In Linear Algebra Done Right, by Sheldon Axler, the proposition 1.45 states "(V_1 + \cdots + V_m) is a direct sum if and only if the only way to write 0 as a sum (v_1 + \cdots + v_m), where each (v_k \in V_k) is by taking each (v_k) equal to 0". However, why can't I take (v_1 = -v_2)? My idea is to write something like "(0 = (-5) + 5 + 0 + \cdots + 0)" but using vectors.
#Direct sum
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In Linear Algebra Done Right, by Sheldon Axler, the proposition 1.45 states "(V_...
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In Linear Algebra Done Right, by Sheldon Axler, the proposition 1.45 states "(V_...
Do you mean something like if you have v_1+v_2…..+v_m=0 (each v_i being in V_i) then you can have v_1=-v_2 with v_1 being non zero ? Well no if the sum is direct then by definition the sum is unique so if you have v_1+….+v_m=0 then v_1+…+v_m=0+….+0 so for all i, v_i=0 necessarily
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