[Math] Ranks of Composition of Linear Transformations vs. Mapped Vector Space

linear algebralinear-transformations

I'm having trouble with this proof:

Let $\mathbb{U, V, W}$ be finite dimensional vector spaces and let $L: \mathbb{V} \to \mathbb{U}$ and $M : \mathbb{U} \to \mathbb {W}$ be linear mappings

a) Prove that rank$(M \circ L) \leq $ rank$(M)$.

b) Prove that rank$(M \circ L) \leq $ rank$(L)$.

I've attempted a), I've gone as far as showing

rank$(M) \leq $ dim$(\mathbb{U})$ and rank$(M \circ L)$ $\leq $ dim$(\mathbb{V})$

with the Rank-Nullity theorem, and but that obviously doesn't get me very far. I'm not sure how else to approach this. Haven't attempted b), as I'd imagine it extends off of a).

Best Answer

Hints:

Consider the linear maps $f$ and $g$ associated to the matrices $L$ and $M$ respectively. $\DeclareMathOperator{\im}{Im}\DeclareMathOperator{\rk}{rank}$ We know $ML$ is associated to the composition $g\circ f$.

We know $\;\rk L=\dim\im f$, $\;\rk M=\dim\im g$, $\;\rk ML=\dim\im (g\circ f)$.

  • How do the subspaces $\im (g\circ f)$ and $\im g$ compare?
  • How do $\dim\im f$ and $\dim V$ compare? And $\dim\im(g\circ f)$ and $\dim V$?
Related Question