2
$\begingroup$

Is the following correct ?

Find the characteristic and minimal polynomial of the matrix:

enter image description here

Is the matrix similar to any diagonal matrices ?

I started this problem as: $$\det(A-\lambda I) = 0$$ And I got: $$\begin{vmatrix} 1 -\lambda & 0 & 1 & -1\\ 1 & -\lambda & 1 & -1\\ 1& 0 & -\lambda & -1\\ 1 & 0 & 1 & -1-\lambda\\ \end{vmatrix} = -\lambda \cdot \begin{vmatrix} 1 -\lambda & 1 & -1 \\ 1 & -\lambda & -1 \\ 1& 1 & -1-\lambda \\ \end{vmatrix} = -\lambda^3(1+\lambda)$$

$-\lambda^3(1+\lambda)$ is the characteristic polynomial and the minimal polynomial. Here I have a question if this really is a minimal polynomial, how to write it etc. ?

Now if there exists a similar diagonal matrix is true if this matrix can be diagonalized, which means that it should have in this case $4$ linearly independent eigenvectors. Let's see if this is true:

$$\lambda_{1,2,3} = 0 \implies \text{Eigenvector should not be trivial}:$$

We subtract the first row from every other and we get the equation: $x+z-w = 0 \iff x = w-z \implies v_1 = \begin{bmatrix} 1 & 0 & 2 & 1 \end{bmatrix}^T$

Now we see that we can also get two other vectors that satisfy the equation $x = w-z$ that are linearly independent from the first one and from each other:

$$v_2 = \begin{bmatrix} 1 & 0 & 1 & 0 \end{bmatrix}^T $$

$$v_3 = \begin{bmatrix} 0 & 0 & 1 & 1 \end{bmatrix}^T $$

However now I see that the first one is dependent on the other two. Does anybody know how to repair this mistake and finish the problem, I got lost here.

$\endgroup$
1
  • 1
    $\begingroup$ Didn't read the whole thing but a $1$ is missing at the beginning of your calculations for the determinant (entry 3,3) $\endgroup$ Commented May 22, 2021 at 20:05

1 Answer 1

Reset to default
3
$\begingroup$

You got $x=w-z$. There is no $y$ here, and therefore $\begin{bmatrix}0&1&0&0\end{bmatrix}^T$ is an eigenvector with eigenvalues $0$. If you take $w=1$ and $z=0$, you get $\begin{bmatrix}1&0&1&0\end{bmatrix}^T$. And, if you take $w=0$ and $z=1$, you get $\begin{bmatrix}-1&0&0&1\end{bmatrix}^T$. And now you have three linearly independent eigenvectors with eigenvalue $0$.

$\endgroup$
2
  • $\begingroup$ I see, so actually I wanted to use this, that because we get 4 linearly independent eigenvectors, we can diagonalise the matrix with: $D = P^{-1}AP$, where $P$ is the matrix that has eigenvectors for columns. Is this correct ? $\endgroup$ Commented May 22, 2021 at 20:22
  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$ Commented May 22, 2021 at 20:23

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.