## 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.