Sylvester's rank inequality states: For matrices A∈Rm×nA \in \mathbb{R}^{m \times n}A∈Rm×n and B∈Rn×pB \in \mathbb{R}^{n \times p}B∈Rn×p,rank(A)+rank(B)−n≤rank(AB)≤min(rank(A),rank(B))\text{rank}(A) + \text{rank}(B) - n \leq \text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B))rank(A)+rank(B)−n≤rank(AB)≤min(rank(A),rank(B))For A=(100100)A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}A=100010 (rank 2, dimensions 3×23 \times 23×2) and B=(100000)B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}B=(100000) (rank 1, dimensions 2×32 \times 32×3), what must rank(AB)\text{rank}(AB)rank(AB) equal?
Can be 0, 1, or 2
Must equal 1
Must equal 2
Must be at least 3