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I am given that $A$ is a $12\times 15$ matrix and the equation $Ax = b$ has a solution for every $b \epsilon \mathbb{R}^{12}$. What are the dimensions of the domain, the range and the kernel of $A$

I know that $Ax = b$ maps vectors from $\mathbb{R}^{15}$ to $\mathbb{R}^{12}$. Does this get me anything?

Secondly, do I know if the set of column vectors in $A$ is linearly independent and does this get me anywhere?

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Recall the rank nullity theorem which tells you that $n=\dim\ker+\dim\text{im}$ if $f:\mathbb{R}^n\to\mathbb{R}^m$. Thus, it suffices to find any two of the three you are asked to find. Now, you tell me, the fact that $Ax=b$ has a solution tells us that our map is surjective--it hits all of $\mathbb{R}^{12}$--so what's it's dimension? The domain is $\mathbb{R}^{15}$, what is this dimension?

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Well, if the domain is $\mathbb{R}^{15}$ then the dimension is 15. And if the transformation's image hits all of $\mathbb{R}^{12}$ then the dimension of the range is 12. And dim(ker[A]) = dim(domain[A]) - dim(range[A]) = 3, correct? – intervade Apr 5 '12 at 1:47
@DaltonConley Yes that is right. The kernel has dimension 3. – user38268 Apr 5 '12 at 1:52
Thank you, this helped me understand the rank nullity theorem better. The book I am using doesn't have a very good explanation, in my opinion. – intervade Apr 5 '12 at 1:54

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