Thus i and ii are satisfied and the transformation IS linear Matrix

# Thus i and ii are satisfied and the transformation is

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Thus (i) and (ii) are satisfied and the transformation IS linear. Matrix multiplication gives a linear transformation IMPORTANT REMARK: Given any m × n matrix A , the transformation from R n to R m given by L ( x ) = A x IS LINEAR. In section 4.2 we will see that all linear transformations can be written in the form L ( x ) = A x for some matrix A . Transformations on vector spaces of matrices EXAMPLES Determine which of the following transformations are linear. 1. T : R 6 × 4 R 4 × 6 defined by T ( A ) = A T . Solution: (i) T ( A + B ) = ( A + B ) T = A T + B T = T ( A ) + T ( B ). (ii) T ( αA ) = ( αA ) T = αA T = αT ( A ). Thus (i) and (ii) are satisfied and the transformation IS linear. 2. T : R 3 × 3 R 3 × 3 define by T ( A ) = A 2 . Solution: (i) T ( A + B ) = ( A + B ) 2 = ( A + B )( A + B ) = A 2 + AB + BA + B 2 . On the other hand: T ( A ) + T ( B ) = A 2 + B 2 . Thus (i) is NOT satisfied in general and the transformation IS NOT linear. Note that (ii) fails as well. In general, if the output involves squares, or products or lengths, x 2 1 or x 1 x 2 or || x || , then L is not linear.
4 4.1 3. Let S = bracketleftbigg 2 4 1 0 bracketrightbigg . We define T : R 2 × 2 R 2 × 2 as T ( A ) = SAS 1 . Solution: (i) T ( A + B ) = S ( A + B ) S 1 = S ( A ) S 1 + S ( B ) S 1 = T ( A ) + T ( B ) . (ii) T ( αA ) = S ( αA ) S 1 = αS ( A ) S 1 = αT ( A ) . Thus (i) and (ii) are satisfied and the operator IS linear. Transformations on vector spaces of functions EXAMPLES Determine which of the following transformations are linear. Assume the transformations are defined from C to C where C denotes the set of all functions whose derivatives of any order are defined and continuous on the real line . 1. L ( f ( t )) = f ( t ) + 3 f ( t ) Solution: (i) Let f and g be two functions in C L ( f + g ) = ( f + g ) + 3( f + g ) = ( f + 3 f ) + ( g + 3 g ) = L ( f ) + L ( g ). (ii) L ( αf ) = ( αf ) + 3( αf ) = αf + α 3 f = α ( f + 3 f ) = αL ( f ) Thus (i) and (ii) are satisfied and the operator IS linear. 2. L ( f ( t )) = 3( f ( t )) 2 + 9 f ( t ) Solution: (i) L ( f + g ) = 3( f + g ) 2 + 9( f + g ) = 3( f 2 + 2 fg + g 2 ) + 9( f + g ). On the other hand: L ( f ) + L ( g ) = (3 f 2 + 9 f ) + (3 g 2 + 9 g ). Thus (i) is obviously not satisfied and the operator IS NOT linear. Note that (ii) is also not satisfied. 3. L ( f ( t )) = f (5) Solution: (i) L ( f + g ) = ( f + g )(5) = f (5) + g (5) = L ( f ) + L ( g ) (ii) L ( αf ) = ( αf )(5) = αf (5) = αL ( f ) . Thus (i) and (ii) are satisfied and the operator IS linear. The Image and Kernel Definition Let L : V W be a linear transformation, then The KERNEL of L , denoted ker ( L ), is defined by ker ( L ) = { v V | L ( v ) = 0 w } = { set of input vectors that are mapped into 0 } The RANGE of L (or IMAGE of L ) is defined by { w W | w = L ( v ) for some v V } = set of all output vectors REMARK: ker ( L ) is a subspace of V .
4.1 5 The Range of L is a subspace of W . EXAMPLE: Let L be the linear operator in R 2 defined by L ( x ) = (2 x 1 + 4 x 2 , 6 x 1 12 x 2 ) T . Find a basis for the Kernel and Image of L . Solution. Ker ( L ) = { ( x 1 , x 2 ) T | L ( x 1 , x 2 ) = 0 } = braceleftbigg ( x 1 , x 2 ) T vextendsingle vextendsingle vextendsingle vextendsingle 2 x 1 + 4 x 2 = 0 6 x 1 12 x 2 = 0 bracerightbigg .
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