# Diagonalizable matrix

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1. Jan 15, 2016

### Linder88

1. The problem statement, all variables and given/known data
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
2. Relevant equations

3. 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!

2. Jan 15, 2016

### Staff: Mentor

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. Jan 15, 2016

### Staff: 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.