Answer to Question #214313 in Linear Algebra for person1

Question

Let V and W be two vector spaces over the field F and T, T1, T2 be linear transformations from V to W. Prove the following:

(a) rank(αT) = rank(T), for all α ∈ F and α 6= 0.

(b) |rank(T1) − rank(T2)| ≤ rank(T1 + T2) ≤ rank(T1) + rank(T2

Answer 1

• For $\alpha\neq 0$ we have $\text{Range} (T)=\text{Range}(\alpha T)$, as for any $y\in \text{Range}(T)$ there is $x\in V$ such that $Tx=y$. Now as $F$ is a field, $\alpha^{-1}$ exists and thus $\alpha T(\alpha^{-1}x)=\alpha \alpha^{-1}Tx=y$ by linearity of $T$ and thus $y\in \text{Range}(\alpha T)$ and $\text{Range}(T)\subseteq \text{Range}(\alpha T)$ . We can apply the same argument to the other inclusion to conclude that $\text{Range}(T)=\text{Range}(\alpha T)$ and thus $rg(T)=rg (\alpha T)$.
• Let us first prove the right inequality. We have $(T_1+T_2)(x)=T_1(x)+T_2(x)$ and thus $\text{Range}(T_1+T_2)\subseteq \text{Range}(T_1)+\text{Range}(T_2)$, where plus on the right is the plus in the sense of vector subspaces of $W$. As $\dim(A+B)\leq \dim(A)+\dim(B)$ for $A,B$ vector spaces, we conclude that $rg(T_1+T_2)\leq \dim(\text{Range}(T_1)+\text{Range}(T_2))\leq rg(T_1)+rg(T_2)$.

Now to find the left inequality, we just need to use the inequality we just found by replacing $T_2$ by $-T_2$ and $T_1$ by $T_1+T_2$ : $rg(T_1+T_2-T_2)\leq rg(T_1+T_2)+rg(-T_2)$. Now as $-1\neq 0$ in $F$, we have $rg(-T_2)=rg(T_2)$ and therefore $rg(T_1)\leq rg(T_1+T_2)+rg(T_2)$ or we can rewrite it as $rg(T_1)-rg(T_2)\leq rg(T_1+T_2)$. Applying the same symmetric argument to $-T_1$ and $T_1+T_2$ we find that $rg(T_2)-rg(T_1)\leq rg(T_1+T_2)$. Combining the two inequalities we conclude that $|rg(T_1)-rg(T_2)|\leq rg(T_1+T_2)$.