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I'm reviewing a little bit of linear algebra, and I got stuck a little bit on this trivial thing: for finite-dimensional vector spaces we know that injectivity follows from surjectivity and vice-versa.

So basically, if I consider a map between vector spaces of different dimensions, it's hopeless to find a linear bijective map. But if I have

$$T:\mathbb{R} \rightarrow \mathbb{R}^5, \;\text{ described by }\; T = \begin{bmatrix}1\\-1\\1\\0\\0\end{bmatrix},$$

then its kernel would be $\text{Ker }T = \{x \in \mathbb{R} \mid Tx = 0\}.$

Then, it would follow that $\begin{bmatrix}1\\-1\\1\\0\\0\end{bmatrix}*x = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \, \, \, \, \Longrightarrow \, \, \, \begin{cases}x=0\\ -x=0\\x=0 \\0=0 \\0=0 \end{cases}$.

Basically this would end up implying that the Kernel is the null vector only? What's the point here?

I know this is trivial but it has been a long time since my basic linear algebra course.

311411
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James Arten
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    Yes. For the transformation you describe, the kernel contains only the zero vector, which in this context is the number $0 \in \Bbb R$. – Ben Grossmann May 24 '21 at 13:21
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    The map is injective but not surjective, which is fine as the source space is of lower dimension than the target space. You can check that a map from a one-dimensional space to a non-trivial space is either injective or the zero map. – Gunnar Þór Magnússon May 24 '21 at 13:23
  • So the dimension of kernel should be 0? The image of this transformation will actually be $\mathbb{R}^3$ – James Arten May 24 '21 at 13:32

2 Answers2

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Indeed, the kernel is only the zero vector - you are correct. One thing you could learn from this is that the kernel of a linear map must be a subspace of the domain, in this case $\mathbb{R}$. There are only two such subspaces, namely $\{0\}$ and $\mathbb{R}$ itself. So any linear map from $\mathbb{R}$ into anything must either map everything to zero or only $0$ to zero.

Wizact
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Your calculation of the kernel is right, but something is wrong with your theorem

for finite dimensional vector spaces we know that injectivity follows from surjectivity and vice-versa.

Your map is not onto 5-space. For example there is no way to hit $(0,0,0,0,1)$.

311411
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  • Thanks, but what I don't understand is the following: based on what I remember there's no chance to find any map that is injective (and hence surjective) between vector spaces of DIFFERENT dimensions. But it also holds that injectivity follows from kernel being 0 vector only. So the map should be supposed to be injective and hence surjective but this can not happen. Where is the mistake? – James Arten May 24 '21 at 13:30
  • An injective linear map of V onto W is a vector space isomorphism, so if you have such a map then dim V = dim W. (this is something like your "there's no chance" statement, but without the "hence".) I think you are mis-remembering this theorem: https://math.stackexchange.com/questions/1861183 – 311411 May 24 '21 at 13:39
  • your example T is linear, injective and between spaces of different dimensions. – 311411 May 24 '21 at 13:44
  • I'm so sorry but I still didn't get the point. QUESTION: is it possible to have a linear map between two finite-dimensional vector spaces with DIFFERENT dimensions that is injective and NOT surjective? (As it turns out from my initial example) – James Arten May 24 '21 at 14:53
  • Don't worry, the example T you invented is a very good one. T is linear, is just matrix multiplication. It has domain/source finite-dim. vec sp., it has co-domain/target of a DIFFERENT finite dim. Because T has trivial kernel, it is injective (you proved this). It is not surjective, because e.g. $Tx = (0,0,0,0,1)$ is true for no $x$. You should picture the range of T as a straight line through the origin of 5-space. Once you are happy with all that, Consider the possibility of using your 5-by-1 matrix to set up a linear bijection of a 1-dimen. space onto a (necessarily) one-dimensional space. – 311411 May 24 '21 at 15:45
  • so YES it is possible. – 311411 May 24 '21 at 15:46
  • Thanks! Now I finally remember how things work! – James Arten May 25 '21 at 07:19