00:00 Introduction

00:25 Cauchy-Schwarz inequality

02:00 Proof

08:15 Triangle inequality for the norm

1 00:00:00,479 –> 00:00:02,369 Hello and welcome back to

2 00:00:02,380 –> 00:00:03,509 functional analysis

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7 00:00:09,180 –> 00:00:10,960 And I can also tell you that

8 00:00:10,970 –> 00:00:12,470 now you find the PDF

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10 00:00:14,359 –> 00:00:15,600 in the description below.

11 00:00:16,430 –> 00:00:18,180 So this is part 10 today

12 00:00:18,190 –> 00:00:19,739 and we will still talk about

13 00:00:19,750 –> 00:00:20,780 inner products.

14 00:00:21,719 –> 00:00:23,010 In particular, I want to

15 00:00:23,020 –> 00:00:24,590 prove the Cauchy-Schwarz

16 00:00:24,610 –> 00:00:25,430 inequality.

17 00:00:26,379 –> 00:00:27,319 Depending where you come

18 00:00:27,329 –> 00:00:27,809 from,

19 00:00:27,829 –> 00:00:29,190 the equality is known by

20 00:00:29,200 –> 00:00:30,280 different names,

21 00:00:30,290 –> 00:00:31,700 but I stay here with the

22 00:00:31,709 –> 00:00:32,598 most common one.

23 00:00:33,270 –> 00:00:34,330 And here it is named after the

24 00:00:34,340 –> 00:00:35,770 French mathematician Cauchy

25 00:00:36,119 –> 00:00:37,659 and a German mathematician

26 00:00:37,669 –> 00:00:38,119 Schwarz.

27 00:00:39,060 –> 00:00:40,569 Indeed, this inequality is

28 00:00:40,580 –> 00:00:42,270 very important, because it

29 00:00:42,279 –> 00:00:44,049 holds for all inner products.

30 00:00:44,560 –> 00:00:45,770 Therefore, let’s choose an

31 00:00:45,779 –> 00:00:47,729 F-vector space X and an

32 00:00:47,740 –> 00:00:48,529 inner product

33 00:00:49,369 –> 00:00:50,790 and we also consider the

34 00:00:50,799 –> 00:00:52,369 corresponding norm, which

35 00:00:52,380 –> 00:00:53,459 is the square root of the

36 00:00:53,470 –> 00:00:54,229 inner product.

37 00:00:54,880 –> 00:00:56,650 And then with these notations,

38 00:00:56,659 –> 00:00:58,119 the following inequality

39 00:00:58,130 –> 00:01:00,020 holds. We

40 00:01:00,029 –> 00:01:01,389 look at the inner product

41 00:01:01,400 –> 00:01:02,630 x with y

42 00:01:03,400 –> 00:01:04,489 and because it could be in

43 00:01:04,500 –> 00:01:05,980 general a complex number,

44 00:01:05,989 –> 00:01:07,120 we need the absolute value

45 00:01:07,129 –> 00:01:07,489 here

46 00:01:08,239 –> 00:01:09,849 and then we can say this

47 00:01:09,860 –> 00:01:11,819 is less or equal than the

48 00:01:11,830 –> 00:01:13,419 two norms multiplied.

49 00:01:14,339 –> 00:01:15,580 And that’s the whole Cauchy-

50 00:01:15,610 –> 00:01:16,919 Schwarz inequality.

51 00:01:17,930 –> 00:01:19,510 Now we can visualize it with

52 00:01:19,519 –> 00:01:21,139 the same picture we had when

53 00:01:21,150 –> 00:01:22,339 we started with the inner

54 00:01:22,349 –> 00:01:24,059 products. In this

55 00:01:24,069 –> 00:01:25,510 multiplication x with

56 00:01:25,519 –> 00:01:27,199 y only the parallel

57 00:01:27,209 –> 00:01:28,900 part of y, the yellow part

58 00:01:28,910 –> 00:01:30,050 here, should matter.

59 00:01:30,779 –> 00:01:32,160 Then this picture tells you,

60 00:01:32,169 –> 00:01:33,319 yeah, of course, the length

61 00:01:33,330 –> 00:01:34,870 of the yellow part is less

62 00:01:34,879 –> 00:01:36,199 or equal than the length

63 00:01:36,209 –> 00:01:37,400 of the red arrow here.

64 00:01:38,209 –> 00:01:39,669 With this in mind, we also

65 00:01:39,680 –> 00:01:41,589 get the result in which cases

66 00:01:41,599 –> 00:01:43,120 the equality here holds.

67 00:01:43,900 –> 00:01:45,290 Indeed, this should only

68 00:01:45,300 –> 00:01:46,730 be the case when the arrows

69 00:01:46,739 –> 00:01:48,169 go into the same direction.

70 00:01:48,989 –> 00:01:50,760 In other words, x and y

71 00:01:50,769 –> 00:01:52,419 are linearly dependent

72 00:01:52,430 –> 00:01:53,099 vectors.

73 00:01:53,739 –> 00:01:54,199 OK.

74 00:01:54,209 –> 00:01:55,419 Now the picture gets us in

75 00:01:55,430 –> 00:01:56,919 the correct direction, but

76 00:01:56,930 –> 00:01:58,230 we don’t have any choice.

77 00:01:58,239 –> 00:01:59,910 We need to prove the inequality

78 00:01:59,919 –> 00:02:00,319 now.

79 00:02:01,160 –> 00:02:01,529 OK.

80 00:02:01,540 –> 00:02:02,709 So let’s start with an easy

81 00:02:02,720 –> 00:02:04,400 case. Let’s call it the first

82 00:02:04,410 –> 00:02:06,389 case where x is the zero

83 00:02:06,400 –> 00:02:06,860 vector.

84 00:02:07,809 –> 00:02:09,270 Of course, there we know

85 00:02:09,288 –> 00:02:10,839 the left-hand side has to

86 00:02:10,850 –> 00:02:11,589 be zero.

87 00:02:12,830 –> 00:02:14,410 simply by the linearity,

88 00:02:14,419 –> 00:02:15,940 we can just pull out the

89 00:02:15,949 –> 00:02:16,990 factor zero

90 00:02:17,660 –> 00:02:18,679 and of course, the right-

91 00:02:18,690 –> 00:02:20,309 hand side is also zero,

92 00:02:20,380 –> 00:02:22,190 because the norm of x has

93 00:02:22,199 –> 00:02:23,089 to be zero here.

94 00:02:23,929 –> 00:02:25,440 In particular, the general

95 00:02:25,449 –> 00:02:26,970 inequality is

96 00:02:26,979 –> 00:02:28,649 obviously fulfilled for this

97 00:02:28,660 –> 00:02:29,369 simple case.

98 00:02:29,990 –> 00:02:31,570 Now you might already guess

99 00:02:31,580 –> 00:02:33,220 the actual interesting case

100 00:02:33,229 –> 00:02:34,559 would be the second one.

101 00:02:35,320 –> 00:02:36,869 Here x is not a zero

102 00:02:36,880 –> 00:02:38,419 vector, which means we can

103 00:02:38,429 –> 00:02:40,270 divide by the norm of x

104 00:02:41,029 –> 00:02:42,789 and that’s what we do immediately.

105 00:02:42,800 –> 00:02:44,660 I want to define x hat

106 00:02:44,710 –> 00:02:46,160 as the normalized vector

107 00:02:46,169 –> 00:02:48,039 X. This means

108 00:02:48,050 –> 00:02:50,039 that x hat has length one.

109 00:02:50,050 –> 00:02:51,539 So it just gives the direction

110 00:02:51,550 –> 00:02:53,300 of the vector x and

111 00:02:53,309 –> 00:02:54,500 then we calculate the inner

112 00:02:54,509 –> 00:02:56,070 product with x hat and

113 00:02:56,080 –> 00:02:57,940 y and go into the

114 00:02:57,949 –> 00:02:59,500 direction of x hat.

115 00:03:00,850 –> 00:03:02,070 Now, you might recognize

116 00:03:02,080 –> 00:03:03,660 that, because if we have

117 00:03:03,669 –> 00:03:05,389 considered the normal Euclidean

118 00:03:05,399 –> 00:03:07,070 geometry in the plane, this

119 00:03:07,080 –> 00:03:08,240 picture would be completely

120 00:03:08,250 –> 00:03:09,009 correct

121 00:03:09,199 –> 00:03:10,860 and this expression is

122 00:03:10,869 –> 00:03:12,830 exactly this orange line.

123 00:03:13,559 –> 00:03:14,699 By our abstraction

124 00:03:14,710 –> 00:03:16,179 this should be the same here.

125 00:03:16,190 –> 00:03:17,520 So let’s call the orange arrow

126 00:03:17,759 –> 00:03:19,009 just y

127 00:03:19,020 –> 00:03:20,899 parallel. In the

128 00:03:20,910 –> 00:03:22,039 Euclidean geometry

129 00:03:22,050 –> 00:03:23,940 this is known as the orthogonal

130 00:03:23,949 –> 00:03:25,800 projection of y onto

131 00:03:25,809 –> 00:03:26,470 x.

132 00:03:27,309 –> 00:03:28,550 So it makes sense to do the

133 00:03:28,559 –> 00:03:30,020 same for a general inner

134 00:03:30,029 –> 00:03:30,619 product.

135 00:03:31,460 –> 00:03:33,119 Now the gray line here is

136 00:03:33,130 –> 00:03:34,639 the orthogonal part, which

137 00:03:34,649 –> 00:03:35,850 we also can calculate

138 00:03:35,860 –> 00:03:37,630 now. This is

139 00:03:37,639 –> 00:03:39,050 simply given by y

140 00:03:39,059 –> 00:03:40,729 minus the parallel part.

141 00:03:41,520 –> 00:03:41,880 OK.

142 00:03:41,889 –> 00:03:43,220 So we defined some vectors

143 00:03:43,229 –> 00:03:44,300 we want to deal with,

144 00:03:44,449 –> 00:03:46,380 but now we need an idea how

145 00:03:46,389 –> 00:03:47,850 to get to the inequality.

146 00:03:48,580 –> 00:03:50,100 Indeed, the whole idea is

147 00:03:50,110 –> 00:03:51,160 that we can calculate the

148 00:03:51,169 –> 00:03:52,940 length or the norm of

149 00:03:52,949 –> 00:03:54,240 this orthogonal part,

150 00:03:55,179 –> 00:03:56,910 simply because we know the

151 00:03:56,919 –> 00:03:58,830 norm can’t be negative.

152 00:03:59,830 –> 00:04:01,399 Now, on the right we can put

153 00:04:01,410 –> 00:04:03,389 in the definition and then

154 00:04:03,399 –> 00:04:05,149 of course, also the definition

155 00:04:05,160 –> 00:04:06,619 of y parallel.

156 00:04:08,070 –> 00:04:09,169 Now if you see something

157 00:04:09,179 –> 00:04:11,089 like this. Calculating norms,

158 00:04:11,100 –> 00:04:12,570 but you have an inner product;

159 00:04:12,600 –> 00:04:14,490 it’s better to use squares.

160 00:04:15,149 –> 00:04:16,769 So square everything and

161 00:04:16,779 –> 00:04:18,309 the square roots will vanish

162 00:04:19,248 –> 00:04:20,878 and now you can see on the

163 00:04:20,889 –> 00:04:22,169 right-hand side, we have

164 00:04:22,178 –> 00:04:23,868 this long vector in the middle,

165 00:04:24,058 –> 00:04:25,658 in both components of the

166 00:04:25,669 –> 00:04:26,378 inner product.

167 00:04:27,299 –> 00:04:28,720 So it looks like this

168 00:04:29,989 –> 00:04:31,170 and in order to simplify

169 00:04:31,179 –> 00:04:32,500 this, we will use the

170 00:04:32,510 –> 00:04:33,410 linearity

171 00:04:34,209 –> 00:04:35,390 and this linearity in the

172 00:04:35,399 –> 00:04:36,950 second component means we

173 00:04:36,959 –> 00:04:38,649 can pull out this minus

174 00:04:38,660 –> 00:04:39,350 sign here.

175 00:04:39,929 –> 00:04:41,149 So we have here the product

176 00:04:41,160 –> 00:04:43,000 with y minus the other

177 00:04:43,010 –> 00:04:43,380 part.

178 00:04:44,119 –> 00:04:45,390 In the next step, we want

179 00:04:45,399 –> 00:04:46,730 to pull out the minus sign

180 00:04:46,739 –> 00:04:48,570 here and here, and we

181 00:04:48,579 –> 00:04:50,510 know we can do that, because

182 00:04:50,519 –> 00:04:51,679 the inner product is

183 00:04:51,690 –> 00:04:53,250 conjugate linear in the

184 00:04:53,260 –> 00:04:54,369 first argument.

185 00:04:54,989 –> 00:04:56,160 So minus signs are not a

186 00:04:56,170 –> 00:04:56,769 problem

187 00:04:56,779 –> 00:04:58,230 and scalars get a complex

188 00:04:58,239 –> 00:04:58,989 conjugation.

189 00:04:59,570 –> 00:05:00,839 So that’s the first part

190 00:05:00,850 –> 00:05:01,750 and the second part would

191 00:05:01,760 –> 00:05:03,480 be this vector with y

192 00:05:04,239 –> 00:05:05,380 and of course, now we do

193 00:05:05,390 –> 00:05:06,470 the same with the second

194 00:05:06,480 –> 00:05:07,480 inner product here.

195 00:05:08,829 –> 00:05:09,179 OK.

196 00:05:09,190 –> 00:05:11,049 Here we have y with the vector

197 00:05:11,059 –> 00:05:12,100 on the right-hand side

198 00:05:12,140 –> 00:05:13,730 and now we have minus minus.

199 00:05:13,739 –> 00:05:15,450 So plus the rest.

200 00:05:16,399 –> 00:05:17,600 In fact, here we have the

201 00:05:17,609 –> 00:05:19,019 same vector left and right

202 00:05:19,029 –> 00:05:19,959 in the inner product.

203 00:05:20,619 –> 00:05:21,679 Therefore, now comes the

204 00:05:21,690 –> 00:05:23,160 part where we can rewrite

205 00:05:23,170 –> 00:05:23,630 everything

206 00:05:23,640 –> 00:05:24,109 again.

207 00:05:24,640 –> 00:05:26,070 The first thing is the norm

208 00:05:26,079 –> 00:05:27,190 of y squared.

209 00:05:27,970 –> 00:05:28,880 For the two parts in the

210 00:05:28,890 –> 00:05:30,549 middle, we see they are almost

211 00:05:30,559 –> 00:05:31,899 the same. The one is the

212 00:05:31,910 –> 00:05:33,260 complex conjugate of the

213 00:05:33,269 –> 00:05:33,660 other one.

214 00:05:34,390 –> 00:05:35,440 So we can write it in this

215 00:05:35,450 –> 00:05:36,839 way with parentheses, where we

216 00:05:36,850 –> 00:05:38,209 put a bar over the second

217 00:05:38,220 –> 00:05:38,640 part

218 00:05:39,459 –> 00:05:41,079 and finally, the last part

219 00:05:41,089 –> 00:05:42,760 is the norm of this vector

220 00:05:42,769 –> 00:05:43,440 squared.

221 00:05:44,380 –> 00:05:45,760 In the next equality here,

222 00:05:45,769 –> 00:05:47,079 I want to simplify that even

223 00:05:47,089 –> 00:05:48,709 more, because you see

224 00:05:48,720 –> 00:05:50,450 here’s a complex number plus

225 00:05:50,459 –> 00:05:51,799 the complex conjugate of

226 00:05:51,809 –> 00:05:53,350 this number. Which means

227 00:05:53,359 –> 00:05:54,910 this is two times the real

228 00:05:54,920 –> 00:05:56,559 part of this complex

229 00:05:56,570 –> 00:05:58,359 number. Which means that

230 00:05:58,369 –> 00:05:59,869 we can write it like this

231 00:06:00,790 –> 00:06:01,970 and now in the third part,

232 00:06:01,980 –> 00:06:03,500 we can pull out the scalar.

233 00:06:03,940 –> 00:06:05,149 Which means it comes out

234 00:06:05,160 –> 00:06:06,850 with the absolute value squared.

235 00:06:07,649 –> 00:06:09,299 You see it remains the norm

236 00:06:09,309 –> 00:06:10,799 of x hat where we already

237 00:06:10,809 –> 00:06:12,279 know, this is 1.

238 00:06:12,529 –> 00:06:14,339 Now, finally, I want to simplify

239 00:06:14,350 –> 00:06:15,769 the middle part here, which

240 00:06:15,779 –> 00:06:16,839 to be honest, we could have

241 00:06:16,850 –> 00:06:17,750 done before.

242 00:06:17,760 –> 00:06:18,970 But then we would have done

243 00:06:18,980 –> 00:06:19,880 it two times.

244 00:06:20,649 –> 00:06:22,190 I want to show you visually

245 00:06:22,200 –> 00:06:23,350 what we want to do.

246 00:06:23,540 –> 00:06:25,230 We pull this scalar out

247 00:06:25,239 –> 00:06:26,799 from the first argument of

248 00:06:26,809 –> 00:06:28,510 the inner product, but

249 00:06:28,519 –> 00:06:29,980 then it gets a complex

250 00:06:29,989 –> 00:06:30,679 conjugation.

251 00:06:31,549 –> 00:06:32,799 So let’s fill in the gaps

252 00:06:32,809 –> 00:06:33,220 again

253 00:06:33,230 –> 00:06:34,529 and then you see we have

254 00:06:34,540 –> 00:06:36,100 the same complex number left

255 00:06:36,109 –> 00:06:36,739 and right.

256 00:06:36,750 –> 00:06:38,130 But the one is the complex

257 00:06:38,140 –> 00:06:39,480 conjugate of the other one.

258 00:06:40,290 –> 00:06:41,510 In other words, it’s the

259 00:06:41,519 –> 00:06:42,970 absolute value of the complex

260 00:06:42,980 –> 00:06:44,149 number squared

261 00:06:44,880 –> 00:06:46,359 and since that is clearly

262 00:06:46,369 –> 00:06:47,940 a real number, we can omit

263 00:06:47,950 –> 00:06:48,959 the real part here.

264 00:06:49,769 –> 00:06:50,989 So in summary, we have the

265 00:06:51,000 –> 00:06:52,899 norm of y squared minus two

266 00:06:52,910 –> 00:06:54,309 times this number,

267 00:06:54,910 –> 00:06:56,500 but also plus one

268 00:06:56,510 –> 00:06:57,869 times the same number.

269 00:06:58,869 –> 00:07:00,089 Therefore, we can put both

270 00:07:00,100 –> 00:07:01,420 things together and have

271 00:07:01,429 –> 00:07:02,600 a simple result here.

272 00:07:03,410 –> 00:07:05,369 The norm of y squared minus

273 00:07:05,380 –> 00:07:06,720 the inner product in absolute

274 00:07:06,730 –> 00:07:07,540 value squared.

275 00:07:08,359 –> 00:07:09,720 So this is our right-hand

276 00:07:09,730 –> 00:07:11,529 side and on the left, we

277 00:07:11,540 –> 00:07:12,339 have the zero.

278 00:07:13,600 –> 00:07:14,739 Hence, in the next step,

279 00:07:14,750 –> 00:07:15,660 I want to bring the inner

280 00:07:15,670 –> 00:07:17,339 product to the other side.

281 00:07:18,380 –> 00:07:20,079 Now, you can see we are almost

282 00:07:20,089 –> 00:07:21,670 there with our inequality.

283 00:07:21,679 –> 00:07:22,859 The only thing missing is

284 00:07:22,869 –> 00:07:23,929 that we have to translate

285 00:07:23,940 –> 00:07:24,959 back x hat

286 00:07:25,820 –> 00:07:27,119 which is not so hard to do,

287 00:07:27,130 –> 00:07:28,920 because we defined it as x

288 00:07:28,929 –> 00:07:30,670 divided by the norm of x.

289 00:07:31,359 –> 00:07:32,399 Of course, we do the same

290 00:07:32,410 –> 00:07:34,190 as always, we pull out one

291 00:07:34,200 –> 00:07:35,799 divided by the norm of x.

292 00:07:36,359 –> 00:07:37,959 Hence, that’s what we get

293 00:07:37,970 –> 00:07:39,239 and you see, we just have

294 00:07:39,250 –> 00:07:40,839 to multiply with the norm

295 00:07:40,850 –> 00:07:42,600 of x squared on both sides

296 00:07:43,679 –> 00:07:45,010 and then taking the square

297 00:07:45,019 –> 00:07:46,579 root on both sides, we get

298 00:07:46,589 –> 00:07:48,070 out our Cauchy-Schwarz

299 00:07:48,079 –> 00:07:48,980 inequality.

300 00:07:49,549 –> 00:07:51,130 Well, that’s the whole proof

301 00:07:51,140 –> 00:07:52,279 of the inequality.

302 00:07:52,510 –> 00:07:54,170 I will skip the part about

303 00:07:54,179 –> 00:07:55,970 the equality now, because

304 00:07:55,980 –> 00:07:57,790 I want to use the time to show

305 00:07:57,799 –> 00:07:58,950 you that the symbol we used

306 00:07:58,959 –> 00:08:00,600 here is indeed a norm.

307 00:08:00,609 –> 00:08:02,059 So it fulfills the triangle

308 00:08:02,070 –> 00:08:03,029 inequality

309 00:08:03,190 –> 00:08:04,700 and we will do that by

310 00:08:04,709 –> 00:08:06,220 using the Cauchy-Schwarz

311 00:08:06,230 –> 00:08:07,059 inequality.

312 00:08:07,920 –> 00:08:09,880 In fact, the triangle inequality

313 00:08:09,890 –> 00:08:11,200 is most of the time, the

314 00:08:11,209 –> 00:08:12,820 hardest part of the proof that

315 00:08:12,829 –> 00:08:13,989 something is a norm.

316 00:08:15,179 –> 00:08:16,779 So let’s do that very quickly.

317 00:08:16,790 –> 00:08:18,059 Of course, we square the

318 00:08:18,070 –> 00:08:18,839 norm again,

319 00:08:19,820 –> 00:08:21,119 because then we don’t need

320 00:08:21,130 –> 00:08:22,369 the square root. we can just

321 00:08:22,380 –> 00:08:24,350 write x plus y, x

322 00:08:24,359 –> 00:08:26,000 plus y in the inner product.

323 00:08:26,420 –> 00:08:28,019 The linearity now gets us

324 00:08:28,029 –> 00:08:29,920 to a similar result as before.

325 00:08:30,809 –> 00:08:32,750 Namely the norm of x and

326 00:08:32,760 –> 00:08:33,619 y squared

327 00:08:33,630 –> 00:08:34,799 and in the middle, we get

328 00:08:34,808 –> 00:08:36,479 two times the real part of

329 00:08:36,489 –> 00:08:37,239 the product.

330 00:08:37,890 –> 00:08:39,210 Now, of course, if we use

331 00:08:39,219 –> 00:08:40,409 the absolute value instead

332 00:08:40,419 –> 00:08:42,020 of the real part, it gets

333 00:08:42,030 –> 00:08:42,489 bigger

334 00:08:43,369 –> 00:08:44,710 and of course, here we now

335 00:08:44,719 –> 00:08:46,599 can use our Cauchy-Schwarz

336 00:08:46,609 –> 00:08:47,510 inequality.

337 00:08:48,080 –> 00:08:49,630 And now knowing the binomial

338 00:08:49,640 –> 00:08:51,109 theorem, you see this is

339 00:08:51,119 –> 00:08:52,669 just the square of the sum

340 00:08:52,679 –> 00:08:53,729 of the two norms.

341 00:08:54,320 –> 00:08:55,419 Taking the square root on

342 00:08:55,429 –> 00:08:56,109 both sides,

343 00:08:56,119 –> 00:08:57,640 you see this is simply the

344 00:08:57,650 –> 00:08:59,130 triangle inequality for the

345 00:08:59,140 –> 00:09:00,820 norm we defined by the inner

346 00:09:00,830 –> 00:09:01,349 product.

347 00:09:02,169 –> 00:09:03,210 The other two properties

348 00:09:03,219 –> 00:09:04,609 for a norm shouldn’t be a problem

349 00:09:04,619 –> 00:09:05,250 to show

350 00:09:05,299 –> 00:09:06,789 and therefore now it makes

351 00:09:06,799 –> 00:09:08,650 sense to use this symbol

352 00:09:08,659 –> 00:09:09,770 and call it a norm in an

353 00:09:09,780 –> 00:09:11,229 inner product space.

354 00:09:11,770 –> 00:09:12,229 OK.

355 00:09:12,239 –> 00:09:13,190 I think that’s good enough

356 00:09:13,200 –> 00:09:13,909 for today.

357 00:09:13,929 –> 00:09:15,169 In the next videos, we will

358 00:09:15,179 –> 00:09:16,590 talk a little bit more about

359 00:09:16,599 –> 00:09:17,669 the geometry

360 00:09:17,679 –> 00:09:18,989 an inner product gives the

361 00:09:19,000 –> 00:09:20,780 space and I

362 00:09:20,789 –> 00:09:22,280 already told you, if you need

363 00:09:22,289 –> 00:09:23,650 it, you can download the

364 00:09:23,659 –> 00:09:25,169 PDF version of this video

365 00:09:25,179 –> 00:09:27,159 in the description and of

366 00:09:27,169 –> 00:09:28,679 course use the comments, if

367 00:09:28,690 –> 00:09:29,599 you have questions.

368 00:09:29,719 –> 00:09:30,979 Otherwise I see you in the

369 00:09:30,989 –> 00:09:31,609 next video.

370 00:09:32,590 –> 00:09:33,609 Have a nice day.

371 00:09:33,700 –> 00:09:34,289 Bye.

Q1: Let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space and $x,y \in X$ with $\langle x, x \rangle = \langle y, y \rangle = 1$. What is always correct?

A1: $|\langle x, y \rangle| \leq 1$

A2: $|\langle x, y \rangle| \geq 1$

A3: $|\langle x, y \rangle| = 1$

A4: $|\langle x, y \rangle|^2 \geq \langle x, x \rangle + \langle y, y \rangle $

A5: $|\langle x, y \rangle|^2 \geq \langle x, x \rangle \cdot \langle y, y \rangle $

A6: $|\langle x, y \rangle|^2 \leq \frac{1}{2} \langle x, x \rangle \cdot \langle y, y \rangle $

Q2: Let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space and $x,y \in X$ with $\langle x, x \rangle = \langle y, y \rangle = 1$. Which statement is not true?

A1: If $\langle x, y \rangle = 1$, then $x$ and $y$ are linearly dependent.

A2: If $x$ and $y$ are linearly dependent, then $\langle x, y \rangle = 1$.

A3: If $x$ and $y$ are linearly independent, then $\langle x, y \rangle = 0$.

A4: If $\langle x, y \rangle = 0$, then $x$ and $y$ are linearly independent.

Q3: Let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space with $\dim(X) \geq 1$. Is the map $x \mapsto \langle x, x\rangle$ a norm?

A1: Yes it is!

A2: No, it’s not positive definite.

A3: No, it’s not homogenous.

A4: One needs more information.