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$A,B$ are subspace of a finite-dimensional vector space $V$. Show that $\dim(A) + \dim(B) = \dim(A+B)$ if and only if $A \cap B = \{0\}$.

It (kind of) seems intuitive but I'm having a hard time putting words to my intuitions.

Chappers
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lars
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    You should check: http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem – YoTengoUnLCD Apr 15 '15 at 01:56
  • @YoTengoUnLCD What does this have to do with Rank-nullity, apart from dimension-counting? There is no linear map here. – Chappers Apr 15 '15 at 02:11
  • If one knows the formula for the dimension of the quotient space, all one has to do, is to show that the canonical map $T : A \longrightarrow \frac {A + B} {B}$ defined by $x \mapsto x + B,$ is a linear surjection with kernel $A \cap B.$ Once we show that we are done by the virtue of first isomorphism theorem. – Anil Bagchi. Jun 02 '23 at 06:57

1 Answers1

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Pick a basis $\{a_i\}_i$ for $A$ and a basis $\{b_j\}_j$ for $B$. Suppose first that $A \cap B \neq \{ 0\}$. This means that there is a nonzero vector, $v$, in $A \cap B$. Then $v \in A$, so $$ v = \sum_i \lambda_i a_i $$ for some scalars $\lambda_i$. Similarly, $v \in B$, so there are scalars $\mu_j$ so $$ v = \sum_j \mu_j b_j. $$ But then, $$ 0 = v-v = \sum_i \lambda_i a_i-\sum_j \mu_j b_j. $$ This is a nonzero linear combination of $a_i$ and $b_j$ that gives zero, so the set $\{ a_i \}_i \cap \{ b_j\}_j$ is linearly dependent. (If the linear combination is zero, then some of the $a_i$ are equal to some of the $b_j$, with the same result.) But $\{ a_i \}_i \cap \{ b_j\}_j$ spans $A + B$, so a basis must have fewer vectors in it than this set; hence the equality does not hold.

Whereas, if we assume the intersection is just the zero vector, there is no such linear combination, and hence the $a_i$ and $b_j$ are linearly independent and so form a basis.

(Alternatively, you could pick a basis for the intersection, then do two extensions of it, to $A$ and to $B$, and then count.)

Chappers
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