rank of f^2 = rank of f

**dwsmith** · September 26th 2011, 04:23 PM

{Let X be a finite dimensional vector space over F, and $f:X\to X$ a linear transformation. If the rank of $f^2=f\circ f$ is equal to the rank of f, prove that $\text{ker} f \cap \text{Im}f=\{0\}$

Since $\text{rank}f\circ f=\text{rank} f\Rightarrow \text{Im}f\circ f=\text{Im} f$ .

???????

**Drexel28** · September 26th 2011, 07:01 PM

Originally Posted by dwsmith

{Let X be a finite dimensional vector space over F, and $f:X\to X$ a linear transformation. If the rank of $f^2=f\circ f$ is equal to the rank of f, prove that $\text{ker} f \cap \text{Im}f=\{0\}$

Since $\text{rank}f\circ f=\text{rank} f\Rightarrow \text{Im}f\circ f=\text{Im} f$ .

???????

Well, I think a direct method may perhaps be a better choice. Suppose that $\text{span}(f(x))\in\ker f\cap\text{im }f$ for some non-zero [/tex]f(x)[/tex]. You can see then that $f(f(x))=0$ now try using this and the ideas behind why $\text{rank}(f^2)\leqslant \text{rank}(f)$ to conclude that $\text{rank}(f^2)\leqslant\text{rank}(f)-1$ .

**dwsmith** · September 26th 2011, 07:08 PM

Originally Posted by Drexel28

Well, I think a direct method may perhaps be a better choice. Suppose that $\text{span}(f(x))\in\ker f\cap\text{im }f$ for some non-zero [/tex]f(x)[/tex]. You can see then that $f(f(x))=0$ now try using this and the ideas behind why $\text{rank}(f^2)\leqslant \text{rank}(f)$ to conclude that $\text{rank}(f^2)\leqslant\text{rank}(f)-1$ .

Why the span(f(x))??

**Drexel28** · September 26th 2011, 07:09 PM

Originally Posted by dwsmith

Why the span(f(x))??

You want to show that $\text{im }f^2$ misses an entire dimension one subspace of [tex]\text{im }f

**dwsmith** · September 26th 2011, 07:17 PM

Originally Posted by Drexel28

You want to show that $\text{im }f^2$ misses an entire dimension one subspace of [tex]\text{im }f

I am still lost. Did you look at the pdf of the notes yesterday? Nothing is in there to help us (the students understand).

**Deveno** · September 27th 2011, 12:35 AM

apply the rank-nullity theorem to the 2nd f in fof, to get:

dim(f(X)) = dim(f(f(X)) + dim(ker(f)∩f(X)).

since dim(f(X)) = dim(f(f(X)), dim(ker(f)∩f(X)) = 0.

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