# Linear Algebra. Maps!!

• May 9th 2008, 09:58 PM
flawless
Linear Algebra. Maps!!
Let T and T* be linear mappings on R^n defined by T(x)=Ax, and T*(x)=(A^T)x (A^T = A transpose). Show that kerT=(range of T*)perpendicular (ie, not the range of T*, but the set of vectors which are perpendicular to the range of T*. THANKS TO ANYONE WHO CAN HELP ME OUT WITH THIS EVEN THE SLIGHTEST!!
• May 10th 2008, 02:13 AM
Opalg
Quote:

Originally Posted by flawless
Let T and T* be linear mappings on R^n defined by T(x)=Ax, and T*(x)=(A^T)x (A^T = A transpose). Show that kerT=(range of T*)perpendicular (ie, not the range of T*, but the set of vectors which are perpendicular to the range of T*. THANKS TO ANYONE WHO CAN HELP ME OUT WITH THIS EVEN THE SLIGHTEST!!

The key fact is that $\langle Tx,y\rangle = \langle x, T^*y\rangle$ for all x,y∈R^n, where the angled brackets denote the inner product. In matrix notation, the same equation can be written $y^{\textsc t}Ax = x^{\textsc t}A^{\textsc t}y$.

In that equation, it's kind of obvious that Tx=0 if and only if x is perpendicular to everything in the range of T*.