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DSC 155 & MATH 182: “WINTER” 2020HOMEWORK 3
Available Friday, February 28 Due Friday, March 6
Turn in the homework on Gradescope before midnight. Late homework will not be ac-cepted.
1. Let P be an orthogonal projection matrix, with column space V . Show that each eigen-value of P is either 0 or 1. What can you say about eigenvectors of of P , in relation toV ?
2. For 1 i, j N , let Eij denote the N ⇥N matrix with a 1 in the (i, j) entry and zeroeselsewhere: [Eij]ab = �ia�jb. The {Eij}1i,jN are called the matrix units.
(a) Show that EijEk` = �jkEi`.(b) Show that, for any matrix X 2 MN⇥N , Tr(XEij) = [X]ji.(c) Use the matrix units to prove the following: if X is an N ⇥ N matrix with the
property thatTr(XAB) = Tr(XBA) for all A,B 2 MN⇥N
then X = �IN for some constant � 2 R. [Hint: apply the identity with A andB various choices of matrix units, and use parts (a) and (b) to conclude that theoff-diagonal entries of X are 0, while the diagonal entries are all equal.]
3. Let C+ = {z = x + iy : y > 0} be the upper half-plane, and consider the functionG : C+ ! C defined by
G(z) =�z +
pz2 � 4
2.
The function G is the Stieltjes transform of a probability density: compute this density.[Hint: Don’t write things out in terms of z = x + iy; just assume the complex squareroot works the way you would hope.]
Problems 4 and 5 are about the Marchenko–Pastur density f%, which is defined as follows.For % > 0, set
a(%) = (1�p%)2, b(%) = (1 +
p%)2.
Then
f%(x) =
p(b(%)� x)(x� a(%))
2⇡x [a(%),b(%)](x).
You may take it for granted that f% is a probability density when % � 1.
God
4. Show that, for any % > 0, f1/%(x) = f%(%x). Use this to show thatZ
Rf%(x) dx = % for % 2 (0, 1).
5. Let Y be a continuous random variable whose density is f1. Show thatpY has a
quarter-circular density:
fpY (x) =1
⇡
p4� x2
[0,2](x).
2
1.P is an orthogonal projection , i .
PEP P .- PTlU
If v. an eigenvector with eigenvalue X,
by a Pv.
= P2 , =P ( Py)= Play ) = XPy = X - by
-
- II.
Since v. is an eigenvector,
it is te , so a ;D .
1=0 or 1=1.
⇐ xx-D -o
ki : P, ⇒ is true for all ee f- GKP)Py -- e ⇒ ye dull (P) = 611 PT )
t
(Fundamental Theoremof Linear Algebra)
Smo Papi
⇒ Eigenspace of e is V'
Eigenspace of l is V
2. la) tfsij Ere lab = § lEijlaclnelcb'- §Siasjcfkcsebw
the only nonzero terms in the| sum have ⇐j and c -k
'
'
-
e Sia . Sjk . Seb= Sin Sia Seb = Sjr ( Eielab
i. Eij Ere = Sik Fie
.
(b) Tr ( XEij) - § EXEijlaa - ↳ ④abtEijlba=? Adab fibs ,a= ( Xlji .
⑥ Suppose X satisfies TrlXABI 'TRIXBA)for all matrices A
,B
.
That means,for all
"DHTrix Eij End = TRIX Ene Iii)
Tr Sheene) try"
- Sei Eng'd by ca)
Sju CX Ili Shi je by (b) .
Thus : Sin CX Ili = Sei Cxljn for all i. joke .
• Look at any j, k with jtk .
Take f- it.Thus
Sfi CX h ,z Si , cxlja i . lxljn --o
'
f j#k .' I
0 I
. Look at je k , and i=L .
i. Sjj LX Iii - Sii lxljji i[
i. all diagonal entries are equal , to somenumber X .
Thus x --= XI
.
3. Typeingueton : Gus . - z- ray
-
Z
i. Glxtig = -H - tvcatiyfhq
Se - ImGlxtiy) - § it Im ( IuetiyfT)Take lysine J toi. lying -¥ImGlxtiD=¥f¥nImHcxtiy
= Im flyingvcatiy= Im WHT)
If 10432 , x'- 4 >od
- : Bett EIR se Im 1=0.
If held 2 , ah- 4 soi.VH - f-147 =i4#
so Ima f- 14¥
i. fog > FYI -*InGlatt'D = ↳htt Dixie .
4. f. glass Vlk-ayDhIz
ACalf, bigot")
-i.fypuet-llx-azl.IM#Dcacyp.bNpllHf:::::i::i:ii: :c:::: : ::&
.
i. fight. VK.NO/ffofebkftf-T-Bcaipg,byxptD=rgt-tgVfx-alp7bIp-pT2€ Deals#
blpilpd.IE#m.o.uore" is:S :&,= fglpu) .
Now,if fat , Yf > I
,
and so
g. y .
fgkt-fyagfxt-f.gl#yn-:ffpiasdx=ffygh4uf.dyx...-asf?jyduW-fdn .- f - ffygluidn
5.
For f- I , all I - H -RT > ebut = ( HRT = 4 .
i. f. ex ) -- Vxn4Y Ice.ua)
= '%I4 Dee,41"" -
Now, if Y has density fy .- f, , then Go with
probability 1, and for 630,
PITY St) ' PH stye fluidic .
i. fry ftp.dqlplrystlzddg.tjflxldx= filly . adfltg
turn elementalTheorem of
= T4¥%,.pro#9kuhsJ= ¥r4tT I ↳ loft)( ble t'%4,⇒ -Lsts 2
but we know Pi 30So ftp.ltko if too .