Let A be an n x n matrix with the property that the intersection of its column space and its null space is the 0 vector. That is: nullA(intersection)colA={0}. Show that A and A^2 have the same rank........(Headbang)

Printable View

- Apr 20th 2008, 02:48 AMflawlessThis is tricky..Linear algebra again
Let A be an n x n matrix with the property that the intersection of its column space and its null space is the 0 vector. That is: nullA(intersection)colA={0}. Show that A and A^2 have the same rank........(Headbang)

- Apr 20th 2008, 04:25 AMflyingsquirrel
Hi

First, you can show that $\displaystyle \ker A \oplus \mathrm{im}\, A = E$, which might be useful. (rank-nullity theorem)

Then, showing that $\displaystyle A$ and $\displaystyle A^2$ share the same rank is equivalent to proving that $\displaystyle \dim \ker A = \dim \ker A^2$ (rank-nullity theorem again) Here, we are in a special case : as $\displaystyle \ker A\subset \ker A^2$, (find a proof :D) you can try to show that $\displaystyle \ker A\supset \ker A^2$ which will give you $\displaystyle \ker A = \ker A^2$ and the equality of the dimensions.

For $\displaystyle \ker A\supset \ker A^2$, you might take $\displaystyle X\in E$ which can be decomposed as $\displaystyle X=X_k+X_i$ with $\displaystyle X_k\in\ker A$ and $\displaystyle X_I\in \mathrm{im}\,A$ (because $\displaystyle \ker A \oplus \mathrm{im}\, A = E$) and develop $\displaystyle A^2X=\ldots$