Rank and Nullity Questions and Answers

Answer: A

The nullity of \(T\) is 0.

The kernel consists of the vectors with \(T(x_{1}, x_{2}) = (0, 0, 0)\): the first and third coordinates force \(x_{1} = 0\) and \(x_{2} = 0\). So the kernel is \(\{0\}\) and the nullity is 0.

By rank-nullity, \(\dim F^{2} = 2 = \text{rank} + \text{nullity}\), and the images \(T(1, 0) = (1, 1, 0)\) and \(T(0, 1) = (0, 1, 1)\) are independent, so the rank is 2, confirming nullity 0. The other options confuse nullity with the dimension of the codomain (3), of the domain (2), or with the codimension of the image (1).

Answer: A

The minimal polynomial is x(x - 3).

The null space of T is a hyperplane in \(\mathbb{R}^{4}\), of dimension 3, so 0 is an eigenvalue with geometric multiplicity 3. Rank(T - 3I) = 3 means T - 3I has nullity 1, so 3 is an eigenvalue with geometric multiplicity 1.

The algebraic multiplicities are at least the geometric ones and add up to 4, so they are exactly 3 and 1. Each eigenvalue then has as many independent eigenvectors as its multiplicity, so T is diagonalizable with eigenvalues 0, 0, 0, 3, and its minimal polynomial has only simple factors: x(x - 3).

The options with higher powers would describe non-diagonalizable maps, which would need fewer eigenvectors than the null space supplies.

Answer: B

The rank of \(AA^{t}\) is 2.

For a real matrix, \(\text{rank}(AA^{t}) = \text{rank}(A)\), because \(AA^{t}x = 0\) gives \(x^{t}AA^{t}x = |A^{t}x|^{2} = 0\), so \(AA^{t}\) and \(A^{t}\) have the same null space.

So find the rank of A. The third row \((0, -6, 4)\) is \(-2\) times the second row \((0, 3, -2)\), so the rows are dependent and the rank is at most 2. The first two rows are clearly independent (only the first has a nonzero first entry). Hence rank A = 2, and \(AA^{t}\) has rank 2 as well.

Rank 3 would need \(\det A \ne 0\), but \(\det A = 1 \cdot (3 \cdot 4 - (-2)(-6)) = 0\). Rank 1 or 0 would need the first two rows to be dependent, which they are not.

Answer: B

The rank of PA is 2.

A \(3 \times 3\) matrix of rank three is invertible, and multiplying by an invertible matrix does not change rank. PA and A have the same null space, since \(PAx = 0\) gives \(Ax = P^{-1}0 = 0\). So rank PA = rank A.

For A, the rows are in arithmetic progression: row 3 minus row 2 equals row 2 minus row 1, both \((3, 3, 3)\). So row 1 - 2 row 2 + row 3 = 0 and \(\det A = 0\), while the first two rows are independent. Hence rank A = 2 and rank PA = 2.

3 would need A invertible. 1 would need all rows proportional. 'Depends upon matrix P' is the trap: it would be right if P could be singular, but rank three rules that out.