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$ \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\\ $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010