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Question regarding "Linear algebra done right", proposition $2.19$:

Proposition $2.19$: Suppose $V$ is finite dimensional and $U_1,...,U_m$ are subspaces of $V$ such that $$V = U_1 + \dots + U_m\tag{2.20}$$ and $$\dim V=\dim \, U_1 + \dots + \dim \, U_m.\tag{2.21}$$ Then $V = U_1 \oplus \dots \oplus U_m.$

Proof: Choose a basis for each $U_j$. Put these bases together in one list, forming a list that spans $V$ (by $2.20$) and has length $\dim V$ (by $2.21$). Thus this list is a basis of $V$ (by $2.16$), and in particular it is linearly independent.

Now suppose that $u_1 \in U_1,...,u_m \in U_m$ are such that $0=u_1+ ··· +u_m$. We can write each $u_j$ as a linear combination of the basis vectors (chosen above) of $U_j$. Substituting these linear combinations into the expression above, we have written $0$ as a linear combination of the basis of $V$ constructed above. Thus all the scalars used in this linear combination must be $0$.

My question

Given $(v_1, \ldots, v_n)$ as the basis vectors: \begin{align} u_1 &= a_{11}v_1 + \ldots + a_{1n}v_n \\ & \ \vdots \\ u_m &= a_{m1}v_1 + \ldots + a_{mn}v_n. \end{align}

It seems the proof only proves $0 = (a_{11} + \ldots + a_{m1}) = \ldots = (a_{1n} + \ldots + a_{mn})$. But how do we know that each $a_{ij}$ is $0$ too?

Thanks.

glS
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user34829
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2 Answers2

2

You need to show that for $i\neq j$, $w\in U_i\cap U_j$ implies $w=0$. Write $w$ as a linear combinations of vectors from basis for $U_i$ first and then $U_j$ ( these vectors are part of the basis for $V$ as in the first part of your proof, so linearly independent). Then $w-w=0$ and all the coefficients must be zero. Hence $w=0$.

More precisely, we have (for exactly the same reason as above), that

$U_i\cap \sum_{i\neq j}U_j=\{0\}$.

AnyAD
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Let $\beta_{i}$ be a basis for $U_{i}$ for $1\leq i\leq m$. Then since $V = \sum_{i}U_{i}$ we have $\text{span}(\beta_{1}\cup\dots\cup\beta_{m}) = V$ and therefore $\#(\beta_{1}\cup\dots\cup\beta_{m})\geq \text{dim}(V)$. On the other hand $\#(\beta_{1}\cup\dots\cup\beta_{m})\leq \sum_{i = 1}^{m}\#(\beta_{i}) = \sum_{i = 1}^{m}\text{dim}(U_{i}) = \text{dim}(V)$ which proves the result.

Karthik Kannan
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