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\def\bmatrix#1{\left[\matrix{#1}\right]}
\def\dmatrix#1{\left|\matrix{#1}\right|}
\def\fr#1#2{{#1\over #2}}
{\bf Section 6.1}
\noindent{\bf 5.}
$$A = \bmatrix{1&b\cr b&9}.$$
{\it (a)}
$a_{11} = 1 > 0,$
and $\det A = 9 - b^2 > 0$ when $-3 < b < 3.$
{\it (b)}
$$A = \bmatrix{1&0\cr b&1}\bmatrix{1&b\cr 0&9-b^2}
= \bmatrix{1&0\cr b&1}\bmatrix{1&0\cr 0&9-b^2}\bmatrix{1&b\cr 0&1}
$$
{\it (c)}
$(x-v)^TA(x-v) + c = x^TAx - 2x^TAv + v^TAv + c$ lets us generate
arbitrary first and zeroth order terms while retaining the same
second-order terms ($v^TAx = x^TAv$ by symmetry) since $x^TAx$ are the
only second order terms.
$$\fr12(x^2+2bxy+9y^2)-y = \fr12(x^TAx - 2x^TAv + v^TAv + c) \Rightarrow
y = x^TAv - v^TAv - c \Rightarrow
y = x^TAv, c = -v^TAv.$$
$$y = x^TAv \Rightarrow \bmatrix{0\cr1} = Av \Rightarrow
v = \bmatrix{-b/(9-b^2)\cr 1/(9-b^2)},$$
by row reduction.
The minimum is
$$\fr12c = -\fr12v^TAv = -\fr12v^T\bmatrix{0\cr1} = \fr1{2(b^2-9)},$$
when $A$ is positive definite ($|b| < 3$).
{\it (d)}
There is no minimum if $b=3$ because $x^2+2bxy+9y^2 = x^2+6xy+9y^2 =
(x+3y)^2$ is zero where $y = -x/3,$ giving nonzero values of $y$ with a
zero quadratic component and therefore arbitrarily small values of
$\fr12(x^2+2bxy+9y^2) - y.$
{\bf Section 6.2}
\noindent{\bf 4.}
Since $A$ is positive definite, $A$ has eigenvalue $\lambda > 0$ and
eigenvector $v,$ $A^2v = \lambda^2v,$ giving that $A^2$ has eigenvalue
$\lambda^2 > 0$ and no others (because the complete set of eigenvectors,
from symmetry of $A^2,$ span $R^n.$)
Similarly, $A^{-1}$ is symmetric ($(A^{-1}A)^T = I \to (A^{-1})^TA = I
\to (A^{-1})^T = A^{-1}$), so its eigenvectors will span the codomain,
and $Av = \lambda v \to A^{-1}Av = A^{-1}\lambda v \to (1/\lambda)v =
A^{-1}v,$ and $\lambda > 0 \to 1/\lambda > 0,$ so it is also
positive-definite.
\noindent{\bf 24.}
Letting $s=0$ gives us a set of eigenvalues with minimum
$\lambda_{\rm min},$ and the set of eigenvalues for arbitrary $s$ has
minimum $\lambda_{\rm min}+s > 0 \to s > -\lambda_{\rm min}.$
$$
\bmatrix{0&-4&-4\cr -4&0&-4\cr -4&-4&0} \to \lambda = 4,-8 \to s > 8.$$
$$
\bmatrix{0&3&0\cr 3&0&4\cr 0&4&0} \to \lambda = -4,0,4 \to s > 4.$$
\noindent{\bf 27.}
$$A = CC^T = \bmatrix{3&0\cr1&2}\bmatrix{3&1\cr0&2} = \bmatrix{10&3\cr
3&5}.$$
$$A = \bmatrix{4&8\cr8&25} = \bmatrix{2&0\cr 4&3}\bmatrix{2&4\cr 0&3}.$$
(This was found by inspection after a first-order approximation, with
$U = \bmatrix{4&8\cr0&9},$ (in the LU decomposition) giving $\sqrt D =
\bmatrix{2&0\cr0&3}$)
\noindent{\bf 31.}
For $A,$ it has eigenvalue $3$ corresponding to $(1,-1,0)^T$ and
$(1,1,-2)^T.$
$$x^TAx = 3/2(x-y)^2+1/2(x+y-2z)^2.$$
$$x^TBx = x^T\bmatrix{x+y+z\cr x+y+z\cr x+y+z} = (x+y+z)^2.$$
\goodbreak
{\bf Section 6.3}
\noindent{\bf 12.}
{\it (a)}
If $A' = 4A,$ $\Sigma' = 4\Sigma,$ because $(A')^TA' = 4A^T\cdot4A =
16A^TA,$ meaning the diagonal of $\Sigma'$ will have values $4$ times
$\Sigma.$
{\it (b)}
The SVD of $A^T$ is $V\Sigma U^T,$ by $(AB)^T = B^TA^T,$ and that
$\Sigma$ is symmetric (and $V$ and $U$ still satisfy orthogonality
because orthogonality is preserved for square matrices across a
transpose).
The SVD of $A^{-1} = A^+$ is $V\Sigma^{-1} U^T,$ because the
pseudoinverse is the complete inverse for an invertible matrix.
\noindent{\bf 15.}
$$A = \bmatrix{1&1&1&1} = \bmatrix{1}\bmatrix{2&0&0&0}
\bmatrix{1/2&1/2&1/2&1/2\cr
-1/\sqrt2&1/\sqrt2&0&0\cr
-1/\sqrt2&0&1/\sqrt2&0\cr
-1/\sqrt2&0&0&1/\sqrt2\cr}$$
$$A^+ = \bmatrix{1/2&-1/\sqrt2&-1/\sqrt2&-1/\sqrt2\cr
1/2&1/\sqrt2&0&0\cr
1/2&0&1/\sqrt2&0\cr
1/2&0&0&1/\sqrt2\cr}\bmatrix{1/2\cr 0\cr 0\cr 0}
\bmatrix{1}
= \bmatrix{1/4\cr1/4\cr1/4\cr1/4}.
$$
$$B = \bmatrix{0&1&0\cr 1&0&0} = \bmatrix{1&0\cr0&1}\bmatrix{1&0&0\cr0&1&0}
\bmatrix{0&1&0\cr1&0&0\cr0&0&1}
$$
$$B^+ = \bmatrix{0&1&0\cr 1&0&0\cr 0&0&1}\bmatrix{1&0\cr0&1\cr0&0}
\bmatrix{1&0\cr0&1}
= \bmatrix{0&1\cr1&0\cr0&0}$$
$$C = \bmatrix{1&1\cr 0&0} =
\bmatrix{1&0\cr0&1}\bmatrix{\sqrt2&0\cr0&0}\bmatrix{1/\sqrt2&1/\sqrt2\cr1/\sqrt2&-1/\sqrt2}$$
$$C^+ =
\bmatrix{1/\sqrt2&1/\sqrt2\cr1/\sqrt2&-1/\sqrt2}\bmatrix{1/\sqrt2&0\cr0&0}
\bmatrix{1&0\cr0&1} =
\bmatrix{1/2&1/2\cr0&0}$$
\noindent{\bf 18.}
With $r_i$ as the $i$th row of $A,$ $A,$ $\hat x = c_1r_1 + c_3r_3.$
$$A^TA\hat x = A^Tb =
A^T\bmatrix{1&0&0\cr1&0&0\cr1&1&1}(c_1\bmatrix{1\cr0\cr0} +
c_3\bmatrix{1\cr1\cr1}) =
\bmatrix{1&1&1\cr0&0&1\cr0&0&1}(c_1\bmatrix{1\cr1\cr1} +
c_3\bmatrix{1\cr1\cr3}) = $$$$
c_1\bmatrix{3\cr1\cr1} + c_3\bmatrix{5\cr3\cr3}
= \bmatrix{1&1&1\cr0&0&1\cr0&0&1}\bmatrix{0\cr2\cr2} =
\bmatrix{4\cr2\cr2} = (1/2)\bmatrix{3\cr1\cr1} +
(1/2)\bmatrix{5\cr3\cr3} \to \hat x = \bmatrix{1\cr1/2\cr1/2}
$$
\bye
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