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21. Orthonormal Bases

The canonical/standard basis

e 1=0 B BBB@1 0 01 C

CCCA; e

2=0 B BBB@0 1 01 C

CCCA; :::; e

n=0 B BBB@0 0 11 C CCCA has many useful properties. Each of the standard basis vectors has unit length: jjeijj=pe ie i=qe

Tiei= 1:

The standard basis vectors areorthogonal(in other words, at right angles or perpendicular). e ie j=eTiej= 0 wheni6=j

This is summarized by

e

Tiej=ij=(1i=j

0i6=j;

whereijis theKronecker delta. Notice that the Kronecker delta gives the entries of the identity matrix. Given column vectorsvandw, we have seen that the dot productvw is the same as the matrix multiplicationvTw. This is theinner producton R n. We can also form theouter productvwT, which gives a square matrix. 1 The outer product on the standard basis vectors is interesting. Set

1=e1eT1

=0 B BBB@1 0 01 C CCCA

1 0:::0

0 B

BBB@1 0:::0

0 0:::0

0 0:::01

C CCCA n=eneTn =0 B BBB@0 0 11 C CCCA

0 0:::1

0 B

BBB@0 0:::0

0 0:::0

0 0:::11

C CCCA

In short,

iis the diagonal square matrix with a 1 in theith diagonal position and zeros everywhere else. 1

Notice that

ij=eieTiejeTj=eiijeTj. Then: ij=( ii=j

0i6=j:

Moreover, for a diagonal matrixDwith diagonal entries1;:::;n, we can write

D=11+:::+nn:

Other bases that share these properties should behave in many of the same ways as the standard basis. As such, we will study:1 This is reminiscent of an older notation, where vectors are written in juxtaposition. This is called a `dyadic tensor,' and is still used in some applications. 2

Orthogonal basesfv1;:::;vng:

v iv j= 0 ifi6=j In other words, all vectors in the basis are perpendicular.

Orthonormal basesfu1;:::;ung:

u iu j=ij: In addition to being orthogonal, each vector has unit length. SupposeT=fu1;:::;ungis an orthonormal basis forRn. SinceTis a basis, we can write any vectorvuniquely as a linear combination of the vectors inT: v=c1u1+:::cnun: SinceTis orthonormal, there is a very easy way to nd the coecients of this linear combination. By taking the dot product ofvwith any of the vectors inT, we get: vu i=c1u1u i+:::+ciuiu i+:::+cnunu i =c10 +:::+ci1 +:::+cn0 =ci; )ci=vu i )v= (vu

1)u1+:::+ (vu

n)un =X i(vu i)ui:

This proves the theorem:

Theorem.For an orthonormal basisfu1;:::;ung, any vectorvcan be ex- pressed v=X i(vu i)ui:

Relating Orthonormal Bases

SupposeT=fu1;:::;ungandR=fw1;:::;wngare two orthonormal bases forRn. Then: 3 w

1= (w1u

1)u1+:::+ (w1u

n)un w n= (wnu

1)u1+:::+ (wnu

n)un )wi=X ju j(ujw i) As such, the matrix for the change of basis fromTtoRis given by

P= (Pj

i) = (ujw i):

Consider the productPPTin this case.

(PPT)j k=X i(ujw i)(wiu k) X i(uTjwi)(wTiuk) =uTj" X i(wiwTi)# u k =uTjInuk() =uTjuk=jk: In the equality () is explained below. So assuming () holds, we have shown thatPPT=In, which implies that P T=P1:

The equality in the line () says thatP

iwiwTi=In. To see this, we examine (P iwiwTi)vfor an arbitrary vectorv. We can nd constantscj such thatv=P jcjwj, so that: X iw iwTi)v= (X iw iwTi)(X jcjwj) X jcjX iw iwTiwj X jcjX iw iij X jcjwjsince all terms withi6=jvanish =v: 4

Then as a linear transformation,

P iwiwTi=Inxes every vector, and thus must be the identityIn.

DenitionA matrixPisorthogonalifP1=PT.

Then to summarize,

Theorem.A change of basis matrixPrelating two orthonormal bases is an orthogonal matrix. i.e. P 1=PT:

ExampleConsiderR3with the orthonormal basis

S=8 >:u 1=0 B B@2p6 1p6 1p6 1 C

CA;u2=0

B B@0 1p2 1p2 1 C

CA;u3=0

B B@1p3 1p3 1p3 1 C CA9 LetRbe the standard basisfe1;e2;e3g. Since we are changing from the standard basis to a new basis, then the columns of the change of basis matrix are exactly the images of the standard basis vectors. Then the change of basis matrix fromRtoSis given by:

P= (Pj

i) = (ejui) =0 B @e 1u 1e1u 2e1u 3 e 2u 1e2u
2e2u
3 e 3u 1e3u
2e3u
31
C A u 1u2u3 =0 B B@2p6 01p3 1p6 1p2 1p3 1p6 1p2 1p3 1 C CA:

From our theorem, we observe that:

P

1=PT=0

B @u

T1uT2uT31

C A 0 B B@2p6 1p6 1p6 01p2 1p2 1p3 1p3 1p3 1 C CA: 5 We can check thatPTP=Inby a lengthy computation, or more simply, notice that (PTP)ij=0 B @u

T1uT2uT31

C A u 1u2u3 0 B @1 0 0 0 1 0

0 0 11

C A: We are using orthonormality of theuifor the matrix multiplication above. Orthonormal Change of Basis and Diagonal Matrices.SupposeDis a diagonal matrix, and we use an orthogonal matrixPto change to a new basis. Then the matrixMofDin the new basis is:

M=PDP1=PDPT:

Now we calculate the transpose ofM.

M

T= (PDPT)T

= (PT)TDTPT =PDPT =M

So we see the matrixPDPTis symmetric!

References

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