math110s-hw7sol - MATH 110: LINEAR ALGEBRA SPRING 2007/08...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 7 SOLUTIONS 1. Let A,B F n n . Define the function T : F n n F n n by T ( X ) = AXB for all X F n n . (a) Show that T End( F n n ). Solution. Let 1 , 2 F and X 1 ,X 2 F n n . Then by the distributive property of matrix multiplication, T ( 1 X 1 + 2 X 2 ) = A ( 1 X 1 + 2 X 2 ) B = 1 AX 1 B + 2 AX 2 B = 1 T ( X 1 ) + 2 T ( X 2 ) . Hence T is linear. (b) Show that T is invertible if and only if A and B are nonsingular matrices. Solution. Suppose A and B are nonsingular. Define the map S : F n n F n n by S ( X ) = A- 1 XB- 1 for all X F n n . By the associativity of matrix multiplication, we have S ( T ( X )) = A- 1 ( AXB ) B- 1 = ( A- 1 A ) X ( BB- 1 ) = IXI = X, T ( S ( X )) = A ( A- 1 XB- 1 ) B = ( AA- 1 ) X ( B- 1 B ) = IXI = X for all X F n n . So S T = I = T S and so T- 1 = S . Suppose T is invertible, then ker( T ) = { O } . We will prove by contradiction. Suppose A or B is singular. Without loss of generality, we may assume that A is singular (the argument for singular B is similarly). Then there exists a non-zero x nullsp( A ), ie. A x = but x 6 = . Define the matrix X F n n all of whose columns are x , ie. X = [ x ,..., x ] . Then X 6 = O but T ( X ) = AXB = A [ x ,..., x ] B = [ A x ,...,A x ] B = [ ,..., ] B = OB = O. So ker( T ) 6 = { O } and so T is not invertible. 2. Let V be a finite dimensional vector space and T End( V ). Let dim( V ) = n . Let v V be such that T n- 1 ( v ) 6 = and T n ( v ) = . (a) Show that the vectors v , T ( v ) , T 2 ( v ) ,..., T n- 1 ( v ) form a basis for V . Solution. Let ,..., n- 1 F be such that v + 1 T ( v ) + 2 T 2 ( v ) + + n- 1 T n- 1 ( v ) = . (2.1) Date : May 7, 2008 (Version 1.0). 1 Applying T n- 1 to both sides of (2.1), we obtain T n- 1 ( v + 1 T ( v ) + 2 T 2 ( v ) + + n- 1 T n- 1 ( v )) = T n- 1 ( ) , T n- 1 ( v ) + 1 T n ( v ) + 2 T n +1 ( v ) + + n- 1 T 2 n- 2 ( v )) = ....
View Full Document

Page1 / 5

math110s-hw7sol - MATH 110: LINEAR ALGEBRA SPRING 2007/08...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online