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Is the following correct ?

Find the characteristic and minimal polynomial of the matrix:

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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.

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

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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 $0$.

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  • $\begingroup$ 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 ? $\endgroup$
    – VLC
    May 22, 2021 at 20:22
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    $\begingroup$ Yes, that is correct. $\endgroup$ May 22, 2021 at 20:23

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