# Thread: Linear mapping

1. 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?

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?

Originally Posted by 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 $i=0, \ldots , k$ ?

Or do we show it in an other way?

3. Hey mathmari!!

Originally Posted by mathmari
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?
Yep.

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.

Quote:
...
Do we do the same again $k$ times to cocnlude that $c_i=0$ for all $i=0, \ldots , k$ ?
Yep.

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.

Originally Posted by I like Serena
Yep.

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.
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?

5. It looks all correct to me.

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

Originally Posted by I like Serena
Formally, I think we're supposed to set up a proof by induction though...
Ah ok. So, is the induction as follows?

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?

7. Originally Posted by mathmari
Ah ok. So, is the induction as follows?

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.

Quote:
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?

Originally Posted by HallsofIvy
"i" is not the index here, k is.
So, do we have to show that for each $k\geq 0$ it holds that $$c_0x+c_1\psi (x)+\ldots +c_{k-1}\psi^{k-1}(x)+c_k\psi^k(x)=0\Rightarrow c_0=\ldots =c_k=0$$ ?

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?

Originally Posted by 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.

10. 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 $c_{i+1}=0$.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•