Let $\left \{u_1, u_2, u_3, v_1, v_2, v_3 \right \}$ be vectors in $\mathbb{R}^4$. Let $U$ be the span of $\left \{u_1, u_2, u_3 \right \}$

Question

Let $\left \{u_1, u_2, u_3, v_1, v_2, v_3 \right \}$ be vectors in $\mathbb{R}^4$. Let $U$ be the span of $\left \{u_1, u_2, u_3 \right \}$ and let $V$ be the span of $\left \{v_1, v_2, v_3 \right \}$

Consider the following statements:

• (I) If the dimension of $U\cap V$ is $2$ and the dimension of $U$ is $3$, then $\left \{v_1, v_2, v_3 \right \}$ is linearly dependent.

• (II) If $U+V = \left \{u+v\space|\space u \in U,v \in V \right \} = \mathbb{R}^4$ , then either $\left \{u_1, u_2, u_3 \right \}$ is linearly independent or $\left \{v_1, v_2, v_3 \right \}$ is linearly independent.

Which of the above statements is/are true?

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Facts that I think may be useful for this question:

1. $\dim(A)= n\text{(elements in the basis of A)}$ and when in $\mathbb{R}^4$ we know that $\max(\dim) = 4$

2. $\dim(A + B) = \dim(A) + \dim(B) - \dim(A\cap B)$

3. For a given set of vectors $\left \{\vec{a}, \vec{b}, \vec{c} \right \}$ if any vector can be expressed as a linear combination of other two, then we say that the given set is linearly dependent.

4. In general, for vectors $v_1,v_2, ..., v_n \in V$ if there exists scalars $a_1,...,a_n$ which are not all $0$ such that $a_1v_1 + a_2v_2 + \cdots + a_nv_n = \vec{0}$ then the given vectors are said to be linearly dependent.

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After researching, I improved upon my thought process. So for (I) using fact (2) $$\dim(U+V) = 3 + \dim(V) - 2$$ also, $\dim(U+V)\leq 4$ from fact(1) so $$1+\dim(V)\leq4 \implies \dim(V) \leq 3$$

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Questions:

1. How do I improve upon this result to see the validity of statement (I)
2. I have no clue yet how to approach (II). Would help if you can guide me in the correct direction.
3. Is there an intuitive way we can solve this problem? (atleast partially)

Context:

This question was asked in GATE 2024

ST_Q14

Answer 1

You can find counterexamples to both statements using sets instead of vector spaces.

(I) Find subsets $X=\{x_1,x_2,x_3\},Y=\{y_1,y_2,y_3\}$ of $\{1,2,3,4\}$ such that $|X\cap Y|=2$ and $|X|=3$ and yet $|Y|=3$.

(II) Find subsets $X=\{x_1,x_2,x_3\},Y=\{y_1,y_2,y_3\}$ of $\{1,2,3,4\}$ such that $X\cup Y=\{1,2,3,4\}$ but both $|X|<3$ and $|Y|<3$.

Just use the corresponding basis vectors to define $U$ and $V$ from $X$ and $Y$. This is because the span of an intersection or union is the intersection or sum of spans. And cardinality (of distinct elements) corresponds to dimension.

You can do this for other similar problems as well, turning counterexamples for sets into ones for vector spaces. (However, if there is no counterexample for sets, there may very well still be one for vector spaces. For example, the inclusion-exclusion principle for two sets has a true analogue for vector spaces, but the analogue of the principle for three sets is false for vector spaces.)

Answer 2

Clearly, dim $U\le 3$, dim $V\le 3$.

1. As dim $U=3$. Using the result
   max(dim $U$, dim $V$)$\le$ dim $(U+V)\le$ dim$(\mathbb R^4)$
   $\implies 3\le$ dim $(U+V)\le4 $
   so that,
   dim$ (U\cap V)=$dim $U+$dim $V-$dim $(U+V)$...........(*)
   $\implies$ dim $V=2$ or $3$
   so, the given statement is NOT always true if dim $V=3$.

2. $U+V=\mathbb R^4\implies $dim$(U+V)=4 $ . In this case (*) gives you
   dim $(U\cap V)=$dim $U+$dim $V-4$.............(#)
   Now (#) holds even when dim $U$=dim$V=2$ which suggests that this statement is NOT always true.