} = 3.343701525\]
,  
\[5^{2/4} = 2.236067977\]
, and   
\[5^{1/4} = 1.495348781\]
, all to 10 significant figures.