Question

Let \( U, V \) and \( W \) be finite dimensional real vector spaces, \( T: U \to V \), \( S: V \to W \) and \( P: W \to U \) be linear transformations. If range \( (ST) = \) nullspace \( (P) \), nullspace \( (ST) = \) range \( (P) \) and rank \( (T) = \) rank \( (S) \), then which one of the following is TRUE?

• A. nullity of \( T = \) nullity of \( S \)
• B. dimension of \( U \neq \) dimension of \( W \)
• C. If dimension of \( V = 3 \), dimension of \( U = 4 \), then \( P \) is not identically zero
• D. If dimension of \( V = 4 \), dimension of \( U = 3 \) and \( T \) is one-one, then \( P \) is identically zero

Hint:

When dealing with linear transformations, use the rank-nullity theorem and the properties of ranges and nullspaces to relate the dimensions of vector spaces.

Answer 1

The Correct Option is C

Step 1: Establish key relationships

From the rank-nullity theorem:

• For $T: U \to V$: $\dim(U) = \text{rank}(T) + \text{nullity}(T)$
• For $S: V \to W$: $\dim(V) = \text{rank}(S) + \text{nullity}(S)$
• For $ST: U \to W$: $\dim(U) = \text{rank}(ST) + \text{nullity}(ST)$
• For $P: W \to U$: $\dim(W) = \text{rank}(P) + \text{nullity}(P)$

From the given conditions:

• $\text{rank}(ST) = \dim(\text{nullspace}(P)) = \text{nullity}(P)$
• $\text{nullity}(ST) = \dim(\text{range}(P)) = \text{rank}(P)$

Therefore: $$\dim(U) = \text{rank}(ST) + \text{nullity}(ST) = \text{nullity}(P) + \text{rank}(P) = \dim(W)$$

So $\dim(U) = \dim(W)$.

Option (A): nullity of $T$ = nullity of $S$

Given: $\text{rank}(T) = \text{rank}(S)$

From rank-nullity:

• $\text{nullity}(T) = \dim(U) - \text{rank}(T)$
• $\text{nullity}(S) = \dim(V) - \text{rank}(S) = \dim(V) - \text{rank}(T)$

For nullity of $T$ to equal nullity of $S$: $$\dim(U) - \text{rank}(T) = \dim(V) - \text{rank}(T)$$ $$\dim(U) = \dim(V)$$

This is not necessarily true from our given conditions. Option (A) is not always TRUE.

Option (B): dimension of $U \neq$ dimension of $W$

From Step 1, we proved $\dim(U) = \dim(W)$.

Option (B) is FALSE.

Option (C): If $\dim(V) = 3$, $\dim(U) = 4$, then $P$ is not identically zero

Given: $\dim(V) = 3$, $\dim(U) = 4$

From Step 1: $\dim(W) = \dim(U) = 4$

Since $\text{nullity}(ST) = \text{rank}(P)$ and $\dim(U) = 4$: $$\text{rank}(ST) + \text{rank}(P) = 4$$

Now, $\text{rank}(ST) \leq \min(\text{rank}(S), \text{rank}(T))$

Since $\text{rank}(T) \leq \dim(V) = 3$ and $\text{rank}(S) = \text{rank}(T) \leq 3$: $$\text{rank}(ST) \leq 3$$

Therefore: $$\text{rank}(P) = 4 - \text{rank}(ST) \geq 4 - 3 = 1$$

So $\text{rank}(P) \geq 1$, which means $P$ is not identically zero.

Option (C) is TRUE. 

Option (D): If $\dim(V) = 4$, $\dim(U) = 3$ and $T$ is one-one, then $P$ is identically zero

Given: $\dim(V) = 4$, $\dim(U) = 3$, $T$ is one-one (injective)

Since $T$ is one-one: $\text{nullity}(T) = 0$, so $\text{rank}(T) = \dim(U) = 3$

Given: $\text{rank}(S) = \text{rank}(T) = 3$

From Step 1: $\dim(W) = \dim(U) = 3$

Now: $\text{rank}(P) = \text{nullity}(ST) = \dim(U) - \text{rank}(ST) = 3 - \text{rank}(ST)$

Since $\text{rank}(ST) \leq \min(\text{rank}(S), \text{rank}(T)) = 3$ and $\text{rank}(ST) \leq \dim(W) = 3$:

We could have $\text{rank}(ST) = 3$, giving $\text{rank}(P) = 0$ (P identically zero), but we could also have $\text{rank}(ST) < 3$, giving $\text{rank}(P) > 0$ (P not identically zero).

Option (D) is not necessarily TRUE.

Answer: (C)