Commit ec85863d authored by Jim Hefferon's avatar 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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