Question

Let \(L\) be a linear transformation defined on the vector space \(\mathbb{R}^6\) over the field \(\mathbb{R}\). If \(\operatorname{Rank}(L)=4\) and \(\operatorname{Nullity}(L^2)=3\), then \(\operatorname{Rank}(L^2)\) is ____ (\(L^2\) denotes the composition of \(L\) with itself).

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Apply rank-nullity directly to L^2 as a map on R^6: rank(L^2) = 6 - nullity(L^2).

Answer 1

The Correct Option is B

Step 1: Note that \(L^2 : \mathbb{R}^6 \to \mathbb{R}^6\) is itself a linear transformation on the same 6-dimensional space, so the rank-nullity theorem applies directly to it. \[\dim(\mathbb{R}^6) = \operatorname{rank}(L^2) + \operatorname{nullity}(L^2)\]

Step 2: Substitute the known value. We are given \(\operatorname{nullity}(L^2) = 3\), so \[6 = \operatorname{rank}(L^2) + 3\]

Step 3: Solve for the rank. \[\operatorname{rank}(L^2) = 6 - 3 = 3\]

Step 4: Check consistency with \(\operatorname{Rank}(L)=4\). It is a standard fact that \(\operatorname{rank}(L^2) \le \operatorname{rank}(L)\), since the image of \(L^2\) is contained in the image of \(L\). Here \(3 \le 4\), so the answer is consistent with the given data.

\[\boxed{\operatorname{Rank}(L^2) = 3}\]