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I am learning about the dimension theorem:

Let $L: V \to W$ be a linear map from the vector space $V$ to the vector space $W$. If $V$ is finite dimensional then $\dim \ker(L) + \dim \text{Range}(L) = \dim V$.

What is the intuition behind this result? I find it counterintuitive that the dimension of a subspace from $V$ (null space of $V$) plus the dimension of a subspace in $W$ (range of $L$) add up to the dimension of $V$, the vector space where we are mapping from. Any insights are appreciated.

Ben Grossmann
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cbakos
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  • All vectors in $V$ have to be mapped somewhere in $W$. Their images can be zero or non-zero. No vector in $V$ can be mapped to both a non-zero and zero vector so the two subsets partition $V$. – John Douma Sep 13 '21 at 19:49

2 Answers2

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You can think of the dimension of a vector space as the amount of information that it contains. In this way of thinking, if $L : V \to W$ then we can think of $\dim(\ker(L))$ as the amount of information destroyed when going from $V$ to $W$, and $\dim(\mathrm{range}(L))$ as the amount of information that is not destroyed. In this context, if $L$ doesn't destroy any information then we would get $\dim(\mathrm{range}(L)) = \dim(V)$. However, if some of the information is destroyed then we can think of this formula as stating something like: the amount of information destroyed ($\dim (\ker(L))$) plus the amount of information that is not destroyed ($\dim(\mathrm{range} (L))$) equals the original amount of information, which is $\dim(V)$.

Pedro Silva
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    In order to get properly formatted names for the operators dim, ker, and range, you can use \dim, \ker, and \mathrm{range}. – Ben Grossmann Sep 13 '21 at 11:42
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Let $n=\dim\operatorname{Range}L$. Then there is a basis $\{w_1,\ldots,w_n\}$ of $\operatorname{Range}L$. Since each $w_k$ is in the range of $L$, there is some $v_k\in V$ such that $L(v_k)=w_k$ and, since $\{w_1,\ldots,w_n\}$ is linearly independent, so is $\{v_1,v_2,\ldots,v_n\}$. Now, if $m=\dim\ker L$, let $\{v_1',v_2',\ldots,v_m'\}$ be a basis of $\ker L$. It turns out that then $\{v_1,v_2,\ldots,v_n\}\cup\{v_1',v_2',\ldots,v_m'\}$ is a basis of $V$ (that's not hard to prove), and therefore$$\dim V=m+n=\dim\ker L+\dim\operatorname{Range}L.$$

José Carlos Santos
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