I am studying linear algebra and I viewed an online NPTEL course yesterday on linear algebra on the topic "Direct Sum Decompositions of Subspaces and Projections" in the Lecture 32. There I found an equivalence of three statements which are as follows :

Let $W_1,W_2,\cdots,W_k$ be subspaces of a finite dimensional vector space $V$. Then the following statements are equivalent $:$

$(1)$ $W_1,W_2,\cdots,W_k$ are independent.

$(2)$ $\forall j$, $2 \leq j \leq k$ $$W_j \cap (W_1+W_2+\cdots+W_{j-1}) = \{0 \}.$$

$(3)$ If $B_1,B_2,\cdots,B_k$ are bases for $W_1,W_2,\cdots,W_k$ then $B=B_1 \cup B_2 \cup \cdots \cup B_k$ is a basis for $W=W_1+W_2+\cdots+W_k$.

The instructor only proved $(1) \implies (2)$ and left other as an exercise. I have proved $(2) \implies (1)$ which proves the equivalence of $(1)$ and $(2)$ and then I proved $(1) \implies (3)$. But I think $(3) \implies (1)$ cannot be proved. For instance if I take $W=\mathbb R^3$ and take $W_1=\mathrm{span} \{(1,0,0),(0,1,0) \}$,$W_2=\mathrm{span} \{(0,1,0),(0,0,1) \}$ and $W_3=\mathrm{span} \{(0,0,1),(1,0,0) \}$. Then clearly the given condition $(3)$ holds. Though $(1,1,0)+(0,-1,1)+(-1,0,-1)=(0,0,0)$ where $(1,1,0) \in W_1$ ,$(0,-1,1) \in W_2$ and $(-1,0,-1) \in W_3$. This clearly shows that $W_1,W_2,W_3$ are not independent subspaces of $\mathbb R^3$.

So what is wrong in statement $(3)$. Please help me by telling this. It will really help me a lot.

Thank you in advance.

  • $\begingroup$ One way that the statement could be made true is to require that the $B_i$ be disjoint. $\endgroup$ Commented Oct 8, 2017 at 5:36
  • $\begingroup$ That's true. But how can this kind of error be made in an online course? That's really confusing to the viewer. If some error was occured that should be made correct. Isn't it? $\endgroup$
    – user251057
    Commented Oct 8, 2017 at 5:39
  • $\begingroup$ What can I do? Would I write it in the comment box in Youtube? $\endgroup$
    – user251057
    Commented Oct 8, 2017 at 5:42
  • $\begingroup$ Can you please add the definition of independent subspaces? $\endgroup$
    – egreg
    Commented Oct 8, 2017 at 12:20

1 Answer 1


There is an implied quantifier. The following statement is indeed equivalent to (1) and (2), so I assume it's what was meant:

(3) For every choice of bases $B_1, \ldots, B_k$ of the subspaces $W_1, \ldots, W_k$ the union $B = B_1 \cup \cdots \cup B_k$ is a basis for the sum $W = W_1 + \cdots + W_k$.

Now for (3) $\Rightarrow$ (1) assume $w_1 + \cdots + w_k = 0$ with $w_i \in W_i$. For each $W_i$ choose a basis $B_i$ such that if $w_i \neq 0$ then $w_i \in B_i$. The union $B$ of the $B_i$ is a basis for $W$ so, in particular, the nonzero $w_i$ (which are all contained in this union) must be independent. If any of the $w_i$ were nonzero the expression $w_1 + \cdots + w_k = 0$ would contradict this, so it must be that $w_i = 0$ for all $i$.

  • $\begingroup$ Would you please prove $(3) \implies (1)$ from your claim. $\endgroup$
    – user251057
    Commented Oct 8, 2017 at 6:27
  • $\begingroup$ youtu.be/… $\endgroup$
    – user251057
    Commented Oct 8, 2017 at 6:32
  • $\begingroup$ please view this link $\endgroup$
    – user251057
    Commented Oct 8, 2017 at 6:32
  • $\begingroup$ See the edit I just made. $\endgroup$
    – Jim
    Commented Oct 8, 2017 at 12:14
  • $\begingroup$ Given the current statement for 3, there's no reason that we can't have $w_i=- w_j$ for some $i, j$, so there's no reason for the conclusion that they must be independent. $\endgroup$ Commented Oct 9, 2017 at 3:28

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