hi everyone, I have a question
My goal is to find the eigenvalues of a square matrix using rank's notion. Let $A\in\mathbb{R}^{n\times n}, x\in\mathbb{R}^n$, the objective is to find nontrivial solutions for the following homogenous system
$$
(\lambda I - A) x =0.
$$
As a result of the following incomplete row reducted echolen form, we get
$$
\begin{bmatrix}
(\lambda -1) & 0 \\ 0 & (\lambda+1)(\lambda-5)
\end{bmatrix}
$$

This is where the problem arises. I must first assume $\lambda \neq 1$ and multiply $(\lambda+1)(\lambda-5)$ by -1 to avoid having $\lambda=-1$ degenerates the rank, henc…

There is no unital ring homomorphism $\phi:\mathbb Q \to \mathbb Z$.

Proof: Suppose $\phi$ is such a unital ring homomorphism. Let $z, z' \in \mathbb Z$. Notice $\phi(z) = z \phi(1) = z$ because $1_{\mathbb Q} \mapsto 1_{\mathbb Z}$. But then $\phi(z) = \phi(z')\phi(\frac{z}{z'})= z$, and therefore $\phi(z') = z'$ is a factor of $z$ in $\mathbb Z$. Since $z'$ was arbitrary, $0$ is a factor of $z$, and therefore $z = 0$. But the map does not have $1_{\mathbb Q} \mapsto 1_{\mathbb Z}$, a contradiction.

sorry my question was missing some info. This is the modified version. My goal is to find the eigenvalues of a square matrix using rank's notion. Let $𝐴\in\mathbb{R}^{n\times n},x\in\mathbb{R}^n$, the objective is to find nontrivial solutions for the following homogenous system $(\lambda 𝐼−𝐴)x=0$ which occurs if and only if $(\lambda 𝐼−𝐴)$ singular. I can use the determinant for this purpose but I need to use the rank. For example, let $A=\begin{bmatrix}1&2\\4&3 \end{bmatrix}$ As a result of the following incomplete row reducted echolen form, we get

@TedShifrin They are not random. Let me clarify my calculations.
$$
(\lambda I - A) = \begin{bmatrix} (\lambda -1) & -2\\-4 &(\lambda-3) \end{bmatrix}
$$
The first step I did $R_2\rightarrow R_2 - \frac{4}{(\lambda-1)}R_1, \lambda\neq 1$, we get
$$
\begin{bmatrix} (\lambda -1) & -2\\0 &(\lambda-3)+\frac{4}{(\lambda-1)}(-2) \end{bmatrix} \rightarrow
\begin{bmatrix} (\lambda -1) & -2\\0 &\frac{(\lambda-3)(\lambda-1)-8}{(\lambda-1)} \end{bmatrix}
$$
The second operation is $R_1 \rightarrow R_1 + \frac{2}{(\lambda-3)(\lambda-1)-8} R_2, \lambda\neq 5,-1$, we get

@TedShifrin did I understand this correctly: We have a surface $S$. Now at a given point $P=\sigma(u(t_0),v(t_0))$ on the surface we can compute it's normal vector, and we can consider a unit vector $v$ which lies in the tangent plane $T_PS$. Then we can consider a plane which is spanned by $v$ and $N_P$, lets call it $N$. This plane intersects the surface and we denote this intersection to be the curve $\gamma_v$. Then if we compute the curvature of $\gamma_v$ at $P$ we get the normal curvature in direction $v$, i.e. $K_n^P(v)$. But now if we want to find out in which direction the surfac…

Proof. Assume $f$ is not monotonic. We have points $s_1<t_1$ with $f(s_1) < f(t_1)$ and points $s_2<t_2$ with $f(s_2) > f(t_2)$. Choose $s=\min(s_1,s_2)$ and $t=\max(t_1,t_2)$. The set $f([s,t])$ has a maximum and minimum value. If $f(s)$ and $f(t)$ do not both attain minimum and maximum values of $f$, then the points for which $f$ attains it lie in $(s,t)$, thus $f((s,t))$ fails to be open.
Otherwise, we know that the segment $(s_i, t_i)$ ($1\leq i \leq 2)$ lies in $(s,t)$. Then the same reasoning applies to $(s_i, t)$ or $(s,t_i)$.