HackMD - Collaborative Markdown Knowledge Base

$$ \newcommand{\miao}[1]{\begin{pmatrix}#1\end{pmatrix}} \newcommand\tr{{\mathrm{tr}}} $$ $$ A=\miao{-3&-4&4\\2&-1&-12\\1&6&13} $$ $$ \tr(A)=-3-1+13=9 $$ $$ \det\miao{-3&-4\\2&-1}=13 $$ $$ \det\miao{-3&4\\1&13}=-43 $$ $$ \det\miao{-1&-12\\6&13}=59 $$ The sum of principal minor of size 2 = 27 Now calculate the determinant of $A$, we see $$ \det\miao{-3&-4&4\\2&-1&-12\\1&6&13}=\det\miao{-3&-4&4\\2&-1&-12\\3&5&1}=\det\miao{0&1&5\\2&-1&-12\\3&5&1} $$ $$ =\det\miao{0&1&5\\2&0&-7\\3&5&1}=50-21-2=27 $$ The characteristic polynomial $$ \lambda^3-9\lambda^2+27\lambda-27=(\lambda-3)^3 $$ This means, eigenvalue is $3,3,3$. It is **not diagonalizable** $$ A-3I=\miao{-6&-4&4\\2&-4&-12\\1&6&10} $$ I saw the result. Didn't you? $$\dim(\ker(A-3I))=1$$ $$\dim(\ker(A-3I)^2)=2$$ $$\dim(\ker(A-3I)^3)=3$$ So for this $A$, it must have only one Jordan block of eigenvalue 3. Now let us find the Zeno's eigenvector. We have to take $\epsilon^2\neq\epsilon^3=0$ $$ A-3I-\epsilon I = $$ $$ \miao{-6-\epsilon&-4&4\\2&-4-\epsilon&-12\\1&6&10-\epsilon} $$ To solve the following linear equation, $$ \miao{-6-\epsilon&-4&4\\2&-4-\epsilon&-12\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ first row + 3 times the second row. Second row minus 2 times the thrid row $$ \miao{-\epsilon&-16-3\epsilon&-32\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ first row + $\epsilon$ times the last row $$ \miao{0&-16+3\epsilon&-32+10\epsilon-\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ First row minus the second row $$ \miao{0&4\epsilon&8\epsilon-\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ multiply the first row by $4$ $$ \miao{0&16\epsilon&32\epsilon-4\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ first row + second row $\times 16$ $$ \miao{0&-\epsilon^2&-2\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ first row $\times 16$ $$ \miao{0&-16\epsilon^2&-32\epsilon^2\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ first row -$\epsilon^2\times$ second row $$ \miao{0&0&0\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{x\\y\\z}=\miao{0\\0\\0} $$ I can solve this equation now. Observation: second column and thrid column is not colinear, which means to find a non-trivial linear dependence relation among columns, I must use the first column. I can observe a solution $$ \miao{0&0&0\\0&-16-\epsilon&-32+2\epsilon\\1&6&10-\epsilon}\miao{*\\-32+2\epsilon\\16+\epsilon}=\miao{0\\{0}\\{\color{red}0}} $$ OK $$ *+6(-32+2\epsilon)+(10-\epsilon)(16+\epsilon)=0 $$ $$ *=-6(-32+2\epsilon)-(10-\epsilon)(16+\epsilon) $$ $$ *=192-12\epsilon-160+6\epsilon+\epsilon^2=32-6\epsilon+\epsilon^2 $$ Because the Zeno-eigenvector for eigenvalue $3+\epsilon$ is $$ \vec v=\miao{32-6\epsilon+\epsilon^2\\-32+2\epsilon\\16+\epsilon} $$ My eigenvalue is $3+\epsilon$ The matrix $M$ such that $M^2+2M=A$, I can just assume that this M also has $\vec v$ as an eigenvector. Suppose the corresponding eigenvalue for $M$ is $\lambda$. $$ \lambda^2+2\lambda=3 $$ $$ \lambda^2+2\lambda-3=(\lambda-1)(\lambda+3) $$ Let's think about $\lambda=1$ We actually want (here $x$ is some nilpotent symbol) $$ (1+x)^2+2(1+x)=3+\epsilon $$ $$ (4x+x^2)=\epsilon $$ For this $x$ the vector $$ \vec v=\miao{32-6\epsilon+\epsilon^2\\-32+2\epsilon\\16+\epsilon} $$ is a vector of eigenvalue $1+x$ $$ \vec v=\miao{32-6(4x+x^2)+(4x+x^2)^2\\-32+2(4x+x^2)\\16+(4x+x^2)} $$ So now we have eigenvalue of $M$ is $1$ and $\vec v$ is the zeno's eigenvector, so we can find matrix $M$ out. **Is that clear?** $$ \vec v=\miao{32-6(4x+x^2)+16x^2\\-32+2(4x+x^2)\\16+(4x+x^2)} $$ $$ \vec v=\miao{32-24x+10x^2\\-32+8x+2x^2\\16+4x+x^2} $$ Canonical basis $$ \left\{ \miao{32\\-32\\16},\miao{-24\\8\\4},\miao{10\\2\\1} \right\} $$ Let $$ P=\miao{32&-24&10\\-32&8&2\\16&4&1} $$ $$ MP=P\miao{1&1&0\\0&1&1\\0&0&1} $$ $$ M=\miao{32&-24&10\\-32&8&2\\16&4&1}\miao{1&1&0\\0&1&1\\0&0&1}\miao{32&-24&10\\-32&8&2\\16&4&1}^{-1} $$ Is that enough? or you want me to calculate it out? ## Method of Lagurange polynomial, You only need to find a polynomial to map $3+\epsilon$ to $1+x$ where $(4x+x^2)=\epsilon$ Then apply this polynomial to the matrix $A$ then you can get the matrix $M$ We need to find a polynomial to map $3+4x+x^2$ to $1+x$. I construct the Taylor polynomial. $$ f(t)=1+a(t-3)+b(t-3)^2 $$ $$ f(3+4x+x^2)=1+a(4x+x^2)+16bx^2=1+4ax+(a+16b)x^2$$ Solve the equation $$ 4a=1\qquad a+16b=0 $$ $$ a=\frac14\qquad b=-\frac1{64} $$ Then the polynomial is $$ f(t)=1+\frac14(t-3)-\frac1{64}(t-3)^2 $$ So we only need to take $$ M = 1+\frac14(A-3)-\frac1{64}(A-3)^2 $$