# We can verify that the columns of t a a given the

• Notes
• 64
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 38 - 41 out of 64 pages.

We can verify that the columns of [ T ] A A given the image of { v 1 , v 2 , v 3 } , written as linear combinations of themselves. For instance, T ( v 2 ) = T (2 , 1 , 0) = (5 , 3 , - 1) = 2 v 1 + 4 v 2 + ( - 1) v 3 . The coefficients 2, 4, and - 1 are indeed the second column of [ T ] A A . Observe, that we now have many different ways of writing down a matrix rep- resenting the same linear transformation, but they are all related by the change of basis matrix. Thus it makes sense to have the following definition. Definition 11. Let A, B M n × n . We say that B is similar to A if there is an invertible matrix Q M n × n such that B = Q - 1 AQ . This definition says that two matrices are similar precisely when they represent the same linear transformation written with respect to two different bases. 5. Determinants Area and mappings from the plane to itself: Recall that in Section 2 we found a linear mapping to take the unit square S = { 0 x 1 , 0 y 1 } to any parallelogram P with one corner at the origin. We can write the parallelogram P as P = { xv + yw : 0 x 1 , 0 y 1 } , where v and w are the two vectors which form the edges of P starting at the origin (0 , 0). Then we can write the linear transformation as T x y = xv + yw, [ T ] = v 1 w 1 v 2 w 2 , where v = ( v 1 , v 2 ) and w = ( w 1 , w 2 ) in components. Notice that the mapping T is invertible precisely when it does not collapse S down to a line segment (or a point), which happens precisely when the area of the parallelogram P is non-zero. You might recall that in MAM1000 you defined an object called the determinant, written det([ T ]) = v 1 w 2 - w 1 v 2 , and were told det([ T ]) 6 = 0 T invertible Area( P ) 6 = 0 . We’ll see next that det([ T ]) is the area of P , up to a sign. 38
This is easiest to see with the shear map we examined in the last set of notes. Start with the sheer map T whose matrix representation is [ T ] = 1 b 0 1 . (In the earlier set of notes we wrote the entry in the upper right corner of [ T ] as a , but it will turn out to be convenient to call it b for our later discussion.) In this case, T maps the unit square S = { 0 x 1 , 0 y 1 } to the parallelogram P spanned by the two vectors v = 1 0 and w = b 1 ; in other words, T ( S ) = P = { xv + yw : 0 x 1 , 0 y 1 } = x 1 0 + y b 1 : 0 x 1 , 0 y 1 . We reproduce a picture here: 1 1 Area = 1 - (1 , 0) (1 , 1) Area = 1 We already know that the unit square S has area 1, but let’s see that P also has area 1. The area of a parallelogram is equal to its base times its height, and the height and base of P are both 1, so the area of P is 1 · 1 = 1. On the other hand, det([ T ]) = det 1 b 0 1 = 1 · 1 - 0 · b = 1 = Area( P ) . Now we can rescale the sheer T by a in the horizontal direction and by d and vertical direction, to have something more general. This time we have [ T ] = a b 0 d , and T ( S ) = P = { xv + yw : 0 x 1 , 0 y 1 } = x a 0 + y b d : 0 x 1 , 0 y 1 , and the picture looks like 39
1 1 Area = 1 - ( a, 0) ( b, d ) Area = ad (In this particular picture a = 1 / 2 and d = 2, but this choice of scaling factors is not important.) This time the height of the parallelogram P is d while its base is a , so Area( P ) = base · height = ad . Again, we have | det([ T ]) | = det a b 0 d = | a