This is tricky..Linear algebra again

**flawless** · April 20th 2008, 02:48 AM

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........

**flyingsquirrel** · April 20th 2008, 04:25 AM

Hi

First, you can show that $\ker A \oplus \mathrm{im}\, A = E$ , which might be useful. (rank-nullity theorem)
Then, showing that $A$ and $A^2$ share the same rank is equivalent to proving that $\dim \ker A = \dim \ker A^2$ (rank-nullity theorem again) Here, we are in a special case : as $\ker A\subset \ker A^2$ , (find a proof

) you can try to show that $\ker A\supset \ker A^2$ which will give you $\ker A = \ker A^2$ and the equality of the dimensions.

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

Thread: This is tricky..Linear algebra again

LinkBack

Thread Tools

Display

This is tricky..Linear algebra again

Similar Math Help Forum Discussions

tricky linear algebra proof

tricky little first order linear DE

A tricky linear equation

Tricky-ish algebra problem

tricky math puzzle - algebra triangle

Search Tags