Line 6: Line 6:
 
-1& -1 & -1
 
-1& -1 & -1
 
\end{bmatrix}</math>
 
\end{bmatrix}</math>
 +
<math>\quad rank=1\ne \mbox 3</math><math>\quad ,Not\quad controllable</math>
 +
 +
<math>\mathbf{b)} \quad The  reachable  Subspace\quad is \begin{Bmatrix}
 +
\begin{bmatrix}
 +
1 \\
 +
1 \\
 +
-1 
 +
\end{Bmatrix}.</math>
 +
 +
<math>\mathbf{c)} \quad 0=\begin{bmatrix}
 +
0 & 1 & 1 \\
 +
0 & 0 & 0 \\
 +
0 & 0 & 0 \\
 +
0 & 0 & 0
 +
\end{bmatrix}
 +
\quad Not \quad observable</math>
 +
 +
 +
<math>\mathbf{d)} \quad X_1=r,X_2=-s,X_3=s  \quad
 +
 +
X=r\begin{bmatrix}
 +
1 \\
 +
0 \\
 +
0
 +
\end{bmatrix}+s\begin{bmatrix}
 +
0 \\
 +
-1 \\
 +
1
 +
\end{bmatrix}\\</math>
 +
<math>Subspace\quad is \begin{Bmatrix}
 +
\begin{bmatrix}
 +
1 \\
 +
0 \\
 +
0
 +
\end{bmatrix} ,& \begin{bmatrix}
 +
0 \\
 +
-1  \\
 +
1
 +
\end{bmatrix}
 +
\end{Bmatrix}.\\</math>
 +
<math>\mathbf{e)} \quad \lambda I-A=\begin{bmatrix}
 +
\lambda-1 & -1 & -1 \\
 +
0 & \lambda & -1 \\
 +
0 & 0 & \lambda+1
 +
\end{bmatrix}\quad \lambda_1=1,\lambda_2=0,\lambda_3=-1\\</math>
 +
<math>\qquad for \;\lambda_1=1 \qquad\begin{bmatrix}
 +
\lambda I-A & B \end{bmatrix}=\begin{bmatrix}
 +
0 & -1 & -1 & 1 \\
 +
0 & 1 & -1 & 1 \\
 +
0 & 0 & 2 & 0
 +
\end{bmatrix}\\</math>
 +
<math>\qquad for \;\lambda_2=0 \qquad\begin{bmatrix}
 +
\lambda I-A & B \end{bmatrix}=\begin{bmatrix}
 +
-1 & -1 & -1 & 1 \\
 +
0 & 0 & -1 & 1 \\
 +
0 & 0 & 1 & 0
 +
\end{bmatrix}\\</math>
 +
<math>\qquad for \;\lambda_3=-1 \qquad\begin{bmatrix}
 +
\lambda I-A & B \end{bmatrix}=\begin{bmatrix}
 +
-2 & -1 & -1 & 1 \\
 +
0 & -1 & -1 & 1 \\
 +
0 & 0 & 0 & 0
 +
\end{bmatrix}
 +
\qquad rank<3 \qquad must\;contain\;\lambda=-1
 +
\qquad so\qquad No.\\</math>
 +
<math>\mathbf{f)} \quad \begin{bmatrix}
 +
\lambda I-A \\
 +
C \end{bmatrix}=\begin{bmatrix}
 +
\lambda-1 & -1 & -1 \\
 +
0 & \lambda & -1 \\
 +
0 & 0 & \lambda+1 \\
 +
0 & 1 & 1
 +
\end{bmatrix}</math>
 +
 +
<math>for\; \lambda_1=1 \quad \begin{bmatrix}
 +
\lambda I-A \\
 +
C \end{bmatrix}=\begin{bmatrix}
 +
0 & -1 & -1 \\
 +
0 & 1 & -1 \\
 +
0 & 0 & 2 \\
 +
0 & 1 & 1
 +
\end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_1=1</math>
 +
 +
<math>for\; \lambda_2=0 \quad \begin{bmatrix}
 +
\lambda I-A \\
 +
C \end{bmatrix}=\begin{bmatrix}
 +
-1 & -1 & -1 \\
 +
0 & 0 & -1 \\
 +
0 & 0 & 1 \\
 +
0 & 1 & 1
 +
\end{bmatrix}\\</math>
 +
 +
<math>for \lambda-3=-1\quad \begin{bmatrix}
 +
\lambda I-A \\
 +
C \end{bmatrix}=\begin{bmatrix}
 +
-2 & -1 & -1 \\
 +
0 & -1 & -1 \\
 +
0 & 0 & 0 \\
 +
0 & 1 & 1
 +
\end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_3=-1\\</math>

Revision as of 23:29, 16 May 2017

$ \mathbf{a)} \quad C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ -1& -1 & -1 \end{bmatrix} $ $ \quad rank=1\ne \mbox 3 $$ \quad ,Not\quad controllable $

$ \mathbf{b)} \quad The reachable Subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 1 \\ -1 \end{Bmatrix}. $

$ \mathbf{c)} \quad 0=\begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \quad Not \quad observable $


$ \mathbf{d)} \quad X_1=r,X_2=-s,X_3=s \quad X=r\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}+s\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}\\ $ $ Subspace\quad is \begin{Bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} ,& \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} \end{Bmatrix}.\\ $ $ \mathbf{e)} \quad \lambda I-A=\begin{bmatrix} \lambda-1 & -1 & -1 \\ 0 & \lambda & -1 \\ 0 & 0 & \lambda+1 \end{bmatrix}\quad \lambda_1=1,\lambda_2=0,\lambda_3=-1\\ $ $ \qquad for \;\lambda_1=1 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} 0 & -1 & -1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 2 & 0 \end{bmatrix}\\ $ $ \qquad for \;\lambda_2=0 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} -1 & -1 & -1 & 1 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix}\\ $ $ \qquad for \;\lambda_3=-1 \qquad\begin{bmatrix} \lambda I-A & B \end{bmatrix}=\begin{bmatrix} -2 & -1 & -1 & 1 \\ 0 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \qquad rank<3 \qquad must\;contain\;\lambda=-1 \qquad so\qquad No.\\ $ $ \mathbf{f)} \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} \lambda-1 & -1 & -1 \\ 0 & \lambda & -1 \\ 0 & 0 & \lambda+1 \\ 0 & 1 & 1 \end{bmatrix} $

$ for\; \lambda_1=1 \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} 0 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_1=1 $

$ for\; \lambda_2=0 \quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} -1 & -1 & -1 \\ 0 & 0 & -1 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}\\ $

$ for \lambda-3=-1\quad \begin{bmatrix} \lambda I-A \\ C \end{bmatrix}=\begin{bmatrix} -2 & -1 & -1 \\ 0 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{bmatrix} \quad rank<3 \quad must\;have\;\lambda_3=-1\\ $

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett