Let V be the vector space of all 2 × 2 matrices over R. Consider the subspaces

\(W_1 = \left\{ \left( \begin{matrix} a & -a \\\ c & d \end{matrix}\right) ; a, c, d \in R\right\}\)

and 

\(W_2 = \left\{ \left( \begin{matrix} a & b \\\ -a & d \end{matrix}\right) ; a, b, d \in R\right\}\)

If m = dim(W₁ ∩ W₂) and n = dim(W₁ + W₂), then the pair (m, n) is

Concept:

The dimension of a vector space V is the cardinality (i.e. the number of vectors) of a basis.

Calculations:

Let V be the vector space of all 2 × 2 matrices over R and 

\(W_1 = \left\{ \left( \begin{matrix} a & -a \\\ c & d \end{matrix}\right) ; a, c, d \in R\right\}\) and \(W_2 = \left\{ \left( \begin{matrix} a & b \\\ -a & d \end{matrix}\right) ; a, b, d \in R\right\}\) are subspaces.

Then dim W₁ = 3 and dim W₂ = 3

∴ 0 ≤ dim W₁ ⋂ dim W₂ ≤ 3

Hence, dim (W₁ + W₂) = dim W₁ + dim W₂ - dim (W₁⋂ W₂)

⇒ dim (W1 + W2) = 3 + 3 - dim (W1⋂ W2)

⇒ dim (W1 + W2) = 6 - dim (W1⋂ W2)

Case 1:

When dim(W1⋂ W2) = 0

Then dim (W1 + W2) = 6 - 0 = 6

Hence, (m, n) = (0, 6).

Case 2:

When dim(W1⋂ W2) = 1

Then dim (W1 + W2) = 6 - 1 = 5

Hence, (m, n) = (1, 5).

Case 3:

When dim (W1⋂ W2) = 2

Then dim (W1 + W2) = 6 - 2 = 4

Hence, (m, n) = (2, 4).

Case 4:

When dim (W1⋂ W2) = 3

Then dim (W1 + W2) = 6 - 3 = 3

Hence, (m, n) = (3, 3).

Hence, the correct answer is option 2)