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diff --git a/li/hw7.tex b/li/hw7.tex new file mode 100644 index 0000000..61bdb6f --- /dev/null +++ b/li/hw7.tex @@ -0,0 +1,173 @@ +\def\bmatrix#1{\left[\matrix{#1}\right]} +\def\fr#1#2{{#1\over #2}} + + {\bf Section 4.2} + +\noindent{\bf 12.} + +$$\det\bmatrix{1&a&a^2\cr1&b&b^2\cr1&c&c^2} = +\det\bmatrix{1&a&a^2\cr0&b-a&b^2-a^2\cr0&c-a&c^2-a^2} = +(b-a)(c-a)\det\bmatrix{1&a&a^2\cr0&1&b+a\cr0&1&c+a} =$$$$ +(b-a)(c-a)\det\bmatrix{1&a&a^2\cr0&1&b+a\cr0&0&c-b} += (b-a)(c-a)(c-b). +$$ + +\noindent{\bf 17.} + +The determinant of $A$ is $4*3 - 2*1 = 10.$ + +$$\det(AA^{-1}) = \det I = 1 = \det(A)\det(A^{-1}) \to \det(A^{-1}) = +1/\det(A) = 1/10.$$ + +$$\det(A-\lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda ++ 10 = (\lambda - 2)(\lambda - 5).$$ +This gives zeroes (meaning $A-\lambda I$ is singular) of $\lambda = 2, 5.$ + +\noindent{\bf 34.} + +By linearity in each row, the determinant of B is the sum of the +determinants of every choice from the original set. However, if any row +is repeated in the matrix being determined, the determinant is zero, so +\def\row{\mathop{\rm row}} +$$\det B = \left|\matrix{\row 1\cr\row 2\cr\row 3}\right| ++ \left|\matrix{\row 2\cr\row 3\cr\row 1}\right| += 6 + 6,$$ +the determinant of the second matrix being equivalent to $\det A$ by two +row switches. + +\iffalse % this is the answer to 35 :'( +Row operations give us +$$\det(I+M) = \det\bmatrix{1+a & b & c & d\cr -1 & 1 & 0 & 0\cr -1 & 0 & +1 & 0\cr -1 & 0 & 0 & 1} = +\det\bmatrix{1+a+b+c+d & 0 & 0 & 0\cr -1 & 1 & 0 & 0\cr -1 & 0 & +1 & 0\cr -1 & 0 & 0 & 1},$$ +giving determinant by product of the diagonal entries $1+a+b+c+d.$ +\fi + + {\bf Section 4.3} + +\noindent{\bf 3.} + +{\it (a)} + +True. +$$\det(S^{-1}AS) = \det(S^{-1})\det(A)\det(S) = \det(A)\det(S^{-1}S) = +\det(A).$$ + +{\it (b)} + +False, the matrix +$$\bmatrix{1&1\cr 1&1}$$ +has determinant $1*1 - 1*1,$ which is a cofactor expansion with every +cofactor either $1$ or $-1,$ but the determinant is still zero. + +{\it (c)} + +False, +$$\left|\matrix{1&1&0\cr1&0&1\cr0&1&1}\right| = -2.$$ + +\noindent{\bf 6.} + +{\it (a)} + +With $a_{ij}$ the entry on $A$ in the $i$th row and $j$th column, and +$c_{ij}$ the determinant of the matrix minor of $A$ (matrix $A$ without +the $i$th row or $j$th column). + +$$D_n = a_{11}c_{11} - a_{12}c_{12} + a_{13}c_{13} + \cdots = c_{11} - +c_{12} + 0 + \cdots = D_{n-1} - D_{n-2}.$$ +$c_{11}$ is clearly $D_{n-1}$ because the remainder after that elimination +is an $n-1 \times n-1$ tridiagonal matrix with 1s on the diagonals. + +$c_{12}$ is an $n-2\times n-2$ tridiagonal with $1,1,0,\ldots$ as the +first row, inserted above it, and $1,0,0,\ldots$ to its left as the +first column (overlapping with the first row). This gives a simple +cofactor expansion (computing on the first column and ignoring the +zeroes) of the determinant of the $n-2\times n-2$ matrix. Thus, $c_{12} += D_{n-2}.$ + +{\it (b)} + +$D_3 = 0 - 1,$ $D_4 = -1 - 0,$ $D_5 = -1 - (-1) = 0,$ $D_6 = 0 - (-1) = +1,$ $D_7 = 1 - 0,$ $D_8 = 1 - 1 = 0.$ +Therefore, the cycle has period 6, and $1000\bmod6 = 4,$ so $D_{1000} += D_4 = -1.$ + +\noindent{\bf 15.} + +$\det A$ is zero because this is a triangular matrix, so the determinant +is the product of the diagonal entries $x\cdot0\cdot x = 0.$ +The rank of $A$ is 2 unless $x = 0,$ in which case it is 0. This is +because the first and second columns are linearly dependent. + +\noindent{\bf 34.} + +{\it (a)} + +Row operations (including permutations) which make $A$ and $D$ diagonal +will be contained within their respective ``block-rows,'' and once we've +got that, the product of the diagonal entries is the same as the product +of the determinants of the new $A$ and $D$ blocks (because those +triangular matrices have determinant equal to the product of diagonal +entries). None of this requires knowledge of $B.$ + +{\it (b)} + +$$ +\det +\bmatrix{ 0 & 0 & 1 & 0\cr + 0 & 0 & 0 & 1\cr + 1 & 0 & 0 & 0\cr + 0 & 1 & 0 & 0 } += 1, +$$ +because this can be permuted to give the identity in two operations +(switch rows 3 and 1 and rows 4 and 2). + +However, $B = C = I_2,$ which would give our block determinant formula +$|A||D| - |C||B| = 0 - 1 \neq 1.$ + +{\it (c)} + +$$ +\det +\bmatrix{ 0 & 1 & 0 & 0\cr + 0 & 0 & 1 & 0\cr + 0 & 0 & 0 & 1\cr + 1 & 0 & 0 & 0 } += -1, +$$ + +whereas +$$\det(AD - CB) = \det\bmatrix{0&1\cr-1&0} = 1.$$ + + {\bf Section 4.4} + +\noindent{\bf 28.} + +The volume is easily found as the absolute determinant of + +$$\left|\matrix{3&1&1\cr 1&3&1\cr 1&1&3}\right| = +\left|\matrix{3&1&1\cr 0&8/3&2/3\cr 0&-2&2}\right| = +1/3\left|\matrix{3&1&1\cr 0&8&2\cr 0&0&5/2}\right| = +20. +$$ + +The area of the parallelogram faces formed by each pair is the same for +each pair by symmetry of the xyz coordinates. + +If two vectors are orthogonal, the area of the parallelogram they form +(the rectangle) is the product of their norms. +And the area computation is also linear. The area of $\{v_1+v_3, v_2\}$ +is the area of $\{v_1, v_2\}$ plus the area of $\{v_3, v_2\},$ and as +was proven for determinants, the area of $\{v_1, av_1\}$ is zero. +This lets us orthogonalize the first two vectors of our set to find +area. +$$\bmatrix{3&1\cr1&3\cr1&1} = +\bmatrix{3&-10/11\cr1&26/11\cr1&4/11}\bmatrix{1&7/11\cr0&1}.$$ + +This isn't a complete QR decomposition, but the norms of the given +vectors are $\sqrt{11}$ and $\sqrt{792}/11,$ giving an area (by their +product) of $\sqrt{72}$. + +\bye diff --git a/li/hw8.tex b/li/hw8.tex new file mode 100644 index 0000000..9b2938b --- /dev/null +++ b/li/hw8.tex @@ -0,0 +1,169 @@ +\def\bmatrix#1{\left[\matrix{#1}\right]} +\def\dmatrix#1{\left|\matrix{#1}\right|} +\def\fr#1#2{{#1\over #2}} + + {\bf Section 5.1} + +\noindent{\bf 4.} + +$P$ has two eigenvalues $c_1=0$ and $c_2=1,$ giving us corresponding +eigenvectors $v_1=(1,-1)$ and $v_2=(1,1).$ +This gives a general solution +$$c_1v_1 + c_2v_2e^t,$$ +and our initial condition $u(0)$ can be written as the combination +$v_1+4v_2,$ +giving our particular solution +$$v_1 + 4v_2e^t.$$ + +\noindent{\bf 14.} + +The matrix of ones has rank 1 because any pair of rows makes a linearly +dependent set (by including a duplicate). +$$\dmatrix{1-\lambda & 1 & 1 & 1\cr + 1 & 1-\lambda & 1 & 1\cr + 1 & 1 & 1-\lambda & 1\cr + 1 & 1 & 1 & 1-\lambda\cr} = 0\Longrightarrow + \dmatrix{1-\lambda & 1 & 1 & 1\cr + \lambda & -\lambda & 0 & 0\cr + \lambda & 0 & -\lambda & 0\cr + \lambda & 0 & 0 & -\lambda\cr} = 0\Longrightarrow + \dmatrix{4-\lambda & 0 & 0 & 0\cr + 1 & -1 & 0 & 0\cr + 1 & 0 & -1 & 0\cr + 1 & 0 & 0 & -1\cr} = 0, +$$ +by row operations (including multiplications), which do not affect a +zero determinant where $\lambda\neq0.$ Because this is triangular, the +determinant is the product of the diagonal, so $\lambda = 4.$ +This eigenvalue corresponds to $(1,1,1,1)^T.$ + +The checkerboard matrix has rank 2 because any triplet of rows makes a +linearly dependent set (again by including a duplicate). + +$$\dmatrix{-\lambda&1&0&1\cr + 1&-\lambda&1&0\cr + 0&1&-\lambda&1\cr + 1&0&1&-\lambda} = 0\Longrightarrow + \dmatrix{-\lambda&0&\lambda&0\cr + 0&-\lambda&0&\lambda\cr + 0&1&-\lambda&1\cr + 1&0&1&-\lambda} = 0\Longrightarrow$$$$ + \dmatrix{-1&0&1&0\cr + 0&-1&0&1\cr + 0&0&-\lambda&2\cr + 0&0&2&-\lambda} = 0\Longrightarrow + \dmatrix{-1&0&1&0\cr + 0&-1&0&1\cr + 0&0&-\lambda&2\cr + 0&0&0&-\lambda+4/\lambda} = 0 +$$ +This gives $\lambda^2 - 4 = 0 \to \lambda = \pm 2.$ +These eigenvalues correspond to $(1,1,1,1)$ and $(1,-1,1,-1).$ + +\noindent{\bf 15.} + +The rank of an $n\times n$ matrix of ones is 1 (giving $\lambda=0$ with +multiplicity $n-1$), and its remaining non-zero eigenvalue of $n$ +(because $A-nI$ has the sum of every row, a linear combination, equal to +zero, therefore it is an eigenvalue). + +The rank of an $n\times n$ checkerboard matrix is 2 because it has only +two unique columns (giving $\lambda=0$ with multiplicity $n-2$) and its +non-zero eigenvalues being $\pm\sqrt n,$ determined from the cofactor +expansion of $A-\lambda I$ where $A$ is the checkerboard matrix. + + {\bf Section 5.2} + +\noindent{\bf 8.} + +{\it (a)} +$Au = uv^Tu = u(v\cdot u) = (v\cdot u)u,$ +giving that $u$ is an eigenvector with corresponding eigenvalue $v^Tu.$ + +{\it (b)} +The other eigenvalues are zero because there is one non-zero eigenvalue +already shown and the dimension of the null space of $A = A - 0I$ is +$n-r = n-1,$ completing the eigenvalue set. + +{\it (c)} +The trace is equal to the sum of eigenvalues, which is $v^Tu + 0 = +v^Tu,$ and it's equal to the sum of the diagonal, of which each entry is +an element-wise product of $v$ and $u.$ + +\noindent{\bf 12.} + +{\it (a)} +False. +$$\bmatrix{1&1&0\cr0&1&1\cr0&0&1}$$ +is invertible but has only the eigenvector set mentioned. + +{\it (b)} +True. Every eigenvalue corresponds to at least one eigenvector (counting +complex eigenvectors), so there must be only one (multiple) eigenvalue +of $A.$ + +{\it (c)} + +A is not diagonalizable because its eigenvector set does not span $R^3.$ + +\noindent{\bf 32.} +$$A = \bmatrix{2&1\cr1&2}.$$ +This matrix has eigenvalues $1$ and $3,$ and they correspond to +eigenvectors $(1,-1)$ and $(1,1)$ respectively, giving a diagonalization +with +$$S = \bmatrix{1&1\cr-1&1} \Longrightarrow S^{-1} = \fr12\bmatrix{1&-1\cr1&1}$$ +and +$$\Lambda = \bmatrix{1&0\cr0&3}$$ +$$A^k = S\Lambda^kS^{-1} = +\bmatrix{1&1\cr-1&1}\bmatrix{1&0\cr0&3^k}\fr12\bmatrix{1&-1\cr1&1} = +\fr12\bmatrix{1&1\cr-1&1}\bmatrix{1&-1\cr3^k&3^k} = +\fr12\bmatrix{1+3^k&-1+3^k\cr-1+3^k&1+3^k}. +$$ + + {\bf Section 5.3} + +\noindent{\bf 4.} + +$$A = \bmatrix{1/2 & 1/2\cr 1 & 0}.$$ + +{\it (a)} +$A$ has characteristic polynomial $\lambda^2 - \lambda/2 - 1/2 = +(\lambda-1)(\lambda+1/2) \to \lambda = 1, -1/2,$ with eigenvectors +respectively $(1,1)$ and $(1/2, -1).$ + +{\it (b)} + +$$A = -\fr23\bmatrix{1&1/2\cr 1&-1}\bmatrix{1&0\cr0&-1/2} +\fr13\bmatrix{2&1\cr2&-2}.$$ +As $k\to\infty,$ +$$A^k = \fr13\bmatrix{1&1/2\cr 1&-1}\bmatrix{1&0\cr0&-1/2}^k +\bmatrix{2&1\cr2&-2} = +\bmatrix{1&1/2\cr 1&-1}\bmatrix{2&1\cr0&0} = +\bmatrix{2&1\cr2&1}.$$ + +{\it (c)} + +$$A\bmatrix{1\cr0} = -\fr13\bmatrix{2&1\cr2&1}\bmatrix{1\cr0} = +\bmatrix{2/3\cr2/3},$$ +giving us the steady state limit of 2/3 from this initial condition. + +\noindent{\bf 18.} + +These matrices are stable/neutrally stable if both eigenvalues are less +than or equal to 1. Therefore, the maximal value will be when the larger +eigenvalue equals 1. + +For the first matrix with characteristic equation +$\lambda^2 - (.2+a)\lambda + .2a + .64 = 0 \to +(\lambda-(.1+a/2))^2 = .01 + .1a + a^2/4 -.2a + .64,$ +$\lambda = .1+a/2 \pm (.65-.1a+a^2/4) = +.75 - .4a + .25a^2 = 1 \to a = 2.081$ + +The eigenvalues of this matrix are $.2$ and $b,$ so the largest +neutrally stable value of $b$ is 1. + +For the third matrix with characteristic equation $\lambda^2 - +2c\lambda + c^2 - .16 = 0,$ $\lambda = c \pm .4,$ giving maximum $c$ of +$0.6.$ + +\bye |