Diagonalizable matrix

  • #1
25
0

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!
 

Answers and Replies

  • #2
35,394
7,271

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?
 
  • #3
35,394
7,271
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.
 

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