Page 358 - 35Linear Algebra

358 Sample Final Exam

Hence, to compute L(v) we use linearity of L

2
1
3
1
3
2
L(v) = L(b v 1 + b v 2 + b v 3 ) = b L(v 1 ) + b L(v 2 ) + b L(v 3 )
1
 
b

= L(v 1 ) L(v 2 ) L(v 3 )  2  .
b
b 3
(f) From the notation of the previous part, we see that we can list linear
3
transformations L : B → B by writing out all possible bit-valued row
vectors

0 0 0 ,

1 0 0 ,

0 1 0 ,

0 0 1 ,

1 1 0 ,

1 0 1 ,

0 1 1 ,

1 1 1 .
3
3
There are 2 = 8 different linear transformations L : B → B, exactly
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the same as the number of elements in B .
(g) Yes, essentially just because L 1 and L 2 are linear transformations. In
3
detail for any bits (a, b) and vectors (u, v) in B it is easy to check the
linearity property for (αL 1 + βL 2 )
(αL 1 + βL 2 )(au + bv) = αL 1 (au + bv) + βL 2 (au + bv)

= αaL 1 (u) + αbL 1 (v) + βaL 1 (u) + βbL 1 (v)

= a(αL 1 (u) + βL 2 (v)) + b(αL 1 (u) + βL 2 (v))
= a(αL 1 + βL 2 )(u) + b(αL 1 + βL 2 )(v) .
Here the first line used the definition of (αL 1 + βL 2 ), the second line
depended on the linearity of L 1 and L 2 , the third line was just algebra
and the fourth used the definition of (αL 1 + βL 2 ) again.
(h) Yes. The easiest way to see this is the identification above of these
maps with bit-valued column vectors. In that notation, a basis is
n o

1 0 0 , 0 1 0 , 0 0 1 .

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