Commit ec85863d by Jim Hefferon

### jc2 edits

parent 27ac2843
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 ... ... @@ -25642,7 +25642,7 @@ ans = 0.017398 \end{ans} \subsection{Subsection Five.II.3: Eigenvalues and Eigenvectors} \begin{ans}{Five.II.3.20} \begin{ans}{Five.II.3.22} \begin{exparts} \partsitem This \begin{equation*} ... ... @@ -25664,7 +25664,7 @@ ans = 0.017398 \end{exparts} \end{ans} \begin{ans}{Five.II.3.21} \begin{ans}{Five.II.3.23} \begin{exparts} \partsitem The characteristic equation is $$(3-x)(-1-x)=0$$. Its roots, the eigenvalues, are $$\lambda_1=3$$ and ... ... @@ -25777,7 +25777,7 @@ ans = 0.017398 \end{exparts} \end{ans} \begin{ans}{Five.II.3.22} \begin{ans}{Five.II.3.24} The characteristic equation \begin{equation*} 0= ... ... @@ -25845,7 +25845,7 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.23} \begin{ans}{Five.II.3.25} The characteristic equation is \begin{equation*} 0= ... ... @@ -25883,7 +25883,7 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.24} \begin{ans}{Five.II.3.26} \begin{exparts} \partsitem The characteristic equation is \begin{equation*} ... ... @@ -26086,7 +26086,7 @@ ans = 0.017398 \end{exparts} \end{ans} \begin{ans}{Five.II.3.25} \begin{ans}{Five.II.3.27} With respect to the natural basis $B=\sequence{1,x,x^2}$ the matrix representation is this. \begin{equation*} ... ... @@ -26173,7 +26173,7 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.26} \begin{ans}{Five.II.3.28} $\lambda=1, \begin{mat} 0 &0 \\ ... ... @@ -26195,7 +26195,7 @@ ans = 0.017398 \end{mat}$ \end{ans} \begin{ans}{Five.II.3.27} \begin{ans}{Five.II.3.29} Fix the natural basis $B=\sequence{1,x,x^2,x^3}$. The map's action is $1\mapsto 0$, $x\mapsto 1$, $x^2\mapsto 2x$, and $x^3\mapsto 3x^2$ and its representation is easy to compute. ... ... @@ -26244,13 +26244,13 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.28} \begin{ans}{Five.II.3.30} The determinant of the triangular matrix $T-xI$ is the product down the diagonal, and so it factors into the product of the terms $t_{i,i}-x$. \end{ans} \begin{ans}{Five.II.3.29} \begin{ans}{Five.II.3.31} Just expand the determinant of $T-xI$. \begin{equation*} \begin{vmatrix} ... ... @@ -26262,12 +26262,12 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.30} \begin{ans}{Five.II.3.32} Any two representations of that transformation are similar, and similar matrices have the same characteristic polynomial. \end{ans} \begin{ans}{Five.II.3.31} \begin{ans}{Five.II.3.33} It is not true. All of the eigenvalues of this matrix are $0$. \begin{equation*} ... ... @@ -26278,7 +26278,7 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.32} \begin{ans}{Five.II.3.34} \begin{exparts} \partsitem Use $$\lambda=1$$ and the identity map. \partsitem Yes, use the transformation that multiplies all ... ... @@ -26286,12 +26286,12 @@ ans = 0.017398 \end{exparts} \end{ans} \begin{ans}{Five.II.3.33} \begin{ans}{Five.II.3.35} If $t(\vec{v})=\lambda\cdot\vec{v}$ then $\vec{v}\mapsto\zero$ under the map $t-\lambda\cdot\identity$. \end{ans} \begin{ans}{Five.II.3.34} \begin{ans}{Five.II.3.36} The characteristic equation \begin{equation*} 0= ... ... @@ -26306,7 +26306,7 @@ ans = 0.017398 (under the $a+b=c+d$ condition) is routine. \end{ans} \begin{ans}{Five.II.3.35} \begin{ans}{Five.II.3.37} Consider an eigenspace $V_{\lambda}$. Any $\vec{w}\in V_{\lambda}$ is the image $\vec{w}=\lambda\cdot\vec{v}$ of some $\vec{v}\in V_{\lambda}$ ... ... @@ -26317,7 +26317,7 @@ ans = 0.017398 and so $1/\lambda$ is an eigenvalue of $t^{-1}$. \end{ans} \begin{ans}{Five.II.3.36} \begin{ans}{Five.II.3.38} \begin{exparts} \partsitem We have $(cT+dI)\vec{v}=cT\vec{v}+dI\vec{v}=c\lambda\vec{v}+d\vec{v} ... ... @@ -26328,7 +26328,7 @@ ans = 0.017398 \end{exparts} \end{ans} \begin{ans}{Five.II.3.37} \begin{ans}{Five.II.3.39} The scalar$\lambda$is an eigenvalue if and only if the transformation$t-\lambda \identity$is singular. A transformation is singular if and only if it is not an isomorphism ... ... @@ -26336,7 +26336,7 @@ ans = 0.017398 nonsingular). \end{ans} \begin{ans}{Five.II.3.38} \begin{ans}{Five.II.3.40} \begin{exparts} \partsitem Where the eigenvalue$\lambda$is associated with the eigenvector$\vec{x}$then ... ... @@ -26348,7 +26348,7 @@ ans = 0.017398 \end{exparts} \end{ans} \begin{ans}{Five.II.3.39} \begin{ans}{Five.II.3.41} No. These are two same-sized, equal rank, matrices with different eigenvalues. ... ... @@ -26365,12 +26365,12 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.40} \begin{ans}{Five.II.3.42} The characteristic polynomial has an odd power and so has at least one real root. \end{ans} \begin{ans}{Five.II.3.41} \begin{ans}{Five.II.3.43} The characteristic polynomial$x^3-5x^2+6x$has distinct roots $$\lambda_1=0$$, $$\lambda_2=-2$$, and $$\lambda_3=-3$$. Thus the matrix can be diagonalized into this form. ... ... @@ -26383,7 +26383,7 @@ ans = 0.017398 \end{equation*} \end{ans} \begin{ans}{Five.II.3.42} \begin{ans}{Five.II.3.44} We must show that it is one-to-one and onto, and that it respects the operations of matrix addition and scalar multiplication. ... ... @@ -26402,7 +26402,7 @@ ans = 0.017398 similar:~$t_P(cT)=P(c\cdot T)P^{-1}=c\cdot (PTP^{-1})=c\cdot t_P(T)\$. \end{ans} \begin{ans}{Five.II.3.43} \begin{ans}{Five.II.3.45} \answerasgiven % If the argument of the characteristic function of $$A$$ is set equal to $$c$$, adding the first $$(n-1)$$ rows (columns) to the
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