Highlights
Math 304
Linear Algebra
From last time:
I
Harold P. Boas
inner products and norms
Today:
I
boas@tamu.edu
orthonormal sets
June 22, 2006
Example in R 3

√1
2
Example continued



Let u1 =  0 , u2 =
− √1
2

√1
3
− √1 
,

3
1
√
3


√1
6
 √2 
 6 .
√1
6

and u3 =
These vectors form an orthonormal set: each vector is
orthogonal to the others, and each vector has norm equal to 1.
The matrix U whose columns are u1 , u2 , and u3 is called an
orthogonal matrix. Notice that U T U = I (the identity matrix), so
U T = U −1 . This property characterizes orthogonal matrices.
An orthogonal matrix preserves the scalar product: namely,
hUx, Uyi = hx, yi. Here is why:
hUx, Uyi = (Ux)T Uy = xT U T Uy = xT y = hx, yi.
Therefore an orthogonal matrix also preserves the norm:
kUxk = kxk.

√1
2



Let y = 2  0  −
1
−√
2

√1
3


3 − √13 
√1
3


√1
6
 
4  √26 
√1
6

+
= 2u1 − 3u2 + 4u3 .
Problem. Determine kyk.


2
Solution. Since y = U −3, and since multiplication by an
4
orthogonal
matrix
preserves
the norm,
 
2
√
√
−3 = 4 + 9 + 16 = 29.
kyk = 4 In general, kc1 u1 + c2 u2 + c3 u3 k2 = c12 + c22 + c32 .
The analogous statement for an arbitrary orthonormal set is
Parseval’s formula.
More on the example: Fourier coefficients
Projection and approximation
(1, 2, 3)T
Problem. Express the vector v =
as a linear
combination of u1 , u2 , and u3 . In other words, find coefficients
c1 , c2 , and c3 such that v = c1 u1 + c2 u2 + c3 u3 or
 1 
 1 
 1 
 
√
√
√
1
3
6
2





1
2 = c1  0  + c2 − √  + c3  √2 
.
3
6
3
√1
√1
− √1
2
3
6
Solution. No row reduction necessary! Take the inner product
+ c2 u2 + c3 u3 , u1 i = c1 by
with u1 : thus hv, u1 i = hc1 u1√
orthonormality, so√c1 = −2/ 2 . Similarly, √
c2 = hv, u2 i = 2/ 3 , and c3 = hv, u3 i = 8/ 6.
In general, any vector v can be represented in terms of an
orthonormal basis u1 , u2 , and u3 via
v = hv, u1 iu1 + hv, u2 iu2 + hv, u3 iu3 .
Problem. Find the projection of the vector v = (2, 1, 1)T onto
the plane spanned by u1 and u2 .
Solution. We can write
v = hv, u1 iu1 + hv, u2 iu2 + hv, u3 iu3 ,
but u3 is orthogonal to the plane, so the projection equals
 1  2
 7
2
3
6
hv, u1 iu1 + hv, u2 iu2 =  0  + − 23  = − 23  .
1
2
− 12
6
3