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I am trying to show that If $S,T$ are mappings from $V \rightarrow V$, where V is finite, then $\operatorname{Rank}(S+T)\leq\operatorname{Rank}(S)+\operatorname{Rank}(T)$ after finishing a chapter entitled "Applications of Rank-Nullity".

First consider two lists of vectors $S(x_1),\ldots,S(x_n)$ and $T(y_1),\ldots,T(y_n)$ where the $x_i, y_i \in V$ the vectors are linearly independent then it seems to me that the basis S+T will have a trimmed down version of the combination of these two lists (i.e. getting rid of any redundant or linearly dependent vectors) thus its rank would be less than or equal to $S(x) + T(x)$, but I am not sure how to write this in a more formal way. Any Hints appreciated including alternative methods of proof if I have gone off the rails here.

Samuel Adrian Antz
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IntegrateThis
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2 Answers2

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Hint:

Since the rank of a linear map is the dimension of its image, simply prove $\;\DeclareMathOperator\im{Im}\im(S+T)\subset\im S+\im T$.

Bernard
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Rank$(S)=\dim S(V)$ by definition. Thus, Rank$(S+T)=\dim (S+T)(V)$. But $(S+T)(V)\subset S(V)+T(V)$, and so $$ \dim(S+T)(V)\le\dim(S(V)+T(V))\le\dim S(V) +\dim T(V)=\operatorname{Rank}(S)+\operatorname{Rank}(T). $$

Vladimir
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