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From this question:

Suppose $V$ is a vector space with dimension $6$. Let A and B be subspaces of V with dimensions 4 and 5 respectively. What are the possible values for the dimension of A intersection B?

The reasoning given in the answer:

The best way is to look at the basis. Any basis for $A$ consists of $4$ elements, say $\{a, b, c, d\}$. Suppose that $e, f$ are vectors of $V$ such that $\{a, b, c, d, e, f\}$ span the whole $V$. Since $B$ has dim $5$, exactly five elements of $\{a, b, c, d, e, f\}$ are in $B$.

$\ldots$


The part in bold bothers me. Why is it necessary that any basis for $B$ must contain elements from basis of $V$ ?

More generally, what is the relation between basis vectors of a vector space to those of its subspace?

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    $\begingroup$ You are right to be worried. It's wrong. None of the basis vectors need to lie in $B$. $\endgroup$ Commented Mar 24, 2014 at 14:28

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Example: $V=\mathbb{R}^2$, with basis $\{(1,0),(0,1)\}$. Let $B=Span((1,1))$, a diagonal line. Neither basis vector from $V$ is in $B$.

In short, the bold statement is incorrect. Try Math.SE next time instead of Yahoo answers.


Hint for question that led to all this: Given finite-dimensional vector space $V$ and subspaces $A,B$, we have $$\dim(A)+\dim(B)=\dim(A+B)+\dim(A\cap B)$$

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  • $\begingroup$ In that case, would the correct answer to the original question be : any of $0,1,2,3,4$ ? $\endgroup$
    – curryage
    Commented Mar 24, 2014 at 14:34
  • $\begingroup$ @curryage No, either $3$ or $4$. $\endgroup$ Commented Mar 24, 2014 at 14:39
  • $\begingroup$ @DanielFischer Using the formula provided by vadim123, that becomes possible when $dim(A+B)$ is $4$ or $5$, which makes sense. Is there a good online reference where I can read up more specifically on such relationships between subspaces ? $\endgroup$
    – curryage
    Commented Mar 24, 2014 at 15:53
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    $\begingroup$ @curryage When $\dim (A+B)$ is $6$ or $5$. I don't know any online references, but every introductory book on linear algebra should help getting familiar with such relations. $\endgroup$ Commented Mar 24, 2014 at 15:56

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