# Doubt about Vector Spaces A∩B

Hello, in my university exam, I came across the following problem:

Find the basis of A∩B for A=lin{ a1, a2, a3} and B=lin{b1, b2, b3}.

My question is, if I prove that: b1, b2 ∈ A, is it undeniably certain that A∩B=lin{b1, b2}?

I demonstrated in the specific case of my exam (where there were indeed numbers) that b3 is linearly independent from a1, a2, and a3, while b1 and b2 are linearly dependent on a1 and a2.

## 1 Answer

If I prove that: $b_1, b_2 \in A$, is it undeniably certain that $A\cap B=lin\{b_1, b_2\}$?

The answer is NO. You can simply take $A=B=\{i,j,k\}$. Then $A\cap B=\{i,j,k\}.$

In general, it is posisble to also $b_3 \in A$, and if $b_3$ is linearly independent from $b_1$ and $b_2$, then

$$b_3 \in lin\{b_1, b_2,b_2\} \text{but} b_3 \notin lin\{b_1, b_2\}.$$

So $A\cap B \neq lin\{b_1, b_2\}.$

However, if $b_3$ is not linearly independent of $b_1$ and $b_2$, then $A\cap B=lin\{b_1, b_2\}$ is true.

- 1 Answer
- 78 views
- Pro Bono

### Related Questions

- Module isomorphism and length of tensor product.
- Does $\lim_{n \rightarrow \infty} \frac{2^{n^2}}{n!}$ exist?
- Step by step method to solve the following problem: find coordinates of B.
- Algebra Word Problem 2
- Differentiate $f(x)=\int_{\sqrt{x}}^{\arcsin x} \ln\theta d \theta$
- Use first set of data to derive a second set
- Algebra 2 help Please find attachment
- Tensor Product