# 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:

1. $$f$$ is injective, $$n=3$$, $$m=4$$.
2. $$\text{rang}(f)=6$$, $$\text{Nullity}(f)=5$$, $$m=11$$.
3. $$f$$ is surjective, $$n=4$$, $$m=3$$.
4. $$f$$ is injective, $$n=2$$, $$\text{rang}(f)=3$$.
5. $$\text{Nullity}(f)=0$$, $$n=3$$, $$m=5$$.



I have done the following:

1. 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?

2. 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 $$m$$ ?

3. 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?

4. 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?

5. Do we apply here the Rank–nullity theorem? But how?

## 2 Answers

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 $$m=n$$.

• At your last sentence you have that f is injective iff n<m. Therefore isn't my proof of 1 correct and there is not such a map? – Mary Star Feb 2 at 8:57
• You wrote $n=3$ and $m=4$ in the question, but then later you wrote $n=4$ and $m=3$. I used the first one. And just to be clear, it is not true that $f$ is injective iff $n<m$. It is important to say $n\leq m$ instead of $n<m$. – pre-kidney Feb 2 at 8:58
• Ahh ok!! Could you explain to me the proposition 2? Why does this f satisfy the desired conditions? – Mary Star Feb 2 at 10:51
• What part of 2. doesn't make sense? By definition, the nullity equals $\dim \ker f$. I explained why this equals $5$. Also, what you are calling rang($f$) is the same as $\dim f(V)$, or in words the dimension of the image of $f$. From the form of the function $f$ I wrote, you can see that the image of $f$ is $6$-dimensional, corresponding to the first $6$ entries of the $11$-tuple. Since I have given what I believe to be a complete explanation already, if something doesn't make sense to you the onus is on you to ask a concrete and specific question that I can respond to. – pre-kidney Feb 2 at 22:48

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 $$A_{5 \times 3}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$