1.

Consider the linear transformation T : R4 → R4 given by T(x, y, z, u) = (x, y, 0, 0) ∀ (x, y, z, u) ∈ R4. Then which one of the following is correct?1. Rank of T > Nullity of T2. Rank of T < Nullity of T3. Rank of T = Nullity of T = 34. Rank of T = Nullity of T = 2

Answer» Correct Answer - Option 4 : Rank of T = Nullity of T = 2

Concept:

Let V and W be a vector space over F and T: V → W is a linear transformation then null space of T is given by Ker T = {v ∈ V | T(v) = 0 where 0 ∈ W}

So, the nullity of T is the dimension of the null space of T.

Range space of T is given by R(T) = {w ∈ W| T(v) = w fo v ∈ V} and Rank of T is the dimension of R(T)

Rank - Nullity Theorem:

Let T be a linear transformation from V to W i.e T: V → W and V is a finite-dimensional vector space then Rank (T) + Nullity (T) = dim V

Analysis:

Given:

T : R4 → R4

T(x, y, z, u) = (x, y, 0, 0)

Let us first find the dimension of null space (nullty)

Null space of T = kernel of T (ker T)

Ker T = { v ∈ V | T(v) = 0}

Ker T = (0, 0, 0, 0) = (x, y, 0, 0)

x = 0, y = 0

The dimension of Ker T = dimension of basis of Ker T = 2

∴ Nullity = 2

dim (V) = 4

∴ 4 = Nullity + Rank (T)

Rank (T)  = 4 – 2 = 2


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