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\textbf{Exercises on singular value decomposition}
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\begin{problem} (Based on 6.7 \#4.  {\em Introduction to Linear Algebra:} Strang) Verify that if we compute the singular value decomposition $A = U\Sigma V^T$ of the Fibonacci matrix $A= \small \twobytwo 1 1 1 0$, $$
\Sigma = \twobytwoc {\frac{1+\sqrt{5}}{2}}{0}{0}{\frac{\sqrt{5}-1}{2}}.$$

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\solution{

$$ A^T A = A A^T = \twobytwo 2111.$$
The eigenvalues of this matrix are the roots of $x^2 -3x +1$, which are $\displaystyle{\frac{3\pm\sqrt{5}}{2}}$.  Thus we have:
$$
\sigma_1^2 = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad \sigma_2^2 = \frac{3 - \sqrt{5}}{2}.
$$
To check that $\Sigma = \small \twobytwoc {\sigma_1}00{\sigma_2}$, we will square the entries of the matrix $\Sigma$ given above.
$$\left(\frac{1 + \sqrt{5}}{2}\right)^2 = \frac{1 + 2\sqrt{5}+5}{4} = \frac{3 + \sqrt{5}}{2}.\; \checkmark$$
$$\left(\frac{\sqrt{5}-1}{2}\right)^2 = \frac{5 - 2\sqrt{5}+1}{4} = \frac{3 - \sqrt{5}}{2}.\; \checkmark$$

}

\begin{problem}(6.7 \#11.)  Suppose $A$ has orthogonal columns $\vb w_1$, $\vb w_2$, ..., $\vb w_n$ of lengths $\sigma_1$, $\sigma_2$, ..., $\sigma_n$. Calculate $A^T A$. What are $U$, $\Sigma$, and $V$ in the SVD? 
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\solution{ Since the columns of $A$ are orthogonal, $A^T A $ is a diagonal matrix with entries ${\sigma_1}^2$, ..., ${\sigma_n}^2$. Since $A^T A = V \Sigma^2 V^T,$ we find that $\Sigma^2$ is the matrix with diagonal entries ${\sigma_1}^2$, ...,${\sigma_n}^2$ and thus that $\Sigma$ is the matrix with diagonal entries ${\sigma_1}$, ..., ${\sigma_n}$. 

Referring again to the equation $\mb A^T \mb A = \mb V \mb \Sigma^2 \mb V^T$, we conclude also that $V = I.$ 

The equation $A = U\Sigma V^T$ then tells us that $U$ must be the matrix whose columns are $\displaystyle{\frac{1}{\sigma_i} \vb w_i}$. 
}

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