Inner Product Spaces1.Length and Dot Product in Rn 2.Inner
Product Spaces3. Orthonormal Bases: Gram-Schmidt Process 1. Length
and Dot Product in Rn Length: The length of a vector in Rn is given
by 2 2 21 2|| ||( || ||is a real number)nv v v = + + + v v
Properties of length (or norm) (1)0(2)1 (3)0iff(4) (proved in
Theoerm 5.1) c c>= = ==vv vv v 0v vis called a unit vector ) , ,
, (2 1 nv v v = v Notes: The length of a vector is also called its
norm Ex 1: (a) In R5, the length ofis given by (b) In R3, the
length ofis given by) 2 , 4 , 1 , 2 , 0 ( = v5 25 ) 2 ( 4 1 ) 2 ( 0
|| ||2 2 2 2 2= = + + + + = v11717173172172|| ||2 2 2= = |.|
\|+ |.|
\| + |.|
\|= v) , , (173172172 = v(If the length of v is 1, then v is a
unit vector) A standard unit vector in Rn: only one component of
the vector is 1 and the others are 0 (thus the length of this
vector must be 1) (1) 0(2) 0cc > < u and v have the same
direction u and v have the opposite directions Notes: Two nonzero
vectors are parallel if{ } ( ) ( ) ( ) { }1 2: , , , 1, 0, , 0 ,
0,1, , 0 , 0, 0, ,1nnR = e e e{ } ( ) ( ) { }21 2: , 1, 0 , 0,1 R =
e e{ } ( ) ( ) ( ) { }31 2 3: , , 1, 0, 0 , 0,1, 0 , 0, 0,1 R = e e
ec = u v Theorem 1: Length of a scalar multiple Let v be a vector
in Rn and c be a scalar. Then || || | | || || v v c c =|| || | ||
|) () ( ) ( ) (|| ) , , , ( || || ||2 22212 222122 22212 1vvcv v v
cv v v ccv cv cvcv cv cv cnnnn=+ + + =+ + + =+ + + ==Pf: ) , , , (2
1 nv v v = v) , , , (2 1 ncv cv cv c = v Theorem 2: How to find the
unit vector in the direction of v If v is a nonzero vector in Rn,
then the vector has length 1 and has the same direction as v. This
vector u is called the unit vector in the direction of v || || vvu
=Pf: v is nonzero 010 > = vvIf=1u vv(u has the same direction as
v) || || | | || ||1|| || || || 1|| || || ||c c == = =v vvu vv v(u
has length 1) Notes: (1) The vector is called the unit vector in
the direction of v (2) The process of finding the unit vector in
the direction of v is called normalizing the vector v || || vv Ex :
Finding a unit vector Find the unit vector in the direction of v =
(3, 1, 2), and verify that this vector has length 1 ( )22 22 2 22 2
2(3 , 1, 2) 3 1 2 14(3 , 1, 2) 1(3 , 1, 2)|| ||143 ( 1) 23 1 2, ,14
14 143 1 2 1411414 14 14 is a unit vector= = + + = = = + + | |=|\ .
| | | | | |+ + = = |||\ . \ . \ .
v vvvvv Sol: Distance between two vectors: The distance between
two vectors u and v in Rn is|| || ) , ( v u v u = d Properties of
distance (1) (2) if and only if u = v (3) 0 ) , ( > v u d0 ) , (
= v u d) , ( ) , ( u v v u d d = (commutative property of the
function of distance) Ex 3: Finding the distance between two
vectors The distance between u=(0, 2, 2) and v=(2, 0, 1) is 3 1 2 )
2 (|| ) 1 2 , 0 2 , 2 0 ( || || || ) , (2 2 2= + + = = = v u v u d
Dot product in Rn: The dot product ofandis a scalar quantity Ex 4:
Finding the dot product of two vectors The dot product of u=(1, 2,
0, 3) and v=(3, 2, 4, 2) is 7 ) 2 )( 3 ( ) 4 )( 0 ( ) 2 )( 2 ( ) 3
)( 1 ( = + + + = v u1 1 2 2 (isa real number)n nu v u v u v = + + +
u v u v) , , , (2 1 nu u u = u ) , , , (2 1 nv v v = vMatrix
Operations in Excel SUMPRODUCT: calculate the inner product of two
vectors(The dot product is defined as the sum of
component-by-component multiplications) Theorem 3: Properties of
the dot product Ifu, v, and w are vectors in Rn and c is a
scalar,then the following properties are true (1) (2) (3) (4)
(5),andif and only if u v v u = w u v u w v u + = + ) () ( ) ( ) (
v u v u v u c c c = = 2|| || v v v = 0 > v v0 = v v= v 0 The
proofs of the above properties follow easily from the definition of
dot product (commutative property of the dot product) (distributive
property of the dot product over vector addition) (associative
property of the scalar multiplication and the dot product)
Euclidean n-space: Rn was defined to be the set of all order
n-tuples of real numbers When Rn is combined with the standard
operations of vector addition, scalar multiplication, vector
length, anddot product, the resulting vector space is called
Euclidean n-space Sol: 6 ) 8 )( 2 ( ) 5 )( 2 ( ) a ( = + = v u) 18
, 24 ( ) 3 , 4 ( 6 6 ) ( ) b ( = = = w w v u12 ) 6 ( 2 ) ( 2 ) 2 (
) c ( = = = v u v u25 ) 3 )( 3 ( ) 4 )( 4 ( || || ) d (2= + = = w w
w) 2 , 13 ( ) 6 8 , ) 8 ( 5 ( 2 ) e ( = = w v22 4 26 ) 2 )( 2 ( )
13 )( 2 ( ) 2 ( = = + = w v u Ex : Finding dot products ) 3 , 4 (
), 8 , 5 ( , ) 2 , 2 ( = = = w v u(a)(b) (c)(d) (e) v u w v u ) ( )
2 ( v u 2|| || w) 2 ( w v u Ex : (Using the properties of the dot
product) Given 39, 3, 79, = = = uu u v v v) 3 ( ) 2 ( v u v u + +
Sol: ) 3 ( 2 ) 3 ( ) 3 ( ) 2 ( v u v v u u v u v u + + + = + +Find.
254 ) 79 ( 2 ) 3 ( 7 ) 39 ( 3 = + + =v v u v v u u u + + + = ) 2 (
) 3 ( ) 2 ( ) 3 () ( 2 ) ( 6 ) ( 3 v v u v v u u u + + + =) ( 2 ) (
7 ) ( 3 v v v u u u + + = Theorem 4: The Cauchy-Schwarz inequality
If u and v are vectors in Rn, then (denotes the absolute value
of)|| || || || | | v u v u s | | v uv uv u v uv v u u v uv us = =
== = 55 5 111 11, 11, 5 = = = uv uu vv Ex : An example of the
Cauchy-Schwarz inequality Verify the Cauchy-Schwarz inequality for
u=(1, 1, 3) and v=(2, 0, 1) Sol: (The geometric proof for this
inequality is shown on the next slide) Dot product and the angle
between two vectors To find the angle between two nonzero vectors u
= (u1, u2) and v = (v1, v2) in R2, the Law of Cosines can be
applied to the following triangle to obtain
) 0 ( t u u s su cos 22 2 2u v u v u v + = u vv uu vuvu v=+= +
=+ = + = 2 2 1 1222122221222 221 12cos ) ( ) (v u v uu uv vv u v uu
You can employ the fact that |cos | s 1 to prove the Cauchy-Schwarz
inequality in R2 Note: The angle between the zero vector and
another vector is not defined (since the denominator cannot be
zero) The angle between two nonzero vectors in Rn: t u u s s= 0 ,||
|| || ||cosv uv ucos 1u tu == 0cos 1uu ==2cos 0tu tu< v v 0 , =
v v= v 0 Inner product: represented by angle brackets Let u, v, and
w be vectors in a vector space V, and let c be any scalar. An inner
product on V is a function that associates a real number with each
pair of vectors u and v and satisfies the following axioms
(commutative property of the inner product) (distributive property
of the inner product over vector addition) (associative property of
the scalar multiplication and the inner product) , u v , u v Note:
dot product (Euclidean inner product for) , general inner product
for a vector space nRV =< >=u vu v Ex 1: The Euclidean inner
product for Rn Show that the dot product in Rn satisfies the four
axioms of an inner product n nv u v u v u + + + = = 2 2 1 1, v u v
u ) , , , ( , ) , , , (2 1 2 1 n nv v v u u u = = v uSol: By
Theorem3, this dot product satisfies the required four axioms.
Thus, the dot product can be a kind of inner product in Rn Ex 2: A
different inner product for Rn Show that the following function
defines an inner product on R2, where and 2 2 1 12 , v u v u + = v
u) , ( ) , (2 1 2 1v v u u = = v u Sol: 1 1 2 2 1 1 2 2(1) , 2 2 ,
u v u v v u v u = + = + = u v v u w u v uw v u, ,) 2 ( ) 2 (2 2) (
2 ) ( ,2 2 1 1 2 2 1 12 2 2 2 1 1 1 12 2 2 1 1 1+ =+ + + =+ + + =+
+ + = +w u w u v u v uw u v u w u v uw v u w v u 1 2(2)( , ) w w =
w Note: Example 2 can be generalized such that can be an inner
product on Rn 0 , ,2 2 2 1 1 1> + + + =i n n nc v u c v u c v u
c v u1 1 2 2 1 1 2 2(3), ( 2 ) ( ) 2( ) , c c u v u v cu v cu v c =
+ = + = u v u v 2 21 2(4), 2 0 v v = + > v v 2 21 2 1 2, 0 2 0 0
( ) v v v v = + = = = = v v v 0 Ex 3: A function that is not an
inner product Show that the following function is not an inner
product on R3
1 1 2 2 3 32 u v u v u v = + u v Sol: Let) 1 , 2 , 1 ( = vThen
(1)(1) 2(2)(2) (1)(1) 6 0 = + = < v v Axiom 4 is not
satisfiedThus this function is not an inner product on R3 Theorem
7: Properties of inner products Let u, v, and w be vectors in an
inner product space V, and let c be any real number (1) (2) (3) , ,
0 = = 0v v0 , , , + = + u vw uw vw , , c c = u v uv To prove these
properties, you can use only the four axioms in the definition of
inner product Hint: (1)(2) (3)0 , 0 , , 0 = = = uv uv 0v , ,and , ,
, = + = uv vu w u v w u wv + , ,and, , c c = = uv vu uv uv u and v
are orthogonal if Distance between u and v: ) ( = = v u v u v u v u
, || || ) , ( d Angle between two nonzero vectors u and v: t u u s
s = 0 ,|| || || ||,cosv uv u Orthogonal: 0 , = v u) ( v u Norm
(length) of u: u u u , || || = The definition of norm (or length),
distance, angle, orthogonal, and normalizing for general inner
product spaces closely parallel to those for Euclidean n-space
Normalizing vectors (1) If, then v is called a unit vector (2) 1 ||
|| = v= v 0Normalizingvv(the unit vector in the direction of v) (if
v is not a zero vector) Ex 6: Finding inner product 2 22Let( ) 1 2
, ( ) 4 2be polynomials inp x x q x x x P = = +0 0 1 1,n np q a b a
b a b + + + is an inner product (a), ? p q = (b)|| || ? q = (c)( ,
) ? dp q =Sol: (a), (1)(4) (0)( 2) ( 2)(1) 2 p q = + + =2 2 2(b)||
|| , 4 ( 2) 1 21 q q q = = + + =22 2 2(c)3 2 3( , ) || || , ( 3) 2
( 3) 22p q x xd p q p q p qp q = + = = ( )= + + =, and For 1 0 1
0nnnnx b x b b q x a x a a p + + + = + + + = and Properties of
norm: (the same as the properties for the dot product in Rn) (1)
(2) if and only if(3) Properties of distance: (the same as the
properties for the dot product in Rn) (1) (2)if and only if(3) 0 ||
|| > u0 || || = u 0 u =|| || | | || || u u c c =0 ) , ( > v u
d0 ) , ( = v u d v u =) , ( ) , ( u v v u d d = Theorem 8 Let u and
v be vectors in an inner product space V (1) Cauchy-Schwarz
inequality:
(2) Triangle inequality:
(3) Pythagorean theorem: u and v are orthogonal if and only if||
|| || || || || v u v u + s +Theorem 5 2 2 2|| || || || || || v u v
u + = +Theorem 6 || || || || | , | v u v u s Theorem 4 For inner
product spaces:Let u and v be two vectors in an inner product space
V. If, then the orthogonal projection of u onto v is given by 0 v
=,proj,( )=( )vuvu vvv2Consider0, cos| || || || cos | || || ||
|| || || ||proja a a aauu> = = = = = = = = = vv v v uu v u v
uv uvv v u v vuv uv uvu vvv vvvuv0 , proj > = a av uvu
Orthogonal projections: For the dot product space in Rn, we define
the orthogonal projection of u onto v to be projvu = av (a scalar
multiple of v), and the coefficient a can be derived as follows Ex
10: Finding an orthogonal projection in R3 Use the Euclidean inner
product in R3 to find the orthogonal projection of u=(6, 2, 4) onto
v=(1, 2, 0) Sol: 10 ) 0 )( 4 ( ) 2 )( 2 ( ) 1 )( 6 ( , = + + = ) (
v u 5 0 2 1 ,2 2 2= + + = ) ( v v, 10proj (1, 2 , 0) (2 , 4 , 0),
5( ) = = = =( ) vu v u vu v vv v v v Theorem 9: Orthogonal
projection and distance Let u and v be two vectors in an inner
product space V, and if v 0, then,( , proj ) ( , ) , ,d d c c(
)< =( )vu vu u u vv v) proj , ( u uvduvv c) , ( v u c duvuvproj
Theorem 9 can be inferred straightforward by the Pythagorean
Theorem, i.e., in a right triangle, the hypotenuse is longer than
both legs 3.Orthonormal Bases: Gram-Schmidt Process Orthogonal set:
A set S of vectors in an inner product space V is called an
orthogonal set if every pair of vectors in the set is orthogonal
Orthonormal set: An orthogonal setin which each vector is a unit
vector is called orthonormal set { }1 22, , ,For, , , 1For, , 0ni j
i i ii jS Vi ji j= _= ( ) = ( ) = == ( ) =v v vv v v v vv v{ }j iV
Sj in= = ) (_ =for, 0 ,, , ,2 1v vv v v Note: If S is also a basis,
then it is called an orthogonal basis or an orthonormal basis The
standard basis for Rn is orthonormal. For example, is an
orthonormal basis for R3 This section identifies some advantages of
orthonormal bases, and develops a procedure for constructing such
bases, known as Gram-Schmidt orthonormalization process { } ) 1 0 0
( ) 0 1 0 ( ) 0 0 1 ( , , , , , , , , S = Ex 1: A nonstandard
orthonormal basis for R3 Show that the following set is an
orthonormal basis )`|.|
\|||.|
\||.|
\|=31,32,32,32 2,62,62, 0 ,21,213 2 1Sv v v Sol: First, show
that the three vectors are mutually orthogonal092 292920 02 322 320
03 23 161612 1= + = = + = = + + = v vv vv vSecond, show that each
vector is of length 1Thus S is an orthonormal set 1 || ||1 || ||1 0
|| ||9194943 3 3983623622 2 221211 1 1= + + = == + + = == + + = =v
v vv v vv v vBecause these three vectors are linearly independent
(you can check by solving c1v1 + c2v2 + c3v3 = 0) in R3 (of
dimension 3), by Theorem 4.12, they form a basis for R3. So S is a
(nonstandard) orthonormal basis for R3 the standard basis is
orthonormal Ex 2: An orthonormal basis for P3(x) In, with the inner
product, 2 2 1 1 0 0, b a b a b a q p + + = ) (} , , 1 {2x x B =)
(3x PSol: , 0 0 121x x + + = v , 0 022x x + + = v , 0 023x x + + =
v1 21 32 3 , (1)(0) (0)(1) (0)(0) 0 , (1)(0) (0)(0) (0)(1) 0 ,
(0)(0) (1)(0) (0)(1) 0( ) = + + =( ) = + + =( ) = + + =v vv vv
vThen ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )1
1 12 2 23 3 31 1 0 0 0 0 10 0 1 1 0 0 10 0 0 0 1 1 1= ( ) = + + ==
( ) = + + == ( ) = + + =v v , vv v, vv v, v Theorem 10: Orthogonal
sets are linearly independent If is an orthogonal set of nonzero
vectors in an inner product space V, then S is linearly
independent{ }nS v , , v , v 2 1=Pf: Sis an orthogonal set of
nonzero vectors, i.e.,0 for,and0i j i ii j ( ) = = ( ) > v , v v
, v1 1 2 21 1 2 2For, , 0n nn n i ic c cc c c i+ + + = ( + + + ) =
( ) = v v v 0v v v v 0 v0
2 2 1 1= ) ( =) ( + + ) ( + + ) ( + ) ( i i ii n n i i i i icc c
c cv , vv , v v , v v , v v , v 00 is linearly independenti i ic i
S ( ) = = v , v(If there is only the trivial solution for cis,
i.e., all ci = 0, S is linearly independent)(because S is an
orthogonal set of nonzero vectors) Corollary to Theorem 10:
IfVisaninnerproductspaceofdimensionn,thenany orthogonal set of n
nonzero vectors is a basis for V 1. By Theorem 5.10, if S = {v1,
v2, , vn} is an orthogonal set of n vectors, then S is linearly
independent 2. According to Theorem 4.12, if S = {v1, v2, , vn} is
a linearly independent set of n vectors in V (with dimension n),
then S is a basis for V Based on the above two arguments, it is
straightforward to derive the above corollary to Theorem 5.10 Ex 4:
Using orthogonality to test for a basis Show that the following set
is a basis for 4R)} 1 , 1 , 2 , 1 ( , ) 1 , 2 , 0 , 1 ( , ) 1 , 0 ,
0 , 1 ( , ) 2 , 2 , 3 , 2 {(4 3 2 1 = Sv v v v Sol:
0 2 2 6 20 2 4 0 20 2 0 0 24 13 12 1= + = = + + = = + + = v vv
vv v1 2 3 4, , , : nonzero vectors v v v v0 1 2 0 10 1 0 0 10 1 0 0
14 34 23 2= + + = = + + + = = + + + = v vv vv vorthogonal is S
4forbasis a is R S (by Corollary to Theorem 10) The corollary to
Thm. 5.10 shows an advantage of introducing the concept of
orthogonal vectors, i.e., it is not necessary to solve linear
systems to test whether S is a basis (e.g., Ex 2 in Section 4.5) if
S is a set of orthogonal vectors Theorem 11: Coordinates relative
to an orthonormal basis If is an orthonormal basis for an inner
product space V, then the unique coordinate representation of a
vector w with respect to B is } , , , {2 1 nB v v v =1Since,
,then0i ji ji j=( ) = =v vV k k kn ne + + + = v v v w 2 2 1
1(unique representation from Thm. 4.9) Pf: isanorthonormalbasisfor
V } , , , {2 1 nB v v v =1 1 2 2, , ,n n= ( ) +( ) + +( ) w wv v wv
v wv v The above theorem tells us that it is easy to derive the
coordinate representation of a vector relative to an orthonormal
basis, which is another advantage of employing orthonormal bases 1
1 2 21 1, ( ),, , ,for = 1 to i n n ii i i i n n iik k kk k kk i n=
+ + += + + + +=wv v v v vv v v v v vn nv v w v v w v v w w ) ( + +
) ( + ) ( = , , ,2 2 1 1 Note: Ifis an orthonormal basis for V and
, } , , , {2 1 nB v v v = V e wThen the corresponding coordinate
matrix of w relative to B is||(((((
) () () (=nBv wv wv ww,,,21 Ex For w = (5, 5, 2), find its
coordinates relative to the standard basis for R3
2 ) 1 , 0 , 0 ( ) 2 , 5 , 5 ( ,5 ) 0 , 1 , 0 ( ) 2 , 5 , 5 ( ,5
) 0 , 0 , 1 ( ) 2 , 5 , 5 ( ,3 32 21 1= = = ) ( = = = ) (= = = ) (v
w v wv w v wv w v w((((
= 255] [Bw In fact, it is not necessary to use Thm. 5.11 to find
the coordinates relative to the standard basis, because we know
that the coordinates of a vector relative to the standard basis are
the same as the components of that vector The advantage of the
orthonormal basis emerges when we try to find the coordinate matrix
of a vector relative to an nonstandard orthonormal basis (see the
next slide) Ex 5: Representing vectors relative to an orthonormal
basis Find the coordinates ofw = (5, 5, 2) relative to the
following orthonormal basis for)} 1 , 0 , 0 ( , ) 0 , , ( , ) 0 , ,
{(53545453 = B3R Sol: 2 ) 1 , 0 , 0 ( ) 2 , 5 , 5 ( ,7 ) 0 , , ( )
2 , 5 , 5 ( ,1 ) 0 , , ( ) 2 , 5 , 5 ( ,3 353542 254531 1= = = ) (
= = = ) ( = = = ) (v w v wv w v wv w v w((((
= 271] [Bw1v2v3v The geometric intuition of the Gram-Schmidt
process to find an orthonormal basis in R2
1v2v{ }22 1forbasis a is , R v v1 1= w v2v1 12 2 2to
orthogonalis proj1v w v v ww= =2w21proj vw222forbasis l orthonorma
anis } , { Rwwww11 Gram-Schmidt orthonormalization process: is a
basis for an inner product space V} , , , {2 1 nB v v v =1 1Let v w
=1span({ }) S =1w2span({ , }) S =1 2w w} , , , { '2 1 nB w w w = 1
21 2'' { , , , }nnB =w w ww w wis an orthogonal basis is an
orthonormal basis 111 projnnn in n S n n iii i== = v , ww v v v ww,
w23 1 3 23 3 3 3 1 21 1 2 2projS= = v , w v , ww v v v w ww, w w,
w12 12 2 2 2 11 1projS= = v, ww v v v ww , wThe orthogonal
projection onto a subspace is actually the sum of orthogonal
projection onto the vectors in an orthogonal basis for that
subspace (I will prove it on Slides 5.67 and 5.68) Sol: ) 0 , 1 , 1
(1 1= = v w) 2 , 0 , 0 ( ) 0 ,21,21(2 / 12 / 1) 0 , 1 , 1 (21) 2 ,
1 , 0 (22 22 311 11 33 3= = = ww ww vww ww vv w Ex 7: Applying the
Gram-Schmidt orthonormalization process Apply the Gram-Schmidt
process to the following basis for R3
)} 2 , 1 , 0 ( , ) 0 , 2 , 1 ( , ) 0 , 1 , 1 {(3 2 1= Bv v v) 0
,21,21( ) 0 , 1 , 1 (23) 0 , 2 , 1 (11 11 22 2 = = = ww ww vv w} 2)
0, (0, 0), ,21,21( 0), 1, (1, { } , , { '3 2 1= = w w w BOrthogonal
basis} 1) 0, (0, 0), ,21,21( 0), ,21,21( { } , , { ' '3322= =
wwwwww11BOrthonormal basis Ex 10: Alternative form of Gram-Schmidt
orthonormalization process Find an orthonormal basis for the
solution space of the homogeneous system of linear equations 0 6 2
20 74 3 2 14 2 1= + + += + +x x x xx x x Sol:
G.-J.E1 1 0 7 0 1 0 2 1 02 1 2 6 0 0 1 2 8 0 (( (( (((((
+(((((
=(((((
+ =(((((
108101228 22
4321t stst st sxxxxThus one basis for the solution space is)} 1
, 0 , 8 , 1 ( , ) 0 , 1 , 2 , 2 {( } , {2 1 = = v v B( )( ) ( )(
)11 1 112 12 2 2 1 1 2 2 1 1 11 12221 2 2 1 and2,2, 1, 0 , , , 03 3
3 3,,(due to and, 1),2 2 1 2 2 11,8, 0, 1 1,8, 0, 1 , , , 0 , , ,
03 3 3 3 3 33, 4,2,1130| |= = = =|\ .( )= ( ) = ( ) =( ) ( | | | |=
|| (\ . \ . = = =ww v uwv uw v v u u w v u u uu uwuw( )3, 4,2,1
)`|.|
\| |.|
\|= 301,302,304,303, 0 ,31,32,32' ' B In this alternative form,
we always normalize wi to be ui before processing wi+1 The
advantage of this method is that it is easier to calculate the
orthogonal projection of wi+1 on u1, u2,, ui Alternative form of
the Gram-Schmidt orthonormalization process: is a basis for an
inner product space V } , , , {2 1 nB v v v =1 111 122 2 2 2 1 1233
3 3 3 1 1 3 2 23111 2,where ,where ,where { , , , } is an
orthonormal basis for nnn n n n i iinnV== == = = = = = w vuw vwu w
v v, u uwwu w v v, u u v, u uwwu w v v , u uwu u u
