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

Think of it purely in terms of subspace dimensions rather than quoting the rank-nullity theorem by name.

The transformation \(L^2\) sends \(\mathbb{R}^6\) into \(\mathbb{R}^6\). Any linear transformation \(T\) on a 6-dimensional space splits that space into two complementary pieces, its kernel and a part isomorphic to its image, so counting dimensions on both sides always gives \[6 = \dim(\ker T) + \dim(\operatorname{im} T)\]

Apply this with \(T = L^2\). We are told \(\dim(\ker(L^2)) = \operatorname{Nullity}(L^2) = 3\). Plugging in, \[6 = 3 + \dim(\operatorname{im}(L^2))\] \[\dim(\operatorname{im}(L^2)) = 3\]

Since \(\operatorname{Rank}(L^2)\) is by definition \(\dim(\operatorname{im}(L^2))\), \[\operatorname{Rank}(L^2) = 3\]

The value \(\operatorname{Rank}(L) = 4\) given in the problem is not actually needed for this computation once \(\operatorname{Nullity}(L^2)\) is known; it only serves as a consistency check, and indeed \(3 \le 4\) as required since composing with \(L\) again can only shrink or preserve the image.

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