From 6ce1db1f867c545ebdb1afb705580514b356f883 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Wed, 5 Jan 2022 00:02:12 -0500 Subject: more math stuff --- li/hw10.tex | 157 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 157 insertions(+) create mode 100644 li/hw10.tex (limited to 'li/hw10.tex') diff --git a/li/hw10.tex b/li/hw10.tex new file mode 100644 index 0000000..2b372de --- /dev/null +++ b/li/hw10.tex @@ -0,0 +1,157 @@ +\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 -- cgit