408 CHAPTER 17. INNER PRODUCT SPACES

= ∑r,s

Rr j ⟨ur,us⟩Rsk

Let G be the matrix whose rs entry is ⟨ur,us⟩ . This is called the Grammian matrix. Thenthe above reduces to the following matrix equation.

I = RT GR

Taking inverses of both sides yields

I = R−1G−1 (RT )−1

Then it follows thatRRT = G−1. (17.7)

Example 17.2.3 Let the real inner product space V consist of the continuous real functionsdefined on [0,1] with the inner product given by

⟨ f ,g⟩ ≡∫ 1

0f (x)g(x)dx

and consider the functions (vectors){

1,x,x2,x3}. Show this is a linearly independent set

of vectors and obtain an orthonormal set of vectors having the same span.

First, why is this a linearly independent set of vectors? This follows easily from Prob-lem 15 on Page 398. You consider the Wronskian of these functions.

det

1 x x2 x3

0 1 2x 3x2

0 0 2 6x0 0 0 6

 ̸= 0.

Therefore, the vectors are linearly independent. Now following the above procedure withmatrices, let

R =

a1 a2 a3 a4

0 a5 a6 a7

0 0 a8 a9

0 0 0 a10

Also it is necessary to compute the Grammian.

∫ 10 dx

∫ 10 xdx

∫ 10 x2dx

∫ 10 x3dx∫ 1

0 xdx∫ 1

0 x2dx∫ 1

0 x3dx∫ 1

0 x4dx∫ 10 x2dx

∫ 10 x3dx

∫ 10 x4dx

∫ 10 x5dx∫ 1

0 x3dx∫ 1

0 x4dx∫ 1

0 x5dx∫ 1

0 x6dx

=

1 1

213

14

12

13

14

15

13

14

15

16

14

15

16

17

You also need the inverse of this Grammian. However, a computer algebra system canprovide this right away.

G−1 =

16 −120 240 −140−120 1200 −2700 1680240 −2700 6480 −4200−140 1680 −4200 2800



408 CHAPTER 17. INNER PRODUCT SPACES= YVR (u,, Us) RskrsLet G be the matrix whose rs entry is (u,;,us). This is called the Grammian matrix. Thenthe above reduces to the following matrix equation.1=R'GRTaking inverses of both sides yields1=R'G"! (RT)!Then it follows thatRR’ =G"!. (17.7)Example 17.2.3 Let the real inner product space V consist of the continuous real functionsdefined on |0, 1] with the inner product given byna)= [ regexand consider the functions (vectors) {1,x,x7,23} . Show this is a linearly independent setof vectors and obtain an orthonormal set of vectors having the same span.First, why is this a linearly independent set of vectors? This follows easily from Prob-lem 15 on Page 398. You consider the Wronskian of these functions.1x * x2det O 1 2x 3x 40.0 0 2 6x0 0 O 6Therefore, the vectors are linearly independent. Now following the above procedure withmatrices, leta, a @% a4R= 0 a5 a6 ay0 0) ag ag0 0 0 aioAlso it is necessary to compute the Grammian.Jodx — Joxdx fy xdx fy dxJgxdx foxrdx fo xtdx fy x4dxfgxdx fix dx foxtdx fy dx0 xdx fo xtdx 0 xdx fo x°dxAIRE WIE NIEOle Ble OIE IeRe Ul BIR WIRNIA AIR le BIEYou also need the inverse of this Grammian. However, a computer algebra system canprovide this right away.16 —120 240 —140—120 1200 —2700 1680240 —2700 6480 -—4200—140 1680 —-—4200 2800Gl=