I was confused about zero dimension vector subspaces. Can you please answer the following questions with details/examples.

  1. What is the dimension and basis for vector space that is just composed of the zero vector.

  2. If the answer to the latter is an empty set, how can I construct a zero vector of say n rows from the empty set?

  3. Say there are three vector spaces [0] [0, 0] [0 ,0, 0]. What are the basis for them. Isn't just saying empty set led to ambiguity about the zero vector subspace, it is a basis for?


(1) A vector space that is composed of just the zero vector is zero dimensional and its basis is the empty set.

(2) You can construct a zero vector because the empty sum is defined to be zero (this is somewhat of a cheat). The sum $\sum_{v_i\in\emptyset}a_iv_i$ is an empty sum, and it is defined to be the zero element of the vector space.

(3) The vector spaces $\{[0]\}$, $\{[0,0]\}$, and $\{[0,0,0]\}$ are all isomorphic, so there really isn't much ambiguity (they are all, in essence the same space). If you want to use the sum from before, the empty sum valuates to the zero element in the vector space in which you're working.

  • $\begingroup$ How can different linear combinations of the zero element to different zero vectors? $\endgroup$ – Abhishek Bhatia Dec 1 '16 at 4:18
  • $\begingroup$ Because you always work in the context of some space, the empty set is a subset of some vector space and you get the zero for that vector space you're working in. $\endgroup$ – Michael Burr Dec 1 '16 at 4:22
  • $\begingroup$ If possible can you add example to explain this in the answer. $\endgroup$ – Abhishek Bhatia Dec 1 '16 at 4:23
  • $\begingroup$ An example isn't very enlightening. Let $V$ be a vector space. The basis $\emptyset\subseteq V$ generates the subspace $\{0\}$ where $0$ is the zero of $V$. $\endgroup$ – Michael Burr Dec 1 '16 at 4:26

Here is a proof:

Let us assume that $\mathscr B$ is a basis of $\{0\}$.

Then since $\mathscr B$ is a subset of $\{0\}$ so $\mathscr B=\{0\}$ or $\mathscr B=\emptyset$. If it has to be non-empty then $\mathscr B=\{0\}$

But $\{0\}$ is not linearly independent how can it be a basis.

  • $\begingroup$ The proof makes sense. Can you explain more on empty sum. $\endgroup$ – Abhishek Bhatia Dec 1 '16 at 4:15
  • $\begingroup$ If a set $S$ is empty then the sum of elements of $S$ is defined to be $0$;For more google empty sum $\endgroup$ – Learnmore Dec 1 '16 at 4:53

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