Let A and B be subspaces of $R^5$ such that A $ \subseteq$ B

Question

Let A and B be subspaces of $R^5$ such that A $ \subseteq$ B.

$ \{a1,a2,a3\} $is a basis for subspace A.

$\{b1,b2,b3,b4\} $ is a linearly dependant set that spans B.

$ \quad $ (1) Dimension of B = 4

$ \quad $ (2) A = B

$ \quad $ (3) $\{b1,b2,b3,b4\}$ is a basis for B.

$ \quad $ (4) Dimension of A = 3

I think that only (4) would be true, (1) and (3) are untrue because the vectors $b_i$ are linearly dependent, but I am not sure if there is a way to prove that B $\subseteq$ A, which would make (2) true? Please correct me if I am wrong. Thank you

Answer 1

1 is false. The $b$ vectors are linearly dependent and span $B$, so you could remove one and still span $B$, so you could span $B$ with less than 4 vectors, so dimension is less than 4.

3 is also false given the same logic as for 1)

4 is true because 3 vectors form a basis of A.

Now, note that since A is a subset of B, B must be dimension 3. Any 3 linearly independent vectors in B then are a basis of B. The $a$ vectors are linearly independent and in B, and there are 3 of them, so they are also a basis for B. Therefore A and B share a basis, so $A = B$, so 2 is true.

Remember the rules for a list of vectors being a basis: Must have two out of three of the following:

• Spanning
• Linearly independent
• List length the same as the dimension of the space

If you have two, the other one is always implied.

TL;DR 2 and 4 are true, 1 and 3 are false, but do try to understand the reasoning why

Answer 2

You are wrong that only $4)$ has to be true, (which it is since the dimension is the number of elements in a basis). There is a way to deduce that $B\subset A$: since the $\{b_1,b_2,b_3,b_4\}$ are dependent, it follows that $\rm{dim}B=3$. Thus $2)$ is also true.

Answer 3

Some very important facts about spanning sets and linearly independent sets (in finite dimensional spaces) are

1. A spanning set has at least as many elements as any linearly independent set.
2. Any spanning set contains a basis.
3. Any linearly independent set of vectors that has as many elements as the dimension is a basis.

You're given that $B$ has dimension at least $3$, because it contains the linearly independent set $\{a_1,a_2,a_3\}$. The dimension of $B$ can therefore be at most $4$, because it has a spanning set with four elements and we can apply fact 1.

However, you're also given that this spanning set is linearly dependent, so the basis it contains (fact 2) cannot be the whole set. It follows that the dimension of $B$ is at most $3$.

Hence the dimension of $B$ is exactly $3$. Apply fact 3 to $\{a_1,a_2,a_3\}$, which is a linearly independent set in $B$ to conclude that $A=B$.