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\textbf{Exercises on positive definite matrices and minima}
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\begin{problem} (6.5 \#33.  {\em Introduction to Linear Algebra:} Strang) When $A$ and $B$ are symmetric positive definite, $AB$ might not even be symmetric, but its eigenvalues are still positive. Start from $AB \vb x = \lambda \vb x$ and take dot products with $B \vb x.$ Then prove $\lambda>0.$ 

\end{problem}

\solution{

\begin{align*} AB\vb x &= \lambda \vb x\\
(AB\vb x)^T B \vb x &= (\lambda \vb x)^T B \vb x\\
(B \vb x)^T A^T B \vb x &= \lambda \vb x^T B \vb x\\
(B \vb x)^T A (B \vb x) &= \lambda (\vb x^T B \vb x).
\end{align*}
where $A^T = A$ because $A$ is symmetric. Since $A$ is positive definite we know $(B \vb x)^T A (B \vb x) > 0$, and since $B$ is positive definite $\vb x^T B \vb x > 0$. Hence, $\lambda$ must be positive as well.

}

\begin{problem}  Find the quadratic form associated with the matrix $\twobytwo 1579$. Is this function $f(x,y)$ always positive, always negative, or sometimes positive and sometimes negative?

\end{problem}

\solution{ To find the quadratic form, compute $\vb{x^T} A \vb x$:
\begin{eqnarray*}
f(x,y) &=& \onebytwo x y \twobytwo 1579 \twobyone x y\\
&=& x(x+5y) + y(7x+9y)\\
&=& \mathbf{x^2 + 12xy + 9y^2}.
\end{eqnarray*}
This expression can be positive, e.g. when $y=0$ and $x\neq 0.$ 

The expression will sometimes be negative because $A$ is not positive definite. For instance, $f(2,-2) = -8$.  Thus the quadratic form associated with the matrix $A$ is \textbf{sometimes positive and sometimes negative}.  Another way to reach this conclusion is to note that $\det A = -26$ is negative and so $\mb A$ is not positive definite. 
}

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