Linear transformation surjective and/ or injective?

Question

I am new to linear algebra, and would like some feedback regarding the following answer to a textbook question:

Let $T:M_{2\times 3}(\mathbb R) \to M_{3\times 3}(\mathbb R)$ be a linear transformation. Answer the following two questions:

1. Is it possible for T to be surjective?
2. Is it possible for T to be injective?

My answer is the following:

1. Yes, because it is possible for each element in $M_{3\times 3}(\mathbb R)$ to be the result of linear transformation $T$ applied to $M_{2\times 3}(\mathbb R)$.

2. No, because there are more rows in $M_{3\times 3}(\mathbf R)$ than in $M_{2\times 3}(\mathbf R)$, and therefore the elements in $M_{3\times 3}(\mathbb R)$ cannot be the exclusive result of the elements in $M_{2\times3}(\mathbb R)$.

I have no idea whether I am on the right track?

Thank you!

Answer 1

Hint: For 1. use the rank-nullity theorem for $T$.

For 2. consider the embedding of $M_{2,3}(K)$ into $M_{3,3}(K)$.

Answer 2

Let $A:U\to V$ be a linear operator between finite-dimensional vector spaces $U$ and $V$.

(1) $\text{rank}(A)\leq\text{dim}(V)$.

(2) $\text{rank}(A)+\text{nullity}(A)=\text{dim}(U)$.

(3) $A$ is injective iff $\text{nullity}(A)=0$.

(4) $A$ is surjective iff $\text{rank}(A)=\text{dim}(V)$

(5) The dimension of $M_{m\times n}(\mathbb{R})$ is $mn$.

That is more than enough to answer both questions.