# Diagonalizable matrix

## 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

## 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
Mentor

## 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

## 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
Mentor
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.