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For a linear map $h: \mathbb R^3 \rightarrow \mathbb R^2$, the kernel of $h$ is a subspace of $ \mathbb R^2$.

For a linear map $h: \mathbb R^3 \rightarrow \mathbb R^2$, the range of $h$ is a subspace of $ \mathbb R^2$.

Can someone help me prove or disprove these statements. I'm pretty sure that the first one is false and the second one is true...

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You are right in both acounts. Now just check again the definitions to make sure you fully understand and then you'll clearly see why you obviously are right. –  DonAntonio Apr 12 '13 at 3:16
    
You are correct that the first is false and the second is true, so the problem is coming up with proofs. Do you know how to disprove a statement? Maybe find an element in the kernel that is not an element of $\mathbb R^2$ (try the 0 in $\mathbb R^3$). Do you know how to prove that a subset of $\mathbb R^2$ is a subspace? There are axioms that you have to check, do you know what they are? –  Jim Apr 12 '13 at 3:16
    
Actually, in the first case, the kernel is a subspace of $\mathbb{R}^3$, not $\mathbb{R}^2$. The second assertion is correct. –  1015 Apr 12 '13 at 3:19
    
Calculate the kernel and range of an example of such a map, and then you will understand the definitions. –  user66345 Apr 12 '13 at 3:27
    
If you got your answer to the question, I recommend you to tick it. –  Metin Y. Apr 12 '13 at 19:30

2 Answers 2

up vote 1 down vote accepted
  • Consider $h$ taking each element of $\mathbb R^3$ to $(0,0)$. Therefore, $Ker(h)=\mathbb R^3$ and is not a subspace of $\mathbb R^2$.

  • Image of a linear map is always a subspace of the codomain.

These two are more than a hint and less than a solution but you had better check that in the first case $h$ is indeed a linear map and in the second one $Im(h)$ is as well.

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By definition, the kernel of $h$ must lie in the domain, not the codomain. However your wording is kind of unclear, because $\mathbb R^2$ (the codomain) can be viewed as a subspace of $\mathbb R^3$ (the domain). Can you think of a map such that the kernel of $h$ is larger than all of $\mathbb R^2$?

Additionally, the range of $h$ must lie in the codomain, so certainly $\text{Im}\,h\subset\mathbb R^2$. However, do you know that the range of $h$ is a subspace of the codomain?

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$\mathbb{R}^2$ cannot be viewed as a subspace of $\mathbb{R}^3$, it is not even a subset. The most you can say is that there are subspaces of $\mathbb{R}^3$ isomorphic to $\mathbb{R}^2$. –  EuYu Apr 12 '13 at 3:19
    
This seems overly pedantic for the circumstances. –  Ian Coley Apr 12 '13 at 3:20
    
I disagree. The question has a clear cut answer. $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$ and that's that. There is nothing ambiguous. You may regard the line "$\mathbb{R}^2$ can be viewed as a subspace of $\mathbb{R}^3$" in terms of isomorphism, but to the OP that's likely to induce the notion that $\mathbb{R}^2$ can be an actual subspace of $\mathbb{R}^3$. It's unnecessary at best and wrong at worst. –  EuYu Apr 12 '13 at 3:25

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