Prove or disprove the statements about linear map Let $m,n\in \mathbb{N}$ and let $V, W$ be $\ \mathbb{R}$-vector spaces with $\dim V=n$ and $\dim W=m$. 
Let $f:V\rightarrow W$ be a linear map. 
I want to prove or disprove the following: 
There is a linear map $f:V\rightarrow W$ with the following properties: 


*

*$f$ is injective, $n=3$, $m=4$. 

*$\text{rang}(f)=6$, $\text{Nullity}(f)=5$, $m=11$. 

*$f$ is surjective, $n=4$, $m=3$. 

*$f$ is injective, $n=2$, $\text{rang}(f)=3$. 

*$\text{Nullity}(f)=0$, $n=3$, $m=5$. 


$$$$ 
I have done the following: 


*

*Since $f$ is injective, then $\text{ker}(f) = \{0\}$ and so $\dim\text{ker}(f) = 0$.  
It holds that $\text{Im}(f)\subseteq W$. Therefore $\dim \text{Im}(f)\leq \dim W$. 
From the Rank–nullity theorem we have that \begin{equation*}\dim V  = \dim \text{ker}(f) + \dim \text{Im}(f)=\dim \text{Im}(f)\leq \dim W\end{equation*} 
Therefore it cannot be that $n=4$ and $m=3$, right?  

*From the Rank–nullity theorem we have that \begin{equation*}\dim V=\text{Defekt}(f)+\text{Rang}(f)=5+6=11 \Rightarrow n=11\end{equation*} How do we use the information of $m$ ?    

*Since $f$ is surjective, it holds that $\text{Rang}(f)=W$. 
From the Rank–nullity theorem we have that \begin{equation*}\text{Rang}(f)=\dim V-\text{Nullity}(f)\leq \dim V \Rightarrow \dim W\leq \dim V \Rightarrow m\leq n\end{equation*} 
Therefore there can be a linear surjective map with $n=4$ and $m=3$, right?    

*Since $f$ is injective, it holds that $\text{ker}(f) = \{0\}$ and so $\dim\text{ker}(f) = 0$.  
It holds that $\text{Nullity}(f)=\dim \text{ker}(f)=0$. 
From the Rank–nullity theorem we have that \begin{equation*}\text{Rang}(f)=\dim V-\text{Nullity}(f) \Rightarrow 3=4-0 \Rightarrow 3=4\end{equation*} 
A contradiction. 
Therefore there cannot be a linear surjective map with $n=2$ and $\text{Rang}(f)=3$, right? 

*Do we apply here the Rank–nullity theorem? But how? 
 A: For 1., you can consider $f\colon \mathbb R^3\to\mathbb R^4$ given by
$$
f(x,y,z)=(x,y,z,0)
$$
which is injective.
For 2., you can consider $f\colon \mathbb R^{11}\to\mathbb R^{11}$ given by
$$
f(x_1,\ldots,x_{11})=(x_1,\ldots,x_6,0,\ldots,0),
$$
since $\ker f$ equals the set of vectors of the form $(0,\ldots,0,x_7,\ldots,x_{11})$, which is $5$-dimensional.
For 3., you can consider $f\colon \mathbb R^4\to\mathbb R^3$ given by
$$
f(x_1,\ldots,x_4)=(x_1,x_2,x_3),
$$
which is surjective.
For 4., you can show it is impossible regardless of the value of $m$ - more generally, $\dim f(V)\leq \dim V$ regardless of the linear map $f$.
For 5., you can consider $f\colon \mathbb R^3\to\mathbb R^5$ given by
$$
f(x_1,\ldots,x_3)=(x_1,x_2,x_3,0,0),
$$
which is injective.

Notice my examples in 1., 3., and 5. had exactly the same form, namely
$$
f\colon \mathbb R^n\to\mathbb R^m,\qquad f(x_1,\ldots,x_n)=(x_1,\ldots,x_{\min(m,n)},0,\ldots,0).
$$
You can see that $\ker f$ consists of all elements of $\mathbb R^n$ whose first $\min(m,n)$ entries are equal to zero, so
$$
\dim\ker f=n-\min(m,n).
$$
Similarly,
$$
\dim f(\mathbb R^n)=\min(m,n).
$$
The rank-nullity theorem holds because
$$
\bigl(n-\min(m,n)\bigr)+\min(m,n)=n.
$$
Moreover $f$ is injective iff $\dim\ker f=0$, i.e. $n=\min(m,n)$ which is equivalent to $n\leq m$. And $f$ is surjective iff $\dim f(\mathbb R^n)=m$, i.e. $\min(m,n)=m$ or equivalently $m\leq n$. Consequently $f$ is bijective if and only if $m=n$.
A: Since these are linear maps between finite dimensional vector spaces $V_n \longrightarrow W_m$ so each of them (if it exists) can be represented by a $m \times n$ matrix $A$.
For (1): Here $A$ must be $4 \times 3$ matrix. For injective map, no free columns in $A$, so take
$$A_{4 \times 3}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{bmatrix}$$
For (2): $\text{dim} V=\text{rank}+\text{nullity}=6+5=11$, so $A$ should have $6$ pivot columns and $5$ free columns. For example,
$$A_{11 \times 11}=\begin{bmatrix}\mathbf{e_1} & \mathbf{e_2} & \ldots & \mathbf{e_6} & \mathbf{0} & \ldots & \mathbf{0}\end{bmatrix}.$$
For (3): Here $A_{3 \times 4}$. So max rank of $A$ is $3$. For surjective map, we need full rank. Thus
$$A_{3 \times 4}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}$$
For (4): Here $A_{m \times 2}$. $\text{rank}=3$, so $m \geq 3$. For injective map $A$ should not have free columns. But maximum rank of this matrix can be $2$ and $m$ being at least three, there will always be a free column. So not possible.
For (5): Here $A_{5 \times 3}$. Nullity $0$ means injective map, so no free columns in $A$. Maximum rank of this matrix can be $3$. Thus
$$A_{5 \times 3}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$
