I have got another question concerning linear algebra. The excercise is:
Let $\DeclareMathOperator{Im}{Im}\DeclareMathOperator{rank}{rank}\DeclareMathOperator{nullity}{nullity}T_1:\mathbb{C}^3\rightarrow\
\mathbb{C}^2$ and $T_2:\mathbb{C}^2\rightarrow \mathbb{C}^3$. Given that $\rank(T_2T_1)=2$, p\
rove that $T_1$ is surjective and $T_2$ is injective.
This is how I tried to do it. Due to rank-nullity theorem:
$$\dim(\mathbb{C}^3)=\rank(T_1)+\nullity(T_1)$$
What is more:
$$\dim(\Im(T_1))=\rank(T_2T_1)+\nullity(T_2T_1)$$
As $\rank(T_2T_1)=2$, then $\dim(\Im(T_1))=2$. So that, as $\Im(T_1) \subseteq\mathbb{C}^2$ an\
d $\dim(\Im(T_1))=\dim(\mathbb{C}^2)$, then $\Im(T_1)=\mathbb{C}^2$ and so $T_1$ is surjective\
.
As $\rank(T_2T_1)=\dim(\Im(T_2T_1))=2$ and $\dim(\Im(T_1))=2$, then also $\Im(T_2T_1)=(\Im(T_1\
))$, so $T_2$ must be injective.
Does it make any sense? Thank you for your help.
Best Answer
The Rank-Nullity theorem implies that a linear map $T:\Bbb C^m\to\Bbb C^n$ is
Additionally, $T$ satisfies $$\rank(T)\leq\min\{m,n\}\tag{1}$$
Finally, recall that the rank of the composition $T_2\circ T_1$ is bounded above by $$ \rank(T_2\circ T_1)\leq\min\{\rank(T_2),\rank(T_1)\}\tag{2} $$ In our case we have \begin{align*} T_1:\Bbb C^3\to\Bbb C^2 && T_2:\Bbb C^2\to\Bbb C^3 \end{align*} with $\rank(T_2\circ T_1)=2$. Equation (1) gives \begin{align*} 0\leq\rank(T_1)\leq 2&&0\leq\rank(T_2)\leq 2 \tag{3} \end{align*} and equation (2) gives $$ 2\leq\min\{\rank(T_2),\rank(T_1)\}\tag{4} $$ But now (3) and (4) combine to imply $$ \rank(T_1)=\rank(T_2)=2 $$ Hence $T_1$ is surjective and $T_2$ is injective.