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I'm reading Axler's Linear Algebra Done Right book and its my first time doing linear algebra. I'm confused on this proof: Suppose V and W are finite-dimensional vector spaces such that dim V < dim W. Then no linear map from V to W is surjective. Proof: Let T ∈ L(V, W).
Then dim range T = dim V - dim null T ≤ dim V < dim W I understand the first equality (fundamental thm of linear maps). the last inequality i understand comes from the assumption. I'm confused on the middle inequality. Any help would be appreciated.

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The dimension of the null space is weakly greater than $0$. $ a - k \leq a$ if $k \geq 0$.

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  • $\begingroup$ i.e., dimension is nonnegative. $\endgroup$ – parsiad Feb 19 at 5:32
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Generally speaking, if $x$ and $y$ are numbers and $y$ is nonnegative, meaning $y\geq0$, then $x-y\leq x$.

In this case, $\dim\mathrm{null}\;T\geq 0$, so $\dim V - \dim\mathrm{null}\;T \leq \dim V$.

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