I came accross an exercise that reads as follows:
What are the values of the parameter k such that the following endomorphism is simple.
Here's the matrix associated with the endomorphism:
$$ \begin{bmatrix} 1 & 0 & k \\ -10 & 0 & 1 \\ 10 & 0 & 3 \\ \end{bmatrix} $$
So I went ahead and solved it, only to come up with the wrong solution. Here are my steps:
1) Find the characteristic polynomial
$-λ (λ^2 - 4 λ - 10 k + 3)$
2) Check if there are n distinct eigenvalues, so n distinct roots, with n = 3 in this case with our 3x3 matrix. Clearly the first is $λ = 0$.
3) Notice that $(λ^2 - 4 λ - 10 k + 3) = 0$ has 2 solutions when $\Delta > 0$.
4) $\Delta = \sqrt{10k-x}$ so $\sqrt{10k-x} > 0$ therefore $k > -\frac{1}{10}$
5) Notice that the polynomial also goes to $0$ for $(λ^2 - 4 λ - 10 k + 3) = 0 $ when $k = \frac{1}{10}(\lambda^2-4\lambda+3) $ .
So now my final solution would be to say, as a condition for the endomorphism being simple,
$k \not= \frac{1}{10}(\lambda^2-4\lambda+3) $ AND $k > -\frac{1}{10}$
However, my book says that the actual solution is just $k > -\frac{1}{10}$ . What am I doing wrong?
Edit: So as it turns out, I wasn't doing anything wrong, the book was. Marking the first answer as solved as it confirms of my own solution.
You have to solve the polynomial for $\lambda$, not for $k$.
Call your matrix $A$. The characteristic polynomial of $A$ is
$$\chi_A(\lambda) = - \lambda \cdot ( (1 - \lambda)(3 - \lambda) - 10k).$$
One root is $0$ and the other two roots are $\lambda = 2 \pm \sqrt{1 + 10k}$.
If $1 + 10k < 0$, then there are two complex conjugate eigenvalues, hence the matrix is not diagonalizable over $\mathbb{R}$.
Let $k = - 1/10$. In this case, the eigenvalue $2$ has multiplicity $2$, but one can easily check that the corresponding eigenspace has dimension $1$. Hence, $A$ is not diagonalizable for $k = - 1/10$.
Suppose $k = 3/10$. Then the eigenvalues are $\lambda_1 = \lambda_2 = 0$ and $\lambda_3 = 4$. But the eigenspace corresponding to $\lambda_1$ is only one-dimensional, hence $A$ is not diagonalizable.
Finally, let $k \in (-1/10, \infty) \setminus \{3 / 10\}$. Then the eigenvalues of $A$ are $\lambda_1 = 0$, and $\lambda_2 = 2 + \sqrt{1 + 10k} \neq 2 - \sqrt{1 + 10k} = \lambda_3$ with $\lambda_1 \neq \lambda_2$ and $\lambda_1 \neq \lambda_3$, hence all three eigenvalues are distinct. For every such $k$ the matrix $A$ is diagonalizable.