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

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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?

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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 $m=n$.

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  • $\begingroup$ 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? $\endgroup$ – Mary Star Feb 2 at 8:57
  • $\begingroup$ 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 $n<m$. $\endgroup$ – pre-kidney Feb 2 at 8:58
  • $\begingroup$ Ahh ok!! Could you explain to me the proposition 2? Why does this f satisfy the desired conditions? $\endgroup$ – Mary Star Feb 2 at 10:51
  • $\begingroup$ 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. $\endgroup$ – pre-kidney Feb 2 at 22:48
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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 $$A_{5 \times 3}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$

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