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I've a problem with the direct sum of more than two subspaces. For example if I have three vector subspaces of $\mathrm R^6 W_1,W_2,W_3$ with $\mathrm dimW_1=1,dimW_2=2,dimW_3=3$ and I want to show that $W_1 \oplus W_2 \oplus W_3 = R ^6$ is it sufficient to verify that the rank of the matrix which has on the rows the components of the vectors of the bases of the three subspaces is 6? In this way do I check also that every possible intersection of the subspaces is empty? And if so is this enough to conclude that they are supplementar and they generate $\mathrm R^6$ ? Thanks a lot in advice

($\mathrm R^6$ is a six dimension vector space on the field of real numbers)

Gianolepo
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The calculation you gave proves that $W_1+W_2+W_3=\mathbb{R}^6$. This is tight, because the dimensions add up to $6$ exactly.

Suppose there were some vector $u\in W_i\cap W_j$ (for $i\neq j$), then we could extend $\{u\}$ to a basis for $W_i$, and then remove it from that basis. The result would be $W_i'$, a space of one dimension smaller than $W_i$. However $W_i'+W_j=W_i+W_j$.

But now $W_i'+W_j+W_k=W_1+W_2+W_3$ would have dimension at most $5$, a contradiction.

Note: The above assumes that $R^6$ means $\mathbb{R}^6$, a six-dimensional vector space.

vadim123
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  • Thanks a lot for your answer! Could you just explain a little deeper the consequences of what you wrote? I mean, since supposing an intersection leads to a contradiction, can I conclude that $\mathrm W_1 \oplus W_2 \oplus W_3=R^6$ or just that W_1 + W_2 + W_3=R^6$ ? My doubts are mainly about the possibility of showing a direct sum just looking at the rank of the matrix – Gianolepo Nov 16 '15 at 07:36
  • You proved $W_1+W_2+W_3=\mathbb{R}^6$. Proving there is no pairwise intersection combines with this to prove $W_1\oplus W_2\oplus W_3=\mathbb{R}^6$. See, e.g., here. – vadim123 Nov 16 '15 at 15:08