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Spectrum of φ - Eigenspace

mathmari

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Apr 14, 2013
4,036
Hey!! 😊

For a prime $p$ we define $\mathbb{F}_p:=\mathbb{Z}/p\mathbb{Z}$.
Let $\mathbb{K}\in \{\mathbb{F}_2, \mathbb{F}_3, \mathbb{F}_5\}$ and let \begin{equation*}a=\begin{pmatrix}0 & 1 & 0 & 5 \\ 1 & 0 & 2 & 3 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0\end{pmatrix}\in M_4(\mathbb{K})\end{equation*}

  1. Determine $\text{spe}(\phi_a)$ and $\dim \text{Eig}(\lambda, \phi_a)$ for all $\lambda \in \text{spe}(\phi_a)$.
  2. Is there a basis $B$ of $\mathbb{K}^4$, such that each basis vector is an eigenvector, i.e. is there a basis $B$ of $\mathbb{K}^4$ such that $M_B(\phi_a)$ is a diagonal matrix.
I have done the following:

  1. The spectrum is the set of all eigenvalues. The spectrrum of $\phi_a$ is the spectrum of $a$. Is that correct? The characteristic polynomial is \begin{align*}\det (a-\lambda u_4)=0 &\Rightarrow \begin{vmatrix}-\lambda & 1 & 0 & 5 \\ 1 & -\lambda & 2 & 3 \\ 0 & 0 & -\lambda & -1 \\ 0 & 0 & 1 & -\lambda\end{vmatrix}=0 \\ & \Rightarrow (-\lambda)\cdot \begin{vmatrix} -\lambda & 2 & 3 \\ 0 & -\lambda & -1 \\ 0 & 1 & -\lambda\end{vmatrix}-1\cdot \begin{vmatrix} 1 & 0 & 5 \\ 0 & -\lambda & -1 \\ 0 & 1 & -\lambda\end{vmatrix}=0 \\ & \Rightarrow (-\lambda)\cdot (-\lambda)\cdot \begin{vmatrix} -\lambda & -1 \\ 1 & -\lambda\end{vmatrix}-1\cdot 1\cdot \begin{vmatrix} -\lambda & -1 \\ 1 & -\lambda\end{vmatrix} =0 \\ & \Rightarrow \lambda^2 \cdot \left [(-\lambda)^2-1\cdot (-1)\right ]- \left [(-\lambda)^2-1\cdot (-1)\right ]=0\\ & \Rightarrow \left (\lambda^2 -1\right )\cdot \left (\lambda^2+1\right )=0\end{align*} Therefore we have that $\text{spe}(\phi_a)=\left \{-1,1, -i,i\right\}$. Is that correct ?
 
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Klaas van Aarsen

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Mar 5, 2012
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Hey mathmari !!

Shouldn't $\lambda$ be an element of $\mathbb K$?
Otherwise its scalar multiplication with a vector is not defined is it? (Worried)

Let's take a look at $\mathbb F_2$. It only has 0 and 1.
Which of those satisfy the characteristic equation? 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,036
Shouldn't $\lambda$ be an element of $\mathbb K$?
Otherwise its scalar multiplication with a vector is not defined is it? (Worried)

Let's take a look at $\mathbb F_2$. It only has 0 and 1.
Which of those satisfy the characteristic equation? 🤔
So do we have to take cases for $\mathbb{K}$ ?

If $\mathbb{K}=\mathbb{F}_2$ then $\lambda=1$ and so $\text{spe}(\phi_a)=\left \{1\right\}$.
If $\mathbb{K}=\mathbb{F}_3$ then $\lambda=1$ or $\lambda=2$ and so $\text{spe}(\phi_a)=\left \{1,2\right\}$.
If $\mathbb{K}=\mathbb{F}_5$ then $\lambda=1$, or $\lambda=2$, or $\lambda=3$ or $\lambda=4$ and so $\text{spe}(\phi_a)=\left \{1,2,3,4\right\}$.

Is that correct? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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So do we have to take cases for $\mathbb{K}$ ?

If $\mathbb{K}=\mathbb{F}_2$ then $\lambda=1$ and so $\text{spe}(\phi_a)=\left \{1\right\}$.
If $\mathbb{K}=\mathbb{F}_3$ then $\lambda=1$ or $\lambda=2$ and so $\text{spe}(\phi_a)=\left \{1,2\right\}$.
If $\mathbb{K}=\mathbb{F}_5$ then $\lambda=1$, or $\lambda=2$, or $\lambda=3$ or $\lambda=4$ and so $\text{spe}(\phi_a)=\left \{1,2,3,4\right\}$.

Is that correct?
Yep. (Nod)

As an additional observation, note that for $\mathbb F_2$ we have $1\equiv -1$. Consequently:
$$ (\lambda^2-1)(\lambda^2+1) =(\lambda^2-1)(\lambda^2-1) =(\lambda-1)^2(\lambda+1)^2 =(\lambda-1)^2(\lambda-1)^2 =(\lambda-1)^4$$
so $\lambda=1$ has algebraic multiplicity 4. :geek:
 

mathmari

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Apr 14, 2013
4,036
Yep. (Nod)

As an additional observation, note that for $\mathbb F_2$ we have $1\equiv -1$. Consequently:
$$ (\lambda^2-1)(\lambda^2+1) =(\lambda^2-1)(\lambda^2-1) =(\lambda-1)^2(\lambda+1)^2 =(\lambda-1)^2(\lambda-1)^2 =(\lambda-1)^4$$
so $\lambda=1$ has algebraic multiplicity 4. :geek:
Ok! So for each case of $\mathbb{K}$ we have to calculate the eigenvectors, right?
 

Klaas van Aarsen

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Mar 5, 2012
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Ok! So for each case of $\mathbb{K}$ we have to calculate the eigenvectors, right?
I think so yes. :unsure:
Although for $\mathbb F_5$ we can observe that we have 4 distinct eigenvalues... 🤔
 

mathmari

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Apr 14, 2013
4,036
I think so yes. :unsure:
Although for $\mathbb F_5$ we can observe that we have 4 distinct eigenvalues... 🤔
So at each case of $\mathbb{K}$ we can also simplify the element of matrix, or not? For example the rightmost element in the first row, $5$, is equivalent to $1$ in $\mathbb{F}_2$, to $2$ in $\mathbb{F}_3$ and to $0$ in $\mathbb{F}_5$. Or do we not have to do that when we calculate the eigenvectors?
 

Klaas van Aarsen

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Mar 5, 2012
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So at each case of $\mathbb{K}$ we can also simplify the element of matrix, or not? For example the rightmost element in the first row, $5$, is equivalent to $1$ in $\mathbb{F}_2$, to $2$ in $\mathbb{F}_3$ and to $0$ in $\mathbb{F}_5$. Or do we not have to do that when we calculate the eigenvectors?
We don't have to, but yes, it does make the calculations easier. 🤔
 

mathmari

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Apr 14, 2013
4,036
I have done the following:

- For $\mathbb{K}=\mathbb{F}_2$ :

For $\lambda=1$ we have:
\begin{equation*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
From Gauss algorithm we get:
\begin{align*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix} & \ \overset{Z_2:Z_2+Z_1}{\longrightarrow } \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\ \underset{Z_4:Z_4-\frac{1}{2}\cdot Z_2}{\overset{Z_3:Z_3+\frac{1}{2}\cdot Z_2}{\longrightarrow }} \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & -5\end{pmatrix} \\ & \ \overset{Z_4:Z_4+\frac{5}{3}\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations: \begin{equation*}\begin{cases}-x+y+5w=0 \\ 2z+8w=0 \\ 3w=0\end{cases} \Rightarrow \begin{cases}x=y \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{\begin{pmatrix}y \\ y \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}=\left \{y\cdot \begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}\end{equation*}
Therefore the dimension of the eigenspace is $\dim \text{Eig}(1, \phi_a)=1$.



- For $\mathbb{K}=\mathbb{F}_3$ :

For $\lambda=1$ we have:
\begin{equation*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
From Gauss algorithm we get:
\begin{align*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix} & \ \overset{Z_2:Z_2+Z_1}{\longrightarrow } \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\ \underset{Z_4:Z_4-\frac{1}{2}\cdot Z_2}{\overset{Z_3:Z_3+\frac{1}{2}\cdot Z_2}{\longrightarrow }} \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & -5\end{pmatrix} \\ & \ \overset{Z_4:Z_4+\frac{5}{3}\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations: \begin{equation*}\begin{cases}-x+y+5w=0 \\ 2z+8w=0 \\ 3w=0\end{cases} \Rightarrow \begin{cases}x=y \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{\begin{pmatrix}y \\ y \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}=\left \{y\cdot \begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}\end{equation*}
Therefore the dimension of the eigenspace is $\dim \text{Eig}(1, \phi_a)=1$.

For $\lambda=2$ we have:
\begin{equation*}\begin{pmatrix}-2 & 1 & 0 & 5 \\ 1 & -2 & 2 & 3 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
From Gauss algorithm we get:
\begin{align*}\begin{pmatrix}-2 & 1 & 0 & 5 \\ 1 & -2 & 2 & 3 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix} & \ \overset{Z_2:Z_2+\frac{1}{2}\cdot Z_1}{\longrightarrow } \ \begin{pmatrix}-2 & 1 & 0 & 5 \\ 0 & -\frac{3}{2} & 2 & \frac{11}{2} \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix} \ \overset{Z_4:Z_4+\frac{1}{2}\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}-2 & 1 & 0 & 5 \\ 0 & -\frac{3}{2} & 2 & \frac{11}{2} \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 0 & -\frac{5}{2}\end{pmatrix}\end{align*}
So the eigenspace is \begin{equation*}\left \{\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \right \}\end{equation*}
Therefore the dimension of the eigenspace is $\dim \text{Eig}(2, \phi_a)=0$.


- For $\mathbb{K}=\mathbb{F}_5$ :

For $\lambda=1$ we have:
\begin{equation*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
From Gauss algorithm we get:
\begin{align*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix} & \ \overset{Z_2:Z_2+Z_1}{\longrightarrow } \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\ \underset{Z_4:Z_4-\frac{1}{2}\cdot Z_2}{\overset{Z_3:Z_3+\frac{1}{2}\cdot Z_2}{\longrightarrow }} \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & -5\end{pmatrix} \\ & \ \overset{Z_4:Z_4+\frac{5}{3}\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 0 & 0 & 2 & 8 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations \begin{equation*}\begin{cases}-x+y+5w=0 \\ 2z+8w=0 \\ 3w=0\end{cases} \Rightarrow \begin{cases}x=y \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{\begin{pmatrix}y \\ y \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}=\left \{y\cdot \begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}\end{equation*}
Therefore the dimension of the eigenspace is $\dim \text{Eig}(1, \phi_a)=1$.

For $\lambda=2$ we have:
\begin{equation*}\begin{pmatrix}-2 & 1 & 0 & 5 \\ 1 & -2 & 2 & 3 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
From Gauss algorithm we get:
\begin{align*}\begin{pmatrix}-2 & 1 & 0 & 5 \\ 1 & -2 & 2 & 3 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix} & \ \overset{Z_2:Z_2+\frac{1}{2}\cdot Z_1}{\longrightarrow } \ \begin{pmatrix}-2 & 1 & 0 & 5 \\ 0 & -\frac{3}{2} & 2 & \frac{11}{2} \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix} \ \overset{Z_4:Z_4+\frac{1}{2}\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}-2 & 1 & 0 & 5 \\ 0 & -\frac{3}{2} & 2 & \frac{11}{2} \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 0 & -\frac{5}{2}\end{pmatrix}\end{align*}
So the eigenspace is \begin{equation*}\left \{\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \right \}\end{equation*}
Therefore the dimension of the eigenspace is $\dim \text{Eig}(2, \phi_a)=0$.

For $\lambda=3$ we have:
\begin{equation*}\begin{pmatrix}-3 & 1 & 0 & 5 \\ 1 & -3 & 2 & 3 \\ 0 & 0 & -3 & -1 \\ 0 & 0 & 1 & -3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
From Gauss algorithm we get:
\begin{align*}\begin{pmatrix}-3 & 1 & 0 & 5 \\ 1 & -3 & 2 & 3 \\ 0 & 0 & -3 & -1 \\ 0 & 0 & 1 & -3\end{pmatrix} & \ \overset{Z_2:Z_2+\frac{1}{3}\cdot Z_1}{\longrightarrow } \ \begin{pmatrix}-3 & 1 & 0 & 5 \\ 0 & -\frac{8}{3} & 2 & \frac{14}{3} \\ 0 & 0 & -3 & -1 \\ 0 & 0 & 1 & -3\end{pmatrix} \ \overset{Z_4:Z_4+\frac{1}{3}\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}-3 & 1 & 0 & 5 \\ 0 & -\frac{8}{3} & 2 & \frac{14}{3} \\ 0 & 0 & -3 & -1 \\ 0 & 0 & 0 & -\frac{10}{3}\end{pmatrix}\end{align*}
So the eigenspace is \begin{equation*}\left \{\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \right \}\end{equation*}
Therefore the dimension of the eigenspace is $\dim \text{Eig}(3, \phi_a)=0$.

For $\lambda=4\equiv -1$ we have:
\begin{equation*}\begin{pmatrix}-(-1) & 1 & 0 & 5 \\ 1 & -(-1) & 2 & 3 \\ 0 & 0 & -(-1) & -1 \\ 0 & 0 & 1 & -(-1)\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}1 & 1 & 0 & 5 \\ 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
From Gauss algorithm we get:
\begin{equation*}\begin{pmatrix}1 & 1 & 0 & 5 \\ 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 1\end{pmatrix} \ \overset{Z_2:Z_2-Z_1}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 5 \\ 0 & 0& 2 & -2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 1\end{pmatrix}\ \underset{Z_4:Z_4-\frac{1}{2}\cdot Z_2}{\overset{Z_3:Z_3-\frac{1}{2}\cdot Z_2}{\longrightarrow }} \ \begin{pmatrix}1 & 1 & 0 & 5 \\ 0 & 0& 2 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2\end{pmatrix}\ \overset{Z_3\leftarrow Z_4}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 5 \\ 0 & 0& 2 & -2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \end{pmatrix}\end{equation*}
So we get the equations \begin{equation*}\begin{cases}x+y+5w=0 \\ 2z-2w=0 \\ 2w=0\end{cases} \Rightarrow \begin{cases}x=-y \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{\begin{pmatrix}-y \\ y \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}=\left \{y\cdot \begin{pmatrix}-1 \\ 1 \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}\end{equation*}
Therefore the dimension of the eigenspace is $\dim \text{Eig}(4, \phi_a)=1$.


Is everything correct? :unsure:

Is didn't used now the equivalences according to $\mathbb{F}$. Do we get in that way the correct results?
 
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Klaas van Aarsen

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Mar 5, 2012
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Is everything correct?

Is didn't used now the equivalences according to $\mathbb{F}$. Do we get in that way the correct results?
We do need to use the equivalences and we cannot use fractions.
Note that $\frac 12$ does not exist in $\mathbb F_2$. That is, 2 is equivalent to 0. And 0 does not have an inverse. :eek:

Consequently we get another eigenvector in $\mathbb F_2$. :oops:

It is easiest and safest to replace each element in the matrix by $0$ or $1$ and make all calculations in $\mathbb F_2$.
That is, apply that $1+1=0$ and $-1=1$.
Furthermore, no divisions, since they are not defined in $\mathbb F_p$, although we can multiply with the inverse of an element - if it exists. 🤔
 
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Klaas van Aarsen

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So we get for $\mathbb F_2$:
\[ \begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}
=\begin{pmatrix}1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{pmatrix}
\ \overset{Z_2:Z_2+Z_1}{\underset{Z_4:Z_4+Z_3}\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix}
\ \overset{Z_2 \leftrightarrow Z_3}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix} \]
🤔
 
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mathmari

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Apr 14, 2013
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Ok! Therefore we get:

- $\mathbb{K}=\mathbb{F}_2$ :

For $\lambda=1$ :
\begin{equation*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
Gauss-Algorithm:
\begin{equation*}\begin{pmatrix}1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{pmatrix} \ \overset{Z_2:Z_2-Z_1}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{pmatrix}\ \overset{Z_4:Z_4- Z_3}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix} \ \overset{Z_3\leftrightarrow Z_4}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{equation*}
So we get the equations: \begin{equation*}\begin{cases}x+y+w=0 \\ z+w=0 \end{cases} \Rightarrow \begin{cases}x=-y-w \\ z=-w \end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{y\cdot \begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix}+w\cdot \begin{pmatrix}-1 \\ 0 \\ -1\\ 1\end{pmatrix} : y,w\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(1, \phi_a)=2$.



- $\mathbb{K}=\mathbb{F}_3$ :

For $\lambda=1$ :
\begin{equation*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}2 & 1 & 0 & 2 \\ 1 & 2 & 2 & 0 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
Gauss-Algorithm:
\begin{align*}\begin{pmatrix}2 & 1 & 0 & 2 \\ 1 & 2 & 2 & 0 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 1 & 2\end{pmatrix} & \ \overset{Z_2:Z_2+Z_1}{\longrightarrow } \ \begin{pmatrix}2 & 1 & 0 & 2 \\ 3 & 3 & 2 & 2 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 1 & 2\end{pmatrix} \longrightarrow \begin{pmatrix}2 & 1 & 0 & 2 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 1 & 2\end{pmatrix}\ \underset{Z_4:Z_4+ Z_2}{\overset{Z_3:Z_3- Z_2}{\longrightarrow }} \ \begin{pmatrix}2 & 1 & 0 & 2 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 4\end{pmatrix} \\ & \longrightarrow \begin{pmatrix}2 & 1 & 0 & 2 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} \ \overset{Z_3\leftrightarrow Z_4}{\longrightarrow } \ \begin{pmatrix}2 & 1 & 0 & 2 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations: \begin{equation*}\begin{cases}2x+y+2w=0 \\ 2z+2w=0 \\ w=0\end{cases} \Rightarrow \begin{cases}y=-x \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{x\cdot \begin{pmatrix}1 \\ 2 \\ 0\\ 0\end{pmatrix} : x\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(1, \phi_a)=1$.

For $\lambda=2$ :
\begin{equation*}\begin{pmatrix}-2 & 1 & 0 & 5 \\ 1 & -2 & 2 & 3 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}1 & 1 & 0 & 1 \\ 1 & 1 & 2 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
Gauss-Algorithm:
\begin{align*}\begin{pmatrix}1 & 1 & 0 & 1 \\ 1 & 1 & 2 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 1\end{pmatrix} & \ \overset{Z_2:Z_2- Z_1}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 2 & -1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 1\end{pmatrix}\rightarrow \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 1\end{pmatrix} \ \underset{Z_3:Z_3+Z_2}{\overset{Z_4:Z_4+Z_2}{\longrightarrow }} \ \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 3 & 3\end{pmatrix}\longrightarrow \begin{pmatrix}1 & 1 & 0 & 1 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations: \begin{equation*}\begin{cases}x+y+w=0 \\ 2z+2w=0 \\ w=0\end{cases} \Rightarrow \begin{cases}y=-x \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{x\cdot \begin{pmatrix}1 \\ 2 \\ 0\\ 0\end{pmatrix} : x\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(2, \phi_a)=1$.
 

mathmari

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Apr 14, 2013
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- $\mathbb{K}=\mathbb{F}_5$ :

For $\lambda=1$ :
\begin{equation*}\begin{pmatrix}-1 & 1 & 0 & 5 \\ 1 & -1 & 2 & 3 \\ 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}4 & 1 & 0 & 0 \\ 1 & 4 & 2 & 3 \\ 0 & 0 & 4 & 4 \\ 0 & 0 & 1 & 4\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
Gauss-Algorithm:
\begin{align*}\begin{pmatrix}4 & 1 & 0 & 0 \\ 1 & 4 & 2 & 3 \\ 0 & 0 & 4 & 4 \\ 0 & 0 & 1 & 4\end{pmatrix}& \ \overset{Z_2:Z_2+Z_1}{\longrightarrow } \ \begin{pmatrix}4 & 1 & 0 & 0 \\ 5 & 5 & 2 & 3 \\ 0 & 0 & 4 & 4 \\ 0 & 0 & 1 & 4\end{pmatrix} \longrightarrow \begin{pmatrix}4 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 4 & 4 \\ 0 & 0 & 1 & 4\end{pmatrix}\ \underset{Z_4:Z_4+2\cdot Z_2}{\overset{Z_3:Z_3+3\cdot Z_2}{\longrightarrow }} \ \begin{pmatrix}4 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 10 & 13 \\ 0 & 0 & 5 & 10\end{pmatrix} \\ & \longrightarrow \begin{pmatrix}4 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{pmatrix} \end{align*}
So we get the equations: \begin{equation*}\begin{cases}4x+y=0 \\ 2z+3w=0 \\ 3w=0\end{cases} \Rightarrow \begin{cases}y=-4x \\ z=0 \\ w=0\end{cases}\Rightarrow \begin{cases}y=x \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{x\cdot \begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix} : x\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(1, \phi_a)=1$.

For $\lambda=2$ :
\begin{equation*}\begin{pmatrix}-2 & 1 & 0 & 5 \\ 1 & -2 & 2 & 3 \\ 0 & 0 & -2 & -1 \\ 0 & 0 & 1 & -2\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}3 & 1 & 0 & 0 \\ 1 & 3 & 2 & 3 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 1 & 3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
Gauss-Algorithm:
\begin{align*}\begin{pmatrix}3 & 1 & 0 & 0 \\ 1 & 3 & 2 & 3 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 1 & 3\end{pmatrix} & \ \overset{Z_2:Z_2+3\cdot Z_1}{\longrightarrow } \ \begin{pmatrix}3 & 1 & 0 & 0 \\ 10 & 6 & 2 & 3 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 1 & 3\end{pmatrix} \longrightarrow \begin{pmatrix}3 & 1 & 0 & 0 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 1 & 3\end{pmatrix} \ \overset{Z_4:Z_4+3\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}3 & 1 & 0 & 0 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 10 & 15\end{pmatrix} \\ & \longrightarrow \begin{pmatrix}3 & 1 & 0 & 0 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 3 & 4 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations: \begin{equation*}\begin{cases}3x+y=0 \\ y+2z+3w=0 \\ 3z+4w=0\end{cases} \ldots\Rightarrow \begin{cases}y=2x \\ z=3x \\ w=4x\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{x\cdot \begin{pmatrix}1 \\ 2 \\ 3\\ 4\end{pmatrix} : x\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(2, \phi_a)=1$.

For $\lambda=3$ :
\begin{equation*}\begin{pmatrix}-3 & 1 & 0 & 5 \\ 1 & -3 & 2 & 3 \\ 0 & 0 & -3 & -1 \\ 0 & 0 & 1 & -3\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}2 & 1 & 0 & 0 \\ 1 & 2 & 2 & 3 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
Gauss-Algorithm:
\begin{align*}\begin{pmatrix}2 & 1 & 0 & 0 \\ 1 & 2 & 2 & 3 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2\end{pmatrix} & \ \overset{Z_2:Z_2+2\cdot Z_1}{\longrightarrow } \ \begin{pmatrix}2 & 1 & 0 & 0 \\ 5 & 4 & 2 & 3 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2\end{pmatrix} \longrightarrow \begin{pmatrix}2 & 1 & 0 & 0 \\ 0 & 4 & 2 & 3 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2\end{pmatrix} \ \overset{Z_4:Z_4+2\cdot Z_3}{\longrightarrow } \ \begin{pmatrix}2 & 1 & 0 & 0 \\ 0 & 4 & 2 & 3 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 5 & 10\end{pmatrix} \\ & \longrightarrow \begin{pmatrix}2 & 1 & 0 & 0 \\ 0 & 4 & 2 & 3 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations: \begin{equation*}\begin{cases}2x+y=0 \\ 4y+2z+3w=0 \\ 2z+4w=0\end{cases} \ldots \Rightarrow \begin{cases}y=3z \\ x=z \\ w=2z\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{z\cdot \begin{pmatrix}1 \\ 3 \\ 1\\ 2\end{pmatrix} : z\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(3, \phi_a)=1$.

For $\lambda=4$ :
\begin{equation*}\begin{pmatrix}-4 & 1 & 0 & 5 \\ 1 & -4 & 2 & 3 \\ 0 & 0 & -4 & -1 \\ 0 & 0 & 1 & -4\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}1 & 1 & 0 & 0 \\ 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 1 & 1\end{pmatrix}\begin{pmatrix}x \\ y \\ z\\ w\end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0\\ 0\end{pmatrix} \end{equation*}
Gauss-Algorithm:
\begin{align*}\begin{pmatrix}1 & 1 & 0 & 0 \\ 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 1 & 1\end{pmatrix} \ & \overset{Z_2:Z_2-Z_1}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 1 & 1\end{pmatrix}\ \underset{Z_4:Z_4+2\cdot Z_2}{\overset{Z_3:Z_3+2\cdot Z_2}{\longrightarrow }} \ \begin{pmatrix}1 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 5 & 10 \\ 0 & 0 & 5 & 7\end{pmatrix} \longrightarrow \begin{pmatrix}1 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2\end{pmatrix}\\ & \ \overset{Z_3\leftarrow Z_4}{\longrightarrow } \ \begin{pmatrix}1 & 1 & 0 & 0 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{align*}
So we get the equations \begin{equation*}\begin{cases}x+y=0 \\ 2z+3w=0 \\ 2w=0\end{cases} \Rightarrow \begin{cases}x=4y \\ z=0 \\ w=0\end{cases}\end{equation*}
So the eigenspace is \begin{equation*}\left \{y\cdot \begin{pmatrix}4 \\ 1 \\ 0\\ 0\end{pmatrix} : y\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(4, \phi_a)=1$.



Is everything correct? :unsure:




For the second question shouldn't the basis $B$ consist of the eigenvectors of $\phi_a$ ? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,739
- $\mathbb{K}=\mathbb{F}_3$ :

So the eigenspace is \begin{equation*}\left \{x\cdot \begin{pmatrix}1 \\ 2 \\ 0\\ 0\end{pmatrix} : x\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(1, \phi_a)=1$.

So the eigenspace is \begin{equation*}\left \{x\cdot \begin{pmatrix}1 \\ 2 \\ 0\\ 0\end{pmatrix} : x\in \mathbb{K}\right \}\end{equation*}
The dimension is $\dim \text{Eig}(2, \phi_a)=1$.
It can't be that both eigenvalues have the same eigenspace, can't it? (Worried)
Do those eigenvectors actually satisfy the relation $av=\lambda v$?

- $\mathbb{K}=\mathbb{F}_5$ :

Is everything correct?
Looks about correct, although I didn't check it carefully.
Either way, since all 4 eigenvalues are distinct, each of them must also have a distinct eigenvector.
Since the dimension of the vector space is 4, that implies that each eigenvalue has an eigenspace of dimension 1, which is what you found as well.
So the result with the dimensions of the eigenspaces is correct. (Nod)

For the second question shouldn't the basis $B$ consist of the eigenvectors of $\phi_a$ ?
Yes.
And it can only be a basis if we have 4 independent eigenvectors. 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
It can't be that both eigenvalues have the same eigenspace, can't it? (Worried)
Do those eigenvectors actually satisfy the relation $av=\lambda v$?
I found a typo. The first one, for $\lambda=1$ it is $\left \{x\cdot \begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix} : x\in \mathbb{K}\right \}$.


Yes.
And it can only be a basis if we have 4 independent eigenvectors. 🤔
So do we consider $\mathbb{K}=\mathbb{F}_5$ and the basis consists of the vectors $\begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix}$, $\begin{pmatrix}1 \\ 2 \\ 3\\ 4\end{pmatrix}$, $\begin{pmatrix}1 \\ 3 \\ 1\\ 2\end{pmatrix}$, $\begin{pmatrix}4 \\ 1 \\ 0\\ 0\end{pmatrix}$ ? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,739
I found a typo. The first one, for $\lambda=1$ it is $\left \{x\cdot \begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix} : x\in \mathbb{K}\right \}$.

So do we consider $\mathbb{K}=\mathbb{F}_5$ and the basis consists of the vectors $\begin{pmatrix}1 \\ 1 \\ 0\\ 0\end{pmatrix}$, $\begin{pmatrix}1 \\ 2 \\ 3\\ 4\end{pmatrix}$, $\begin{pmatrix}1 \\ 3 \\ 1\\ 2\end{pmatrix}$, $\begin{pmatrix}4 \\ 1 \\ 0\\ 0\end{pmatrix}$ ?
Yes and yes. (Nod)