Rank f+g

**dwsmith** · September 26th 2011, 06:02 PM

Let $f,g:X\to Y$ be a linear transformation. Prove that $\text{rank}(f+g)\leq \text{rank} \ f+\text{rank} \ g$

Is this done by using that fact that $\text{dim}(X+Y)=\text{dim}X+\text{dim}Y-\text{dim}(X\cap Y)$ since $\text{rank}f\leq\text{dim}X\mbox{?}$

**Drexel28** · September 26th 2011, 07:05 PM

Originally Posted by dwsmith

Let $f,g:X\to Y$ be a linear transformation. Prove that $\text{rank}(f+g)\leq \text{rank} \ f+\text{rank} \ g$

Is this done by using that fact that $\text{dim}(X+Y)=\text{dim}X+\text{dim}Y-\text{dim}(X\cap Y)$ since $\text{rank}f\leq\text{dim}X\mbox{?}$

Note that if $y\in \text{im }f+g$ then $y=(f+g)(x)=f(x)+g(x)\in\text{im }f+\text{im }g$ and so $\text{im }(f+g)\subseteq\text{im }f+\text{im }g$ . Thus, $\text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g)$ .

**dwsmith** · September 26th 2011, 07:14 PM

Originally Posted by Drexel28

Note that if $y\in \text{im }f+g$ then $y=(f+g)(x)=f(x)+g(x)\in\text{im }f+\text{im }g$ and so $\text{im }(f+g)\subseteq\text{im }f+\text{im }g$ . Thus, $\text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g)$ .

Can you explain the top line in more detail?

Why can say $\text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g)$ ?

Go from looking at rank and then say it is less than the dim and back to rank?

**Drexel28** · September 26th 2011, 08:18 PM

Originally Posted by dwsmith

Can you explain the top line in more detail?

Why can say $\text{rank }(f+g)\leqslant \dim(\text{im }f+\text{im }g)\leqslant \text{rank}(f)+\text{rank}(g)$ ?

Go from looking at rank and then say it is less than the dim and back to rank?

Does it make more sense to recall that $\text{rank}(f)=\text{dim }\text{im }f$ ?

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