- #1
Gregg
- 459
- 0
[tex] \left(
\begin{array}{ccc}
0 & 1 & 0 \\
-1 & -1 & 1 \\
-1 & 0 & 1
\end{array}
\right) [/tex]
The answer is
[tex] \left\{\left(
\begin{array}{ccc}
1 & -1 & 0 \\
0 & 1 & -1 \\
1 & 0 & 0
\end{array}
\right),\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)\right\} [/tex]
The first one [tex] \left( \begin{array}{c}
1 \\
0 \\
1
\end{array} \right) [/tex] which is an eigenvector.
The eigenvalues are 0,0,0.
Looking at [tex] \text{Ker}(M-\lambda I)^2) =\text{Ker}(M^2)=\text{Ker}(\left(
\begin{array}{ccc}
-1 & -1 & 1 \\
0 & 0 & 0 \\
-1 & -1 & 1
\end{array}
\right))[/tex]
[tex]z=x+y[/tex]
For the first eigenvector [tex] \left( \begin{array}{c}
1 \\
0 \\
1
\end{array} \right) [/tex] is in this generalised eigenspace [itex] E_1 \subset E_2 [/itex] I thought about trying to find two vectors that produce the normal vector on that plane when crossed but that doesn't work and I'm not sure it's even the right thing to do. Can choose [tex] \left( \begin{array}{c}
-1 \\
1 \\
0
\end{array} \right) [/tex] for z=x+y. But [tex]\text{Ker}(M^3)=\text{Ker}(0) [/tex]
I know there is the method [itex] (M-\lambda I)\vec{v}_2=\lambda \vec{v_1} [/itex] ? Do I have to use that? I'd prefer to be able to use kernels.
\begin{array}{ccc}
0 & 1 & 0 \\
-1 & -1 & 1 \\
-1 & 0 & 1
\end{array}
\right) [/tex]
The answer is
[tex] \left\{\left(
\begin{array}{ccc}
1 & -1 & 0 \\
0 & 1 & -1 \\
1 & 0 & 0
\end{array}
\right),\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)\right\} [/tex]
The first one [tex] \left( \begin{array}{c}
1 \\
0 \\
1
\end{array} \right) [/tex] which is an eigenvector.
The eigenvalues are 0,0,0.
Looking at [tex] \text{Ker}(M-\lambda I)^2) =\text{Ker}(M^2)=\text{Ker}(\left(
\begin{array}{ccc}
-1 & -1 & 1 \\
0 & 0 & 0 \\
-1 & -1 & 1
\end{array}
\right))[/tex]
[tex]z=x+y[/tex]
For the first eigenvector [tex] \left( \begin{array}{c}
1 \\
0 \\
1
\end{array} \right) [/tex] is in this generalised eigenspace [itex] E_1 \subset E_2 [/itex] I thought about trying to find two vectors that produce the normal vector on that plane when crossed but that doesn't work and I'm not sure it's even the right thing to do. Can choose [tex] \left( \begin{array}{c}
-1 \\
1 \\
0
\end{array} \right) [/tex] for z=x+y. But [tex]\text{Ker}(M^3)=\text{Ker}(0) [/tex]
I know there is the method [itex] (M-\lambda I)\vec{v}_2=\lambda \vec{v_1} [/itex] ? Do I have to use that? I'd prefer to be able to use kernels.