• Tow cl.I.tu J 1.t11 cuunnn.

,; • • 1

~a) Show that foi any t,vo n x n mat~1ces ft .and B, tr(AB) = tr(.l3:A}~,
C

1
1
'
11

(b) Read th~ definition of Sill}~!~r matrices from the Lectl~re I

1~1li,9~:~~t:il(1~, ~~;.~
11

algebra basics ) . Use (a) above to-c show that-if X ai1d Y are .simiiar~ lptat~~~~~, 1

then t r ( X ) = tr(Y). - , i' 11

i ii1l,
11. , ,
1 1
'.1i/''1,1 11 1'/' 1·,,ui".:',>::.·, ,,,',
11 ,II , I1 I II 11 Ill I 11 ,1111 / II ii'1 I 111 I
" 1 jl Ill t1 I j ii j
t, I ,I ti!, I

(4)
tson
,I

(5) As a follow-up to question (4), show that Wand have q:nly ,~11ti ·~a~PJ~n
1
W..L
I

vector, 0.
To show that the orthogonal complement
1
W 1_ of a subspace W of an inner
product s:pace V is itself a subspace of V, we need to verify that W 1_ satisfi,es the
conditions required to be a subspace. Specifically, we need to show that:
I I

• I

1. W 1_ is ·non-emp,ty (contains the zero vector),

2. W 1_ is closed u,nder vector addition, and

3. W ...L is closed under scalar multiplication.

Definition of Orthogonal Complemen t

The orthogonal complement W 1- of a subspace W C Vis defined as:

w- 1
== { V E V I (v, w) == 0 for all w E W}

where(·,•) denotes the inner product on V.

~
 W J_• == { v E V I (v, w) == 0 for all w
•
E W}

where (·, ·) denotes the inner product on V.

Proof
1. W 1_ Contains the Zero Vector:

The zero vector O E V is orthogonal to every vector in V, incfuding every vector

in W. Therefore:

(0, w) == 0 for all w EW

Thus, 0 E W J_. This shows that W J_ is non-empty.

2. W J_ is Closed Under Vector Addition:

Let u, VE WJ_. We need to show that u .J_, , E w.1.. For any w E W:

2. wl_ is Closed Under Vect,or Addition: ' 11
II \ 1
, 11 1l
I I I 11, till
1
I ' 11 '
11 l111 ii 1
I 11 I 11 '

Let u, v E W 1-. We need to show thatu '+'tb 1

111E W 1_. For any w E W:

I n 1I ' ''
I
, 1, ,II,
1 1
1

( u+v,w -:- ul,w

)
, I ( ,i1 1li )

~
I (

v,w )

I I

..
Since u E WJ_ and VE w1-, we have:

(u, w) == 0 and (v, w) == Q,

Thus:

(u + v, w) ::::: (u, w) + (v, w) ::::: 0 + 0 ::::: 0

Since (u + v, w) == 0 for all w E W, it follows that u +v E W J_. Therefore,

W .1_ is closed under vector addition.

3. W 1- is Closed Under Scalar Multiplica .J,, :

Since_ (u + v, w ) = 0 fo r all w E W , it fo llo ws th at u +v E w1 ._ Therefore,

W J_ is clo se d un de r ve ct or ad dit ion .

ltiplication:
3. W J. is ·closed Un de r Scalar Mu

ed to show th at av E W 1.. For any

v E W
Le t l. an d le t o be a scalar. We ne
w E W:

(a v, w) == a( v, w)

Si nc e v E W __1_, we ha ve :

(v, w) == 0

Th us :

(a v, w ) == 0 ;,v , w) == a • 0 == 0

..L::rt Th PrP fnr P

~W
lil/" .L ;~
,J,,
~ W" it fnl lnw ~ t_h_rit __,.,_·_. :P:
\ r\'1 1 l=-~
~in cP ( O'? L 1fWPi - 0- fo r ~11 111
-.
¥€ 4..~~
 I

(v, w) == 0

Thus:

(av, w) == n(v, w) ==a• Q, == 0

I

Since {av, w) == 0 for all w E Writ folJows that av E W_L. Therefor e, W_L is
closed under scalar multiplic ation.

Conclusion
Since W _L contains the zero vector, is closed under vector addition , and is closed
under scalar multiplic ation, it satisfies all the conditio ns to be a subspace of V.
Therefore, w.1. is indeed a subspace of V.
 '- ,, ' ~

(b) Read the definition of shnilar matrices fron1 the Lecture Slides (linear
algebra basics ). Use (a) above to show that if X and Y are shnilar matrices ,
=
then tr(X) tr(Y).
(4) Let V be an inner product space and let W be a subspace of V. Show that
its orthogon al con1plement (defined in lin algebra 2 slides) 1,,V ..L is a subspace of
V.
(5) ~s a follow-up to question (4} 2 show that Wand W...L have only one common
fvector, 0.
(6) As a fu1·ther follov..1 up, show that dim(V) = dim(W) + dim(W_1_ ).
(7) Nov..1 suppose that vV is a subset of the vector space V, but not necessari ly
a subspace. The definition of HI _1_ can still be
To sho w tha t the onl y vec tor in com mo n betw.een a subspace
1
Wa nd 'its
orth ogon~J com ple me nt W J_ is the z.ero ,ve,ctor, we· need to
1

pro ve tha t if V E 1
1

w nw fr the n V ·must be the zero vector.·

I; • I

ij I

Pro of· l
I
I I

I, I ,,

1. Defieition Recap
l I
1
: •

• S~bspace W: A sub set of V tha t is dos ed und er addition and

scalar
•multipli1cation and contains the zero vector.

• Orthogo,nal Com ple men t W_1 : Defined as

W _1 == { v E V I (v, w) == 0 for all w E W}

• n W contains only the zero vector.
We wan t to show tha t W _L

2. Let v be an Arbitrary Vector in W n Tit/ _1:

~
2. let v be an Ar1bitrary Vector in W n W J_:
• By definition of intersection, v E Wand v E W J_. Therefo re: 1

( V' w) = 0 f~7 all w E w

. . I ,1 ,!Iii ·11 11

3. Cq~sider th e Vector V in Both Su bspacest I: I

1 1

·11' '
11 I I
I~ I
I I

• S.1nce V E w , V .IS a. vector .1,n t he ISlllur w

pace • .
11 I

I 1 ' 1
•
1
1 1
11 1

• , v -Is o rth ogona 1, t1m1ev,ery vect or In

•
1 1 1
1 1 11 1

•· •
S Ince v E wj_ 1 , w· . , n • rt·Icu 1ar, v Is
pa •
1
111 I I I '
~
1
orth,ogona,I to itself: - 1 -

(v,v) =·O

4. Inner Product Properties:

• The inner product ( v, v) is equal to the square of the norm of v, so:

(v, v) = llvll 2
• Since ( v, v) == 0, it folloyvs that ~
◄
 11111

4. l_nner Produ ct Properties: 1

1' II
11

• The inner produ ct (v, v) i~ equal to the square of the norm of v, so:
I

(v, Iv) == llvU2

• Since ( v ,, v) = 0, it follow s that~' 1

I
11
' 11 l I I
1
1 I

I !111t,111~. I 0
• The norm of a vecto r is zero if ahd' bhly if th~ vect~r itself 'Is the ;zero
vecto r. There fore:
1
11 II

llvll == 0 impli~s v == 0
5. Con cl usio n:
I
• Since v must be the zero vector, the only vecto r that is in both W and
W ..l is the zero vector. Therefore:

W n 1v..1 == {O}
~
 llv II~ == U
• The norm, of a vector is zero if and only if the vector itself is the zero
1' I
1 11, 1 •
·,1 II
., vector. Therefore:
1 I•
I'
,.

I'
'I·

I1
II
111
I
I 111
I '
II I·
I
1

''

• I
llvll'·== o' i:mplies v ==' .0

5. Conclusion:

• ~ince v must be the zero vector, the only vector that is in bo th 1

W and
W _1_ is the zero vector. Therefore:

. W n W_1_ == {O}

This completes the proof that the only vector common to both a subspace Wand
its orthogonal complement W 1_ is the zero vector.
V.
(5) As a follo\v-1111 to ql1estio11 (4), sl1ow tl1at l-l·.,. and, lil' ..l l1ave 011ljt o,ir1e con1n1011
,rector, 0.
(6) ~s a further follow ue_, show that dini(V}i : di111(W) + dim[W~J]
(7) No\\r suppose tl1at lll is a sul1set of tl1e vector SJ>ace V, bi1t not 11ecessa.ril}'
a subspace. Tl1e defi11itio11 of l•fl ..l ca11 still be

lll 1- = {v E ll I(-v, u.,) = 0 V ·w E Ht'}

 To show that dim (V) = dim (W) + dim( W _1_ ), wher e W is a subsp ace of an
inner pr:c;Muct space V and W 1_ is its ortho gona l comp leme nt, we can use the
I I I
• I
I

fol lowi n1gi11argt,1,r;t·Jient: I '

I ' ' I I I '

I ' '
1' j'1 1,[, I . I I
I I ' I

Proof , .. I " , I
'I I

'1·1 I, · I , ,

I I

1. Dime•n~ion1 of 10,rthogonal Complement:

• By the, defin itjon of ortho gona l comp leme nt,

W _1_ consists of all vecto rs
in V that are ortho gona l to every vecto r in W.

2. Direct Sum of Subspaces:

• The key prope rty to use is that V can be expressed as a direc t sum of W
and W..L:

V =WE B w1- ~
I,'
• This mean s every vecto r v E V "~ be uniqu ely writte n as 'V == w + 1u1_;
 • The key prope rty to ·use is that V can be expressed as a direct sum of W
and WJ_:

V ==WE B WJ_

• This means every vector v E V can be uniqu ely writte n as v == w + wj_,

where w E Wan d w1_ E. W1-.

3. Dimensions of Direct Sum:

• For subspaces W and W 1- of V, the dimen sion of their dti'rect stJ1m is

given by:

dim(V) == dim{W) + dim(WJ_)

4. Proof by Basis:

• Basis for W: Let { W1, W2, ... , wk} be a basis for W.

• Basis for W 1-: Since W and W 1- ~re orthog onal compl ement s, vector s
in W ..L are orthog onal to all ven ....-.1,.., in W. Let { wf, wf, ... , w;J be a
 I I

4. Proo f by Basis:

• Basis for W: Let { W1, w2, ... , wk} be a basis for W~

I I

I I I
I
Basis for W J_: Since W and
1 I I

• W J_ are ortho gona l comp leme nts, 1~ectors

I
,I 11,11111 i'1l11l11l1

in W J_ are ortho gona l to all vecto rs in ,W. Let { wf, wf, ... , mii1}i', be a 1

. f WJ_
bas1s
• 1 '1, 11 I ,1 ,Iiii' ,, '

or • . ,I' I I

'.I
,i'
I
11'i/i111,li'li
11111 i
IJ I

1,! · 1 ,
I
':! 1\ti ,11 ! ' · II'~ 11 1 1 ,, 1 1 1 1
IIII I
' 'J Ii'
• I 1·1111111 I ·111
• ·, I ' i
I I 11 I' I
• Cons truct Basis for V: Com.bini1~gq1th~se,tw,o ,sets, we form ·ai ~e-ti111'il ~ 1
1

• • • I I
1 • • • 1 1 I I II I I
, •
,I
111 1 I
{w· .1. ·; w 2 , • • • , w·· k ,,w_1_ 2 , • • • , w1-}
- I , ·w_1_ ' -•m ,
,
•
i
1
11 '1 11 1 ' ·
l 1
1

!1 I•

I I I
. •
which spans V and is linea rly indep ende nt (by the properti,es o,f
I

ortho gona l comp leme nts).

• Dime nsion Calculation: The -num ber of vecto rs in this comb ined basis is:

dim{V) == k +m .
• ~
wher e k == dim (W) anci m == L..~il l(W _1_ ).
 orthogonal complements) .

• Dimension Calculation: The number of vectors in this combined basis is:

dim(V) ~ k + rn
W'here k == dim(W) and_m == dim(W ..L ).

5. Conclusion:

• Therefore:
dim(V) == dim(W) + dim(W_i_)
This completes the proof that the dimension of V is equal to the sum of the
dimensions ot·W and W J_.
 -=
""" .--:- •
• . -- " - -
.
- · c:-- -
• ".
- _
-_ ~-
_-
•• ,
-· " T F.
•
,,
,I•- t,• :1 I'r I -1· p " 11 I' I
- - - , I ·I .
J
'
l
' ,I I
(5) As a. follow-up to' question (4), ;how that lV a11d W 1 ,']1ave orily 011~ ~on1m~n/
- .o.
vect or, i , ,J, i , , , ,I ,
, .! , . i I ,/ I ' ,! "
1:
,!' I 'I
1
, :· ,,
I' '
',1 1,1
I
1· 111 1
1 1
0 - "
I I 11

(6) As a further follow up, show tl1at dim(V) ~ dim(W) + dim(}ll J_) ., 1 ' '
· · !· ' __ , ·1·
'II
II "
1 1
11
11

(7) Now suppose that W is a subset of the vector space V~ but not necessarUYi •
~ subs~ace~ The definition of WI can still b~ .•
11

, , II' '
1
1 1
1
I 1' i 1

,w.:t: = {v E Vl{v,w) = 0 V w E W,}

lket span(WJ refer to the set of all linear combinations of a finite nun1ber 0~
~ectors take~from W. Show that WI= span_(W) 1 , '

(8) Suppose u, v E V, \\rhere 11 is a vector space, such that 11-nH = 3, 11-u.+vll = -1,
and 11 u - v 11 = 6. \J\That n11mber n1ust 11 v 11 equal?
 To show that if W is a subset of the ve<;:~q~ space 1
V,. but not necessarily a
, ii , ' I II
' I':
W J_ == span(W) _1_, follo~, these steps:
I' 11
subspace, then
I' II I : I

1•,
1,:111 '
I l11111
I
Defin'itions ll''il I
1, 11
'
I

'I jlll 1'11 I I

1. Orthogonal Complement W _L.:

W_1_ = {v EV I (v,~) = Oforallw E W} [.

' '

This is the set of all vectors in V that are orthogonal to e•very vector in W.
2. Span of W (denoted as span(W)):

span(W) == {linear combination s of vectors in W}

This is the subspace of V generated by W, which includes aJI possible linear

combinations of vectors from W.

.....--....--
3. Orthoqonal Complement of span(h ~ (denoted as span(W)_1): - - -
; ; , ift iYCMtitfflftrr:'. if p t
 • I ! ~11
I
I I I ~
I I I I I

C0,mbi.natio ns
1
1

of vectorrs frorl111w.ii
II I1
1 i
I'
1
.
' II I I I I

I 111 I 111111 Ill I

3. Orthogonal Complement ~.~i· ~~~b' W) (denoted as span(W)-1): 1

1
1 1
• • ' ',I ,l11111!Jli',1111llj1\!ll1 ' I •

• span(W)-1 = { v ,E l~.U ·;(;U, u) = 0 for all u E span(W)}

1

I ,I
. I I I .
. • ,1,: 1·r1 , , 'ii i ·I I • 1
II 1 1· , . 1 I II 1
I
I ' :, f! IH I'
I. ' ' • I I

This t~ the set of all vectors iri:11V that ~re •orthogonal to every vectqr in
~p.an(W).
I·•

Proof I •

1. Show' W C span(W):

By definiti·on, every vector in W is a vector in the span of W because W

itself is a set of vectors whose span is span(W). Therefore:

WC span(W)
2. Show span(W) C W 1..:
 WC span(W)
2. Show span(W) C W1..:

Consider any vecfor v E span(W)<lfihis means v c.an be w11itten as:

II
I •. I
' ' ' I '

I ·1

•
I I I
In.
1 I
- I
'_: i V ' ~
-, ~ '. -fi, J i. 1

11
0.f1_J1 I 1111
17'1r1· :1 - ,1 I,
. 1 , ii l11
i= 11 I
11

I
I
I II

w,here w·i E Wand ni ar~ scalars. ' I

For any vector u E W l.., we have:

n n
(u,v) (u, L o:iwi) = L o:i(u, wi)
'i=l 'i=l

Since u E W..L, (u, wi) = 0 for all wi E W. Thus:

~
c-
 I ' I
I'

Since u E w1-, (u, Wt) d10 fbr ,au Wi E w. Thus: 1

I
I '
: ,1

I I ,, I I ,I t'11 '
n
'J,I, 1"':, ,' 1',~---, (t,; •
.'!~1~lt1f~l:},I; .:,i{~ ~
I I II ,ll'j'
"
I II
0 -o
j I1 !. 1..l11 ill'~,

\Ii 1111·11::·1!:_U(1! ·111'.,I,,:

i=l

!,I:'
1l,1,1,. •I .

.. , i• l I' I i'I 11 : I
I I, j!F'
' I
'
l_
!
• I I ,- ' 11 I '

He1nce, v E. W
1:_! '

I •I '
1

II
. Thus:. _,
11
.I • '' ,l,t
I. ,
,I I' Jl
. I
I I
I
' !l;! I
, ' l r• • ,

·Sp1 an
1, '('w',
l
.
.,,.
I ,
I I
I

,·r,:1•·
I I
''
I

11
I I
I )'llc::I.
, ,11,
1w
I',I
1

,
I
I ,,

1
I II I
\I
-11-1111111
I1: I I 1111 I I 11'I '
I I I ,I I I 'I I ~
I I ' I I I II 11 I I' I 'I

3. Sh,ow W J_ C span (W)~ : 1

1
I !1
I I I
1

I I

Let ·u E W J_. By defini tion:

(u, w) 0 for all w· E W

To be in span( W)J_ , u must satisfy:

(u,v) 0 for - 11 v E span (W)

~
 .
, u, w) =: U tor a11
1
w E w

To be in span(W)J_, u ~ust satisfy:

'

(u, v)'· = 0 for all v E span(W)

I

Sine€: any vector v E span(i,v} ,can 1be written as a linear oombilnati'on of

1

vectors in W, we have:

n. [

V - ~ OiWi
i=l

for Wi E W and Oi are scalars. Therefore:

n n n

i=l i=l i=l

Thus:
 n n n

(u, v) = (u, L niwi) = L ni(u, wi) =Lai· 0 = 0

i=l i=l
i=l

Thus:

(u, v) - 0 for all v E spa n(W )

H,ence, u E spa n(W )1_ . Thus: •

W 1_ C span(.W)1_

Conclusion
Since we hav e ·sho wn bot h incl usio ns:

W C spa n(W ) and spa n(W ) C W_1_ and W_1_ C span(W)_1_

It follo ws that :
 H,ence, u E span (W) 1-. Thus:
w1- C spa n(W )1-

Conclusion
Since we have show n both inclu sion s:
[

W C spa n(W ) and span (W- ) C WJ_ and w1- C spa n(vV )1-

It follo ws that:

W1_ ::=: span(W)_1_