Solving Linear Mapping Problems with Matrix $M$ in $\mathbb K^2$

[mathmari]

Hey!

Let $\mathbb{K}$ be a field.

1. Find a matrix $M\in \mathbb{K}^{2\times 2}$ such that for the linear mapping $f:\mathbb{K}^2\rightarrow \mathbb{K}^2, x\mapsto Mx$ it holds that $f\neq 0$ and $f^2:=f\circ f=0$.
2. Let $V$ be a $\mathbb{K}$-vector space and $\psi:V\rightarrow V$ be a linear mapping, with $\psi^k\neq 0$ and $\psi^{k+1}=0$ for some $k>0$. Show that there is an element $x\in V$ such that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly independent.

I have done the following:

1. Since $f\neq 0$ it holds that $M$ is not the zero matrix.
   We have that $f^2(x):=f(f(x))=f(Mx)=MMx=M^2x$.
   So, we have to find a matrix $M$ such that $M^2$ is the zero matrix.
   Let $M=\begin{pmatrix}m_1&m_2 \\m_3&m_4\end{pmatrix}$.
   Then we have the following:
   $$M^2=0 \Rightarrow \begin{pmatrix}m_1&m_2 \\m_3&m_4\end{pmatrix}\begin{pmatrix}m_1&m_2 \\m_3&m_4\end{pmatrix}=\begin{pmatrix}0&0 \\0&0\end{pmatrix} \Rightarrow\begin{pmatrix}m_1^2+m_2m_3&m_1m_2+m_2m_4 \\m_3m_1+m_4m_3&m_3m_2+m_4^2\end{pmatrix}=\begin{pmatrix}0&0 \\0&0\end{pmatrix}$$
   So, we are looking for a solution for the following system:
   $$m_1^2+m_2m_3=0 \\ m_1m_2+m_2m_4=0 \Rightarrow m_2(m_1+m_4)=0\\m_3m_1+m_4m_3=0\Rightarrow m_3(m_1+m_4)=0 \\ m_3m_2+m_4^2=0$$
   When we take from the second and third equation that $m_1+m_4=0 \Rightarrow m_1=-m_4$, let $m_1=-m_4=1$, then we get from the first and fourth equation that $1+m_2m_3=0\Rightarrow n_2m_3=-1$. Let $m_2=-m_3=1$.
   So, we get the matrix: $$M=\begin{pmatrix}1&1 \\-1&-1\end{pmatrix}$$
   
   Is this correct? (Wondering)
2. Let $x\in V$. Suppose that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly dependent, i.e., there are $c_i$'s not all zero such that $c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
   $$\psi (c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+ c_k\psi^k(x))=\psi (0)=0 \\ (\text{ it holds that } \psi (0)=0 \text{ since it is a linear mapping, or not? }) \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)=0$$
   
   Is this correct so far? How could we continue? (Wondering)

 

[mathmari]

2. Let $x\in V$. Suppose that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly dependent, i.e., there are $c_i$'s not all zero such that $c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
$$\psi (c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+ c_k\psi^k(x))=\psi (0)=0 \\ (\text{ it holds that } \psi (0)=0 \text{ since it is a linear mapping, or not? }) \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)=0$$

$$\psi (c_0x+ c_1\psi (x)+ \ldots + c_{k-1}\psi^{k-1}(x)+ c_k\psi^k(x))=\psi (0)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-1}\psi^k(x)=0 \\ \Rightarrow \psi (c_0\psi (x)+ c_1\psi^2 (x)+ \ldots + c_{k-2}\psi^{k-1}(x)+c_{k-1}\psi^k(x))=\psi (0) \\ \Rightarrow c_0\psi^2 (x)+ c_1\psi^3 (x)+ \ldots + c_{k-2}\psi^k(x)+c_{k-1}\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi^2 (x)+ c_1\psi^3 (x)+ \ldots + c_{k-2}\psi^k(x)=0 \\ \ldots \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0$$ Since $\psi^k(x)\neq 0$ we conclude that $c_0$.

Do we do the same again $k$ times to cocnlude that $c_i=0$ for all $i=0, \ldots , k$ ? (Wondering)

Or do we show it in an other way? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

1. Find a matrix $M\in \mathbb{K}^{2\times 2}$ such that for the linear mapping $f:\mathbb{K}^2\rightarrow \mathbb{K}^2, x\mapsto Mx$ it holds that $f\neq 0$ and $f^2:=f\circ f=0$.
2. Let $V$ be a $\mathbb{K}$-vector space and $\psi:V\rightarrow V$ be a linear mapping, with $\psi^k\neq 0$ and $\psi^{k+1}=0$ for some $k>0$. Show that there is an element $x\in V$ such that the set $\{x, \psi (x), \ldots , \psi^k(x)\}$ is linearly independent.

...

$$M=\begin{pmatrix}1&1 \\-1&-1\end{pmatrix}$$
Is this correct? (Wondering)

Yep. (Nod)

Alternatively, we can observe that $M$ must be similar to a Jordan normal form.
Suppose for $v\ne 0$ we have $Mv=\lambda v$. Then $M^2v=\lambda^2 v$. Therefore $\lambda = 0$.
Since $M \ne 0$, it follows that $M$ is similar to the Jordan normal form:
$$M \sim \begin{pmatrix}0&1 \\ 0&0\end{pmatrix}$$

Your matrix $M$ is indeed similar to that form. (Nerd)

...
Do we do the same again $k$ times to cocnlude that $c_i=0$ for all $i=0, \ldots , k$ ? (Wondering)

Yep. (Nod)

Now let's add the initial choice for $x$ such that $\psi^k(x) \ne 0$. Such an $x$ must exist, since $\psi^k \ne 0$.
Then it follows that we have a contradiction.
Therefore, with that choice of $x$, it follows that $x, \psi(x), ..., \psi^k(x)$ are linearly independent. (Thinking)

 

[mathmari]

Yep. (Nod)

Now let's add the initial choice for $x$ such that $\psi^k(x) \ne 0$. Such an $x$ must exist, since $\psi^k \ne 0$.
Then it follows that we have a contradiction.
Therefore, with that choice of $x$, it follows that $x, \psi(x), ..., \psi^k(x)$ are linearly independent. (Thinking)

Could we formulate it also as follows?

Let $x\in V$ such that $\psi^k(x) \ne 0$. The set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent if and only if it holds that $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ for $c_0=c_1=\ldots =c_{k-1}=c_k=0$.

Since $\psi$ is a linear mapping we have that $\psi (0)=0$.

Since $\psi^{k+1}(x)=0$ and $\psi (0)=0$ we have that $\psi^n(x)=0, \ \forall n\geq k+1$.

We have the following:
$$\psi^k (c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^k (0) \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1} (x)+\ldots +c_{k-1}\psi^{2k-1}(x)+c_k\psi^{2k}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0 \\ \Rightarrow c_0=0$$

So, we get $c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.

We have that
$$\psi^{k-1} (c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^{k-1} (0) \\ \Rightarrow c_1\psi^k (x)+\ldots +c_{k-1}\psi^{2k-2}(x)+c_k\psi^{2k-1}(x)=0 \\ \Rightarrow c_1\psi^k(x)=0 \\ \Rightarrow c_1=0$$

So, we get $c_2\psi^2 (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.

When we repeat this $(k-2)$-times (or not?) we get that $c_i=0, \ \forall i=1, \ldots k$. Is everything correct? Could I improve something? (Wondering)

 

[I like Serena]

It looks all correct to me. (Nod)

Formally, I think we're supposed to set up a proof by induction though... (Thinking)

 

[mathmari]

Formally, I think we're supposed to set up a proof by induction though... (Thinking)

Ah ok. So, is the induction as follows? (Wondering)

Let $x\in V$ such that $\psi^k(x) \ne 0$. We wil show that the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent, i.e., that it holds it when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$.

Base case: For i=0 it holds the following:

Since $\psi$ is a linear mapping we have that $\psi (0)=0$.
$$\psi^k (c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^k (0) \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1} (x)+\ldots +c_{k-1}\psi^{2k-1}(x)+c_k\psi^{2k}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0 \\ \Rightarrow c_0=0 \ \ \checkmark$$ Inductive hypothesis: We suppose that it holds that for $i\leq m$ : $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$$ Inductive step: We want to shw that it holds for $i=m+1$:
From the inductive hypothesis we get that $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$. So $c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
Then we have the following:
$$\psi^{k-m-1} (c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^{k-1} (0) \\ \Rightarrow c_{m+1}\psi^k (x)+\ldots +c_{k-1}\psi^{2k-m-2}(x)+c_k\psi^{2k-m-1}(x)=0 \\ \Rightarrow c_{m+1}\psi^k(x)=0 \\ \Rightarrow c_{m+1}=0$$
So, we have that when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$, i.e., the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent. This is the only way to show that the set is inearly independent, right? (Wondering)

 

[HallsofIvy]

Ah ok. So, is the induction as follows? (Wondering)

Let $x\in V$ such that $\psi^k(x) \ne 0$. We wil show that the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent, i.e., that it holds it when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$.

Base case: For i=0 it holds the following:

"i" is not the index here, k is.

Since $\psi$ is a linear mapping we have that $\psi (0)=0$.
$$\psi^k (c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^k (0) \\ \Rightarrow c_0\psi^k(x)+c_1\psi^{k+1} (x)+\ldots +c_{k-1}\psi^{2k-1}(x)+c_k\psi^{2k}(x)=0 \\ \Rightarrow c_0\psi^k(x)=0 \\ \Rightarrow c_0=0 \ \ \checkmark$$ Inductive hypothesis: We suppose that it holds that for $i\leq m$ : $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$$ Inductive step: We want to shw that it holds for $i=m+1$:
From the inductive hypothesis we get that $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0 \Rightarrow c_i=0, \forall 1\leq i\leq m$. So $c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$.
Then we have the following:
$$\psi^{k-m-1} (c_{m+1}\psi^{m+1}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x))=\psi^{k-1} (0) \\ \Rightarrow c_{m+1}\psi^k (x)+\ldots +c_{k-1}\psi^{2k-m-2}(x)+c_k\psi^{2k-m-1}(x)=0 \\ \Rightarrow c_{m+1}\psi^k(x)=0 \\ \Rightarrow c_{m+1}=0$$
So, we have that when $c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0$ then $c_i=0, \forall i$, i.e., the set $\{x, \psi (x), \ldots , \psi^k\}$ ist linearly independent. This is the only way to show that the set is inearly independent, right? (Wondering)

 

[mathmari]

"i" is not the index here, k is.

So, do we have to show that for each $k\geq 0$ it holds that $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0\Rightarrow c_0=\ldots =c_k=0$$ ? (Wondering)

Base case: For $k=0$ we have that $\psi (x)\neq 0$ and then $$c_0x=0\Rightarrow \psi(c_0x)=\psi(0)\Rightarrow c_0\psi(x)=0\Rightarrow c_0\ \checkmark$$

Inductive hypothesis: We suppose that it holds for $k=n$:
$$c_0x+c_1\psi (x)+\ldots +c_{n-1}\psi^{n-1}(x)+c_n\psi^n(x)=0\Rightarrow c_0=\ldots =c_n=0$$

Inductive step: We want to show that it holds for $k=n+1$:
$$c_0x+c_1\psi (x)+\ldots +c_{n}\psi^{n}(x)+c_{n+1}\psi^{n+1}(x)=0\Rightarrow c_0=\ldots =c_{n+1}=0$$
What can we do here? How can we use the inductive hypothesis? We could apply it only if we would have $c_0x+c_1\psi (x)+\ldots +c_{n-1}\psi^{n-1}(x)+c_n\psi^n(x)=0$, or not? (Wondering)

 

[mathmari]

Inductive step: We want to show that it holds for $k=n+1$:
$$c_0x+c_1\psi (x)+\ldots +c_{n}\psi^{n}(x)+c_{n+1}\psi^{n+1}(x)=0\Rightarrow c_0=\ldots =c_{n+1}=0$$

Tp apply here the inductive hypothesis, we have to show first that $c_{n+1}=0$, or not? But how? When we apply any map $\psi^i$ the term $c_{n+1}\psi^{n+1}(x)$ will get zero. (Wondering)

 

[I like Serena]

I think we should proof for an arbitrary $k$ that $c_0 = 0, ..., c_i=0, ..., c_k=0$.

Then the base case is to proof that $c_0 = 0$.
And the induction step is to assume that $c_0=...=c_i=0$, and to proof that $c_{i+1}=0$. (Thinking)

 

[mathmari]

I tried now the following:

Let $x\in V$ such that $\psi^k(x) \ne 0$. We will show that the set $\{\psi^{k-n}(x),\psi^{k-(n-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent for $0\leq n\leq k$, i.e., that it holds it when $c_0\psi^{k-n}(x)+c_1 \psi^{k-(n-1)}(x)+\ldots +c_k\psi^{k}(x)=0$ then $c_0=\ldots c_k=0$. Base case: For $n=0$ we have the set $\{\psi^k(x)\}$. A non-zero element set is linealry independent.

Inductive hypothesis: We suppose that it holds for $n=i$:
The set $\{\psi^{k-i}(x),\psi^{k-(i-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent, i.e., $$c_0\psi^{k-i}x+\ldots +c_k\psi^k(x)=0\Rightarrow c_0=\ldots =c_k=0$$

Inductive step: We want to show that it holds for $n=i+1$:
$$c_0\psi^{k-(i+1)}(x)+c_1 \psi^{k-i}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^{k}(x)=0 \\ \Rightarrow \psi (c_0\psi^{k-(i+1)}(x)+c_1 \psi^{k-i}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^{k}(x))=\psi (0) \\ \Rightarrow c_0\psi^{k-i}(x)+c_1 \psi^{k-(i-1)}(x)+\ldots +c_{k-1}\psi^{k}(x)+c_k\psi^{k+1}(x)=0 \\ \Rightarrow c_0\psi^{k-i}(x)+c_1 \psi^{k-(i-1)}(x)+\ldots +c_{k-1}\psi^{k}(x)=0$$ then by the inductive hypothesis we get that the set $\{\psi^{k-i}(x),\psi^{k-(i-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent, and so $c_0=\ldots c_{k-1}=0$

So, from $c_0\psi^{k-(i+1)}(x)+c_1 \psi^{k-i}(x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^{k}(x)=0$ we get that $c_k\psi^{k}(x)=0$ and since $\psi^{k}(x)\neq 0$, it follows that $c_k=0$.

Therefore, $c_0=c_1=\ldots =c_{k-1}=c_k=0$.

So, the set $\{\psi^{k-n}(x),\psi^{k-(n-1)}(x),\dots,\psi^{k}(x)\}$ is linearly independent for each $n$.

For $n=k$ we get that the set $\{x,\psi(x),\dots,\psi^{k}(x)\}$ is linearly independent. Is this correct? (Wondering)