Determine Diagonalizability of LTI System A

[Linder88]

Homework Statement

Consider the LTI (A,B,C,D) system
$$
\dot{x}=
\begin{pmatrix}
0.5&0&0&0\\
0&-2&0&0\\
1&0&0.5&0\\
0&0&0&-1
\end{pmatrix}
x+
\begin{pmatrix}
1\\
1\\
0\\
0
\end{pmatrix}
u
$$
$$
y=
\begin{pmatrix}
0&1&0&1
\end{pmatrix}
x
$$
Determine if A is diagonalizable

Homework Equations

The Attempt at a Solution

The characteristic equation is given by
$$
\lambda I-A=
\begin{pmatrix}
\lambda-0.5&0&0&0\\
0&\lambda+2&0&0\\
1&0&\lambda-0.5&0\\
0&0&0&\lambda+1
\end{pmatrix}
$$
The determinant is given by
$$
det(\lambda I-A)=(\lambda-0.5)(\lambda+2)(\lambda-0.5)(\lambda+2)
$$
What do I have to do next to determine if A is diagonalizable? I am waiting hopefully for an answer!

 

[Mark44]

Homework Statement

Consider the LTI (A,B,C,D) system
$$
\dot{x}=
\begin{pmatrix}
0.5&0&0&0\\
0&-2&0&0\\
1&0&0.5&0\\
0&0&0&-1
\end{pmatrix}
x+
\begin{pmatrix}
1\\
1\\
0\\
0
\end{pmatrix}
u
$$
$$
y=
\begin{pmatrix}
0&1&0&1
\end{pmatrix}
x
$$
Determine if A is diagonalizable

Homework Equations

The Attempt at a Solution

The characteristic equation is given by
$$
\lambda I-A=
\begin{pmatrix}
\lambda-0.5&0&0&0\\
0&\lambda+2&0&0\\
1&0&\lambda-0.5&0\\
0&0&0&\lambda+1
\end{pmatrix}
$$
The determinant is given by
$$
det(\lambda I-A)=(\lambda-0.5)(\lambda+2)(\lambda-0.5)(\lambda+2)
$$
What do I have to do next to determine if A is diagonalizable? I am waiting hopefully for an answer!

You have a mistake in your work. The last line above should be $| \lambda I - A| = (\lambda-0.5)(\lambda+2)(\lambda-0.5)(\lambda + 1)$.
Set the determinant to 0 to find the three eigenvalues (one is repeated).
The matrix is diagonalizable if these eigenvalues yield four linearly independent eigenvectors.

BTW, what does LTI (A,B,C,D) mean?

 

[Mark44]

I should also mention that since your matrix $\lambda I - A$ is lower triangular (all entries above the main diagonal are zero), its determinant is the product of the entries on the diagonal.