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**Unformatted text preview: **4. 8. 5.3 Math 27‘] 19W? $8, I6,2'-|, 2%. 35.. 3?,39H0, 42 Not orthogonal, the ﬁrst and third column vectors fail to be perpendicular to each other. A + B will not necessarily be orthogonal, because the columns may not be unit vectors.
For example, if A = B = n, then A + B = 2L“, which is not orthogonal. 16. A + B is symmetric, since (A + B)T : AT + BT = A + B.- 24. Not necessarily symmetric. (ATBA)T = AT (ATB)T = ATBTA. 28. Write L(:E) = A55; by Deﬁnition 5.3.1, A is an orthogonal n x 71 matrix, so that ATA : In, 35. by Fact 5.3.7. Now L(17)-L(1Ei) = (Am-(A13) = (At—OTA'LD' =17TATA1IJ' = “JFIH'LU 2 v u? =—
17 - If}, as claimed. Note that we have used Facts 5.3.6 and 5.3.93. Let us ﬁrst think about the inverse L = T—1 of T. 2
. _, _, _, _, _. _' _ 5
Write Lt”) = A113 = [111 112 v3 ] :8. It 13 required that L(é'3) : {1‘3 = g
.1.
3
Furthermore, tilegvectors 111, v2, '03 must form an orthonormal basis of R3. By inspection,
3
we ﬁnd 171 = %
2
3
1
‘5 —2 —1 2
Then 172 2 171x63 = g does thejob. In summary, we have LU?) : § 1 2 2 5.".
_§ 2 —2 1 Since the matrix of L is orthogonal, the matrix of T = L” is the transpose of the matrix
of L: H2 1 2
T(;E)=% —1 2 —2 5;“
2 2 1 There are many other answers (since there are many choices for the vector 171 above). 2 —3 3 2
37. No, since the vectors 3 and 2 are orthogonal, whereas 0 and —3 are not
0 0 2 0
(see Fact 5.3.2). 39. By Fact 5.3.10, the matrix of the projection is WIT; the ijth entry of this matrix is ui'uj. 0.5 -—0.1
40 An ortho l " ' ” — 0'5 " 0'7 '
. norina basis of W IS ul — 0 5 ,U2 = _0 7 (see Exercrse 5.2.9). 0.5 0.1 By Fact 5.3.10, the matrix of the projection onto W is QQT, where Q = [1'51 172 26 18 32 24 1s 74 —24 32
T__1_
QQ ‘100 32 -24 74 18 24 32 18 26 42. a. Suppose we are projecting onto a subspace W of 13.“. Since Arif is in W already, the
orthogonal projection of Ari? onto W is just Ai" itself: A(A:E) : Af, or A2 " = Airf. Since this equation holds for all if, we have A2 = A. b. A = QQT, for some matrix Q with orthonormal columns 111, . . . , 11m. Note that QTQ =
1..., since the ijth entry of QTQ is n;- - 113-. Then A2 = QQTQQT = Q(QTQ)QT =
QImQT = QQT = A- ...

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