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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 nonzero 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 )) = ....
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra

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