AMU 2026 — Mathematics PYQ
AMU | Mathematics | 2026Let be a linear transformation from a -dimensional vector space into a -dimensional vector space . Then,
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Let be a linear transformation from a -dimensional vector space into a -dimensional vector space . Then,
For any linear transformation , the fundamental relationship between the dimensions of the domain, the kernel (null space), and the image (range space) is given by the Rank-Nullity Theorem:
From the problem statement, we are given:
Dimension of domain :
Dimension of codomain :
Substituting these values into the theorem gives:
A linear transformation is injective if and only if its kernel contains only the zero vector, which means:
Let's test if this is possible using our Rank-Nullity equation:
However, the image space is a subspace of the codomain . This means the rank cannot exceed the dimension of :
Since cannot be , can never be . Therefore, cannot be injective.
A linear transformation is surjective if its image fills the entire codomain space, which means:
Let's test if this is possible using our Rank-Nullity equation:
Since a nullity of is completely valid and possible (), we can easily construct a transformation where . Therefore, can be surjective.
cannot be injective because the domain is larger than the codomain (\text{dim}(V) > \text{dim}(W)).
can be surjective.
This directly matches option (c): Can be surjective but cannot be injective.