00:00 Introduction

00:39 Change-of-basis matrix

01:58 Example - explanation

02:50 Example in R²

05:10 First transition matrix

05:54 Second transition matrix

06:39 Composition of change-of-basis

07:51 Gaussian elimination for product

11:41 Credits

1 00:00:00,419 –> 00:00:02,400 Hello and welcome back to

2 00:00:02,410 –> 00:00:04,239 abstract linear algebra,

3 00:00:04,360 –> 00:00:05,579 the video series where we

4 00:00:05,590 –> 00:00:07,539 talk a lot about some advanced

5 00:00:07,550 –> 00:00:09,239 topics in linear algebra.

6 00:00:09,640 –> 00:00:11,479 And now in today’s part nine,

7 00:00:11,489 –> 00:00:13,130 I want to talk more about

8 00:00:13,140 –> 00:00:14,680 the so-called change of

9 00:00:14,689 –> 00:00:15,340 basis.

10 00:00:15,689 –> 00:00:17,030 More concretely, we will

11 00:00:17,040 –> 00:00:18,760 look at a common example.

12 00:00:19,319 –> 00:00:20,940 However, as always before

13 00:00:20,950 –> 00:00:22,360 we start, I really want to

14 00:00:22,370 –> 00:00:23,760 thank all the nice people

15 00:00:23,770 –> 00:00:24,770 who support this channel

16 00:00:24,780 –> 00:00:26,340 on Steady, here on youtube

17 00:00:26,350 –> 00:00:27,549 or via Patreon.

18 00:00:28,100 –> 00:00:29,920 And moreover, as a reward,

19 00:00:29,930 –> 00:00:31,370 you find a lot of additional

20 00:00:31,379 –> 00:00:33,200 material for all the videos

21 00:00:33,209 –> 00:00:34,880 with the link in the description.

22 00:00:35,470 –> 00:00:36,799 So if you’re interested in

23 00:00:36,810 –> 00:00:38,560 learning mathematics, please

24 00:00:38,569 –> 00:00:39,319 have a look.

25 00:00:39,900 –> 00:00:41,189 OK, then let’s immediately

26 00:00:41,200 –> 00:00:42,599 start with the topic of today

27 00:00:42,610 –> 00:00:44,220 by recalling the so

28 00:00:44,229 –> 00:00:45,659 called change of basis

29 00:00:45,669 –> 00:00:46,340 matrix.

30 00:00:46,840 –> 00:00:48,290 We also know it has some

31 00:00:48,299 –> 00:00:50,020 other names like transformation

32 00:00:50,029 –> 00:00:51,340 matrix or transition

33 00:00:51,349 –> 00:00:52,099 matrix.

34 00:00:52,470 –> 00:00:53,959 However, the important fact

35 00:00:53,970 –> 00:00:55,830 is it’s a square matrix

36 00:00:55,840 –> 00:00:57,790 that transforms vectors given

37 00:00:57,799 –> 00:00:59,479 in the basis C into

38 00:00:59,490 –> 00:01:01,220 vectors given in the basis

39 00:01:01,229 –> 00:01:01,700 B.

40 00:01:02,349 –> 00:01:03,869 Hence a column vector that

41 00:01:03,880 –> 00:01:05,480 comes in here from the right

42 00:01:05,489 –> 00:01:06,769 hand side should be

43 00:01:06,779 –> 00:01:08,529 interpreted as a vector

44 00:01:08,540 –> 00:01:10,150 represented with respect

45 00:01:10,160 –> 00:01:11,300 to the basis C.

46 00:01:11,839 –> 00:01:12,889 And then the column vector

47 00:01:12,900 –> 00:01:14,120 that comes out is

48 00:01:14,129 –> 00:01:15,809 represented with respect

49 00:01:15,819 –> 00:01:16,800 to the basis B.

50 00:01:17,480 –> 00:01:18,940 So please recall they are

51 00:01:18,949 –> 00:01:20,879 just the images with respect

52 00:01:20,889 –> 00:01:22,839 to the basis isomorphisms.

53 00:01:23,339 –> 00:01:24,720 So usually we say they

54 00:01:24,730 –> 00:01:26,339 represent the same vector

55 00:01:26,349 –> 00:01:27,870 V in our abstract vector

56 00:01:27,879 –> 00:01:28,529 space.

57 00:01:28,540 –> 00:01:30,199 However, with respect to

58 00:01:30,209 –> 00:01:31,959 different chosen bases.

59 00:01:32,629 –> 00:01:34,529 So now we know if the abstract

60 00:01:34,540 –> 00:01:36,250 vector space V has dimension

61 00:01:36,260 –> 00:01:38,129 N, then this change of base

62 00:01:38,139 –> 00:01:40,129 matrix is an N times N

63 00:01:40,139 –> 00:01:40,849 matrix.

64 00:01:41,230 –> 00:01:43,000 Moreover, we also know

65 00:01:43,010 –> 00:01:44,849 it’s an invertible matrix.

66 00:01:45,440 –> 00:01:47,089 Of course, this makes sense

67 00:01:47,099 –> 00:01:48,730 because we should be able

68 00:01:48,739 –> 00:01:50,290 to translate back again.

69 00:01:51,019 –> 00:01:51,510 OK.

70 00:01:51,519 –> 00:01:53,010 And now in this video, we

71 00:01:53,019 –> 00:01:54,419 will look at a very concrete

72 00:01:54,430 –> 00:01:56,300 example for such a change

73 00:01:56,309 –> 00:01:57,559 of basis matrix.

74 00:01:58,180 –> 00:01:59,699 Indeed, it’s a very common

75 00:01:59,709 –> 00:02:01,389 one and you can always use

76 00:02:01,400 –> 00:02:02,889 it if you want to change

77 00:02:02,900 –> 00:02:04,540 the basis in RN.

78 00:02:05,389 –> 00:02:06,849 This means in our picture

79 00:02:06,860 –> 00:02:08,258 here, the abstract vector

80 00:02:08,270 –> 00:02:10,130 space V is now given

81 00:02:10,139 –> 00:02:11,210 by R two.

82 00:02:12,039 –> 00:02:13,580 And at this point, you might

83 00:02:13,589 –> 00:02:15,509 argue it’s not needed to

84 00:02:15,520 –> 00:02:16,979 have this abstract picture

85 00:02:16,990 –> 00:02:18,979 here because the vector space

86 00:02:18,990 –> 00:02:20,690 is already given by the

87 00:02:20,699 –> 00:02:22,600 very concrete one R two.

88 00:02:23,220 –> 00:02:24,770 However, the basis

89 00:02:24,779 –> 00:02:26,619 isomorphism are still

90 00:02:26,630 –> 00:02:28,490 helpful to understand what

91 00:02:28,500 –> 00:02:29,410 is happening here.

92 00:02:30,149 –> 00:02:31,720 This means if we have two

93 00:02:31,729 –> 00:02:33,389 bases given we can

94 00:02:33,399 –> 00:02:35,320 still translate them to R

95 00:02:35,330 –> 00:02:35,929 two again.

96 00:02:36,679 –> 00:02:38,559 Hence we land in R two

97 00:02:38,570 –> 00:02:39,130 again.

98 00:02:39,139 –> 00:02:40,800 But now with the canonical

99 00:02:40,809 –> 00:02:42,490 basis and as

100 00:02:42,500 –> 00:02:44,050 always the corresponding

101 00:02:44,059 –> 00:02:46,000 basis isomorphisms are

102 00:02:46,009 –> 00:02:47,919 denoted by Phi B and

103 00:02:47,929 –> 00:02:49,479 Phi C respectively.

104 00:02:50,179 –> 00:02:50,539 OK.

105 00:02:50,550 –> 00:02:51,800 Now let’s look at a very

106 00:02:51,809 –> 00:02:53,679 concrete example, let’s say

107 00:02:53,690 –> 00:02:55,070 B is given by two

108 00:02:55,080 –> 00:02:56,699 vectors namely

109 00:02:56,710 –> 00:02:58,029 1 2 and

110 00:02:58,039 –> 00:02:59,000 3 4.

111 00:02:59,529 –> 00:03:00,949 And on the other hand, we

112 00:03:00,960 –> 00:03:02,720 also define C by

113 00:03:02,729 –> 00:03:03,839 two vectors.

114 00:03:04,479 –> 00:03:05,720 And here we choose

115 00:03:05,729 –> 00:03:07,660 1 0 2 2.

116 00:03:08,199 –> 00:03:09,259 The first thing you should

117 00:03:09,270 –> 00:03:10,710 note is that both are

118 00:03:10,720 –> 00:03:12,500 bases in R two.

119 00:03:13,190 –> 00:03:14,850 And maybe you chose spaces

120 00:03:14,860 –> 00:03:16,809 B because it’s helpful to

121 00:03:16,820 –> 00:03:18,490 solve a particular problem,

122 00:03:18,500 –> 00:03:20,119 but then you have to solve

123 00:03:20,130 –> 00:03:21,289 another problem.

124 00:03:21,309 –> 00:03:22,770 And then it turns out that

125 00:03:22,779 –> 00:03:23,990 the basis C is

126 00:03:24,000 –> 00:03:24,800 beneficial.

127 00:03:25,399 –> 00:03:26,929 Therefore, the only thing

128 00:03:26,940 –> 00:03:28,520 you need is the change of

129 00:03:28,529 –> 00:03:29,990 basis matrix here.

130 00:03:30,000 –> 00:03:31,130 On the lower level

131 00:03:31,809 –> 00:03:33,619 with this matrix T, you can

132 00:03:33,630 –> 00:03:35,619 do all the translations between

133 00:03:35,630 –> 00:03:36,500 both sides.

134 00:03:37,190 –> 00:03:38,759 However, now, in order to

135 00:03:38,770 –> 00:03:40,699 calculate this matrix, it’s

136 00:03:40,710 –> 00:03:42,270 helpful to introduce a

137 00:03:42,279 –> 00:03:44,050 third basis as well.

138 00:03:44,679 –> 00:03:46,380 The reason for this is that

139 00:03:46,389 –> 00:03:48,259 we already have a very nice

140 00:03:48,270 –> 00:03:49,619 basis in R two,

141 00:03:50,429 –> 00:03:52,229 namely we have the canonical

142 00:03:52,240 –> 00:03:53,580 basis, the standard

143 00:03:53,589 –> 00:03:54,339 basis.

144 00:03:54,429 –> 00:03:56,160 And I will denote this with

145 00:03:56,169 –> 00:03:57,130 a curved E

146 00:03:57,970 –> 00:03:59,789 now as a reminder E is

147 00:03:59,800 –> 00:04:00,679 just given by

148 00:04:00,690 –> 00:04:02,669 1 0 0 1.

149 00:04:03,490 –> 00:04:05,240 So you know, this is exactly

150 00:04:05,250 –> 00:04:07,160 the basis we want to have

151 00:04:07,169 –> 00:04:08,589 here on the lower level.

152 00:04:09,330 –> 00:04:10,919 Therefore, our basis

153 00:04:10,929 –> 00:04:12,830 isomorphism in this case

154 00:04:12,839 –> 00:04:14,389 does not change anything.

155 00:04:15,050 –> 00:04:16,950 So Phi E is simply

156 00:04:16,959 –> 00:04:18,670 given by the identity map.

157 00:04:19,410 –> 00:04:20,890 And exactly this fact

158 00:04:20,899 –> 00:04:22,790 implies that we have two

159 00:04:22,799 –> 00:04:24,750 very simple transformation

160 00:04:24,760 –> 00:04:25,850 matrices here.

161 00:04:26,630 –> 00:04:28,149 And please note the first

162 00:04:28,160 –> 00:04:29,790 one here sends B to

163 00:04:29,799 –> 00:04:31,359 E and the second one

164 00:04:31,369 –> 00:04:33,000 here sends E to

165 00:04:33,010 –> 00:04:33,440 C.

166 00:04:34,089 –> 00:04:35,950 And in addition, the composition

167 00:04:35,959 –> 00:04:37,890 of both matrices is

168 00:04:37,899 –> 00:04:39,890 exactly our TCB

169 00:04:40,690 –> 00:04:42,309 and exactly with this matrix

170 00:04:42,320 –> 00:04:43,640 product, we want to

171 00:04:43,649 –> 00:04:45,410 calculate this transformation

172 00:04:45,420 –> 00:04:46,019 matrix.

173 00:04:46,920 –> 00:04:47,410 OK.

174 00:04:47,420 –> 00:04:49,040 Then I would say let’s write

175 00:04:49,049 –> 00:04:50,730 down what we already

176 00:04:50,739 –> 00:04:52,489 know, namely,

177 00:04:52,500 –> 00:04:54,130 we already know the

178 00:04:54,140 –> 00:04:56,040 matrix T EB

179 00:04:56,480 –> 00:04:58,130 because it reads that the

180 00:04:58,140 –> 00:04:59,920 basis B is to be

181 00:04:59,929 –> 00:05:01,570 expressed in the standard

182 00:05:01,579 –> 00:05:02,209 basis.

183 00:05:02,869 –> 00:05:04,390 Indeed, if you look at the

184 00:05:04,399 –> 00:05:06,279 basis B, you see the

185 00:05:06,290 –> 00:05:08,000 vectors are already given

186 00:05:08,010 –> 00:05:09,660 with respect to our standard

187 00:05:09,670 –> 00:05:11,630 basis more precisely.

188 00:05:11,640 –> 00:05:13,619 And by definition T E B

189 00:05:13,630 –> 00:05:15,359 is given by Phi

190 00:05:15,369 –> 00:05:17,049 E of b1

191 00:05:17,059 –> 00:05:19,049 in the first column and

192 00:05:19,059 –> 00:05:21,019 Phi E of b two in

193 00:05:21,029 –> 00:05:22,070 the second column.

194 00:05:22,529 –> 00:05:24,290 However, please don’t forget

195 00:05:24,299 –> 00:05:25,809 Phi E is just the

196 00:05:25,820 –> 00:05:26,769 identity map.

197 00:05:27,570 –> 00:05:28,709 In other words, we have a

198 00:05:28,720 –> 00:05:30,279 matrix here with b

199 00:05:30,290 –> 00:05:31,829 one and b two in the

200 00:05:31,839 –> 00:05:32,470 columns.

201 00:05:32,820 –> 00:05:34,559 Hence, we can just take the

202 00:05:34,570 –> 00:05:36,350 basis vectors as column

203 00:05:36,359 –> 00:05:38,260 vectors and put them into

204 00:05:38,269 –> 00:05:39,739 a two times two matrix.

205 00:05:40,250 –> 00:05:41,769 So for our example here,

206 00:05:41,779 –> 00:05:42,989 it means we have

207 00:05:43,000 –> 00:05:45,119 1 2 3 4.

208 00:05:46,179 –> 00:05:47,730 In other words, by writing

209 00:05:47,739 –> 00:05:49,570 down the basis B like this,

210 00:05:49,619 –> 00:05:51,440 we already know we have

211 00:05:51,450 –> 00:05:53,420 this change of basis matrix.

212 00:05:54,140 –> 00:05:55,929 And in the same way, we

213 00:05:55,940 –> 00:05:57,799 also have this for the basis

214 00:05:57,809 –> 00:05:59,559 C represented in the

215 00:05:59,570 –> 00:06:01,040 canonical basis as well.

216 00:06:01,790 –> 00:06:03,570 So this is exactly the inverse

217 00:06:03,579 –> 00:06:05,130 of the matrix we actually

218 00:06:05,140 –> 00:06:06,170 need in the end.

219 00:06:06,839 –> 00:06:08,260 However, this is also

220 00:06:08,269 –> 00:06:09,940 something we should already

221 00:06:09,950 –> 00:06:10,820 write down.

222 00:06:11,329 –> 00:06:13,320 So let’s write down T E C

223 00:06:13,510 –> 00:06:15,459 and now with the basis vectors

224 00:06:15,470 –> 00:06:17,429 of C, which

225 00:06:17,440 –> 00:06:18,910 as you might recall were

226 00:06:18,920 –> 00:06:19,709 given as

227 00:06:19,720 –> 00:06:21,859 1 0 2 2.

228 00:06:22,570 –> 00:06:23,070 OK.

229 00:06:23,079 –> 00:06:24,630 And with that, you see, we

230 00:06:24,640 –> 00:06:26,010 have all the information

231 00:06:26,019 –> 00:06:27,540 we need to actually

232 00:06:27,549 –> 00:06:29,329 calculate the matrix we

233 00:06:29,339 –> 00:06:29,790 want.

234 00:06:30,480 –> 00:06:31,799 And now I want to show you

235 00:06:31,809 –> 00:06:33,600 the most efficient way to

236 00:06:33,609 –> 00:06:34,140 do it.

237 00:06:34,929 –> 00:06:36,179 Indeed, it’s simply a

238 00:06:36,190 –> 00:06:37,820 composition where we have

239 00:06:37,829 –> 00:06:39,630 one inverse involved.

240 00:06:39,950 –> 00:06:41,470 However, first you should

241 00:06:41,480 –> 00:06:43,339 see that the change of basis

242 00:06:43,350 –> 00:06:45,220 matrices nicely fit

243 00:06:45,230 –> 00:06:46,970 together with their indices.

244 00:06:47,720 –> 00:06:49,630 For example, here, if B

245 00:06:49,640 –> 00:06:51,200 goes in from the right hand

246 00:06:51,209 –> 00:06:52,929 side, it also has to

247 00:06:52,940 –> 00:06:54,679 go in from the right here

248 00:06:54,690 –> 00:06:55,880 on the other side.

249 00:06:56,679 –> 00:06:58,519 And moreover, if the basis

250 00:06:58,529 –> 00:07:00,390 C comes out on the left

251 00:07:00,399 –> 00:07:02,350 here, it also has to come

252 00:07:02,359 –> 00:07:03,200 out here.

253 00:07:04,049 –> 00:07:05,459 So whenever you want to put

254 00:07:05,470 –> 00:07:07,019 in a third matrix in the

255 00:07:07,029 –> 00:07:08,850 middle, you just have a matrix

256 00:07:08,859 –> 00:07:10,739 product where E is

257 00:07:10,750 –> 00:07:12,339 also in the middle here.

258 00:07:13,160 –> 00:07:14,450 However, now for our

259 00:07:14,459 –> 00:07:16,200 calculation, we see that

260 00:07:16,209 –> 00:07:18,100 this matrix here is the

261 00:07:18,109 –> 00:07:20,070 inverse of a matrix we already

262 00:07:20,079 –> 00:07:20,470 know.

263 00:07:20,950 –> 00:07:22,809 Hence, we just have to calculate

264 00:07:22,820 –> 00:07:24,190 the inverse here and

265 00:07:24,200 –> 00:07:25,790 multiply the result

266 00:07:25,799 –> 00:07:27,070 with this matrix

267 00:07:27,570 –> 00:07:27,829 T E B.

268 00:07:28,540 –> 00:07:30,359 Now, of course, in this particular

269 00:07:30,369 –> 00:07:31,929 example here, it’s no

270 00:07:31,940 –> 00:07:33,799 problem at all to calculate

271 00:07:33,809 –> 00:07:35,410 the inverse of this two times

272 00:07:35,420 –> 00:07:36,200 two matrix.

273 00:07:36,920 –> 00:07:38,079 However, in higher

274 00:07:38,089 –> 00:07:40,010 dimensions, an inverse can

275 00:07:40,019 –> 00:07:41,529 make a lot of problems

276 00:07:41,540 –> 00:07:43,429 because calculating it is

277 00:07:43,440 –> 00:07:44,309 a lot of work.

278 00:07:45,000 –> 00:07:46,700 And actually, we don’t need

279 00:07:46,709 –> 00:07:48,519 to know the inverse because

280 00:07:48,529 –> 00:07:50,190 we just need to know this

281 00:07:50,200 –> 00:07:51,059 product here.

282 00:07:51,799 –> 00:07:53,209 Therefore, the efficient

283 00:07:53,220 –> 00:07:54,959 way is to calculate this

284 00:07:54,970 –> 00:07:56,529 product in one go.

285 00:07:57,260 –> 00:07:58,609 Indeed you might already

286 00:07:58,619 –> 00:08:00,309 know how to do that because

287 00:08:00,320 –> 00:08:02,279 it’s just a Gaussian elimination.

288 00:08:02,769 –> 00:08:04,170 However, for the sake of

289 00:08:04,179 –> 00:08:06,070 completeness, I want to show

290 00:08:06,079 –> 00:08:07,250 you how to do it.

291 00:08:08,089 –> 00:08:09,420 Now, the first thing you

292 00:08:09,429 –> 00:08:10,709 should note here is that

293 00:08:10,720 –> 00:08:12,109 in order to calculate this

294 00:08:12,119 –> 00:08:14,010 product, we need to solve

295 00:08:14,019 –> 00:08:14,670 an equation.

296 00:08:15,410 –> 00:08:16,940 In fact, the solution of

297 00:08:16,950 –> 00:08:18,619 this matrix equation should

298 00:08:18,630 –> 00:08:19,850 be exactly this

299 00:08:19,859 –> 00:08:21,459 transformation matrix here.

300 00:08:22,149 –> 00:08:23,450 And maybe let’s call this

301 00:08:23,459 –> 00:08:23,910 solution.

302 00:08:23,920 –> 00:08:25,730 We want simply X.

303 00:08:26,470 –> 00:08:27,850 And now if we bring this

304 00:08:27,859 –> 00:08:29,410 matrix here to the left hand

305 00:08:29,420 –> 00:08:31,130 side, we simply have

306 00:08:31,140 –> 00:08:32,859 T E C times X

307 00:08:32,869 –> 00:08:33,929 is equal to

308 00:08:33,940 –> 00:08:35,599 T E B so

309 00:08:35,609 –> 00:08:37,130 exactly this is the

310 00:08:37,140 –> 00:08:38,880 matrix equation we want to

311 00:08:38,890 –> 00:08:39,429 solve.

312 00:08:40,090 –> 00:08:41,169 And now the important thing

313 00:08:41,179 –> 00:08:42,690 to note is that this is

314 00:08:42,700 –> 00:08:44,369 simply a system of

315 00:08:44,380 –> 00:08:45,770 linear equations.

316 00:08:46,169 –> 00:08:47,630 More precisely these are

317 00:08:47,640 –> 00:08:49,530 more systems packed together

318 00:08:49,599 –> 00:08:51,280 because we have more right

319 00:08:51,330 –> 00:08:52,130 hand sides.

320 00:08:53,030 –> 00:08:54,520 In this case here, you see

321 00:08:54,530 –> 00:08:56,179 we have exactly two

322 00:08:56,190 –> 00:08:57,729 right-hand sides because

323 00:08:57,739 –> 00:08:59,390 we have two column vectors.

324 00:09:00,059 –> 00:09:01,619 But we can simply do the

325 00:09:01,630 –> 00:09:03,460 Gaussian elimination for

326 00:09:03,469 –> 00:09:04,849 all right hand sides

327 00:09:04,859 –> 00:09:06,010 simultaneously.

328 00:09:06,799 –> 00:09:08,359 And indeed, this immediately

329 00:09:08,369 –> 00:09:09,969 saves a lot of work.

330 00:09:10,719 –> 00:09:11,119 OK.

331 00:09:11,130 –> 00:09:12,380 Now let’s write down what

332 00:09:12,390 –> 00:09:13,380 we have to solve.

333 00:09:13,390 –> 00:09:13,869 We have

334 00:09:13,880 –> 00:09:15,820 1 0 2 2 on the

335 00:09:15,830 –> 00:09:17,750 left hand side and

336 00:09:17,760 –> 00:09:19,469 1 2 3 4

337 00:09:19,479 –> 00:09:20,989 on the right hand side.

338 00:09:21,690 –> 00:09:22,179 OK.

339 00:09:22,190 –> 00:09:23,679 And now we can do the Gaussian

340 00:09:23,690 –> 00:09:25,549 elimination as always, which

341 00:09:25,559 –> 00:09:26,940 means we bring the left hand

342 00:09:26,950 –> 00:09:28,469 side into a row echelon

343 00:09:28,479 –> 00:09:29,049 form.

344 00:09:29,070 –> 00:09:30,530 But then we can also do the

345 00:09:30,539 –> 00:09:31,909 backward substitution.

346 00:09:32,669 –> 00:09:34,159 And indeed, in this example,

347 00:09:34,169 –> 00:09:35,919 we already have the row echelon

348 00:09:35,929 –> 00:09:36,460 form.

349 00:09:36,469 –> 00:09:37,950 So we just need to do the

350 00:09:37,960 –> 00:09:39,400 backward substitution.

351 00:09:39,890 –> 00:09:41,650 However, we want to do this

352 00:09:41,659 –> 00:09:43,289 inside this matrix

353 00:09:43,299 –> 00:09:44,960 notation, which

354 00:09:44,969 –> 00:09:46,630 essentially means in order

355 00:09:46,640 –> 00:09:48,419 to solve this system here,

356 00:09:48,429 –> 00:09:50,000 we need to generate an

357 00:09:50,010 –> 00:09:51,909 identity matrix on the left

358 00:09:51,919 –> 00:09:53,609 hand side because

359 00:09:53,619 –> 00:09:55,559 then the components of the

360 00:09:55,570 –> 00:09:57,349 solution X can be read

361 00:09:57,359 –> 00:09:58,719 on the right hand side.

362 00:09:59,580 –> 00:10:01,380 In fact, this is what produces

363 00:10:01,390 –> 00:10:03,330 a nice algorithm for us.

364 00:10:04,070 –> 00:10:04,469 OK.

365 00:10:04,479 –> 00:10:05,820 Here, our first step should

366 00:10:05,830 –> 00:10:07,559 be that we multiply the

367 00:10:07,570 –> 00:10:09,099 second row by one

368 00:10:09,109 –> 00:10:11,039 half because then

369 00:10:11,049 –> 00:10:13,030 we already have the one here

370 00:10:13,039 –> 00:10:14,520 in the right lower corner.

371 00:10:15,159 –> 00:10:16,500 And then in the next step,

372 00:10:16,510 –> 00:10:18,140 we want to generate a zero

373 00:10:18,150 –> 00:10:18,609 here.

374 00:10:19,369 –> 00:10:21,229 Hence, from the first row,

375 00:10:21,239 –> 00:10:23,000 we subtract two times the

376 00:10:23,010 –> 00:10:24,190 new second row.

377 00:10:24,840 –> 00:10:26,340 And then we have it, we have

378 00:10:26,349 –> 00:10:27,989 the identity matrix here

379 00:10:28,000 –> 00:10:29,330 on the left hand side

380 00:10:30,000 –> 00:10:31,419 and on the right hand side,

381 00:10:31,429 –> 00:10:33,119 we find minus one minus

382 00:10:33,130 –> 00:10:35,099 one here and there

383 00:10:35,109 –> 00:10:36,429 we have it, the system is

384 00:10:36,440 –> 00:10:37,719 completely solved.

385 00:10:37,729 –> 00:10:38,900 And with this backward

386 00:10:38,909 –> 00:10:40,609 substitution in this matrix

387 00:10:40,619 –> 00:10:42,460 notation, we find all the

388 00:10:42,469 –> 00:10:43,840 components of the solution

389 00:10:43,849 –> 00:10:44,260 here.

390 00:10:45,070 –> 00:10:46,340 And you see, as we’ve

391 00:10:46,349 –> 00:10:47,989 expected, we find a

392 00:10:48,000 –> 00:10:49,250 unique solution

393 00:10:50,010 –> 00:10:51,669 and this solution X is

394 00:10:51,679 –> 00:10:53,390 exactly our change of

395 00:10:53,400 –> 00:10:54,590 basis matrix.

396 00:10:55,000 –> 00:10:56,770 So it’s exactly TCB

397 00:10:56,869 –> 00:10:58,710 and this answers our question

398 00:10:58,719 –> 00:10:59,590 from the beginning.

399 00:11:00,239 –> 00:11:02,059 So with this matrix now

400 00:11:02,070 –> 00:11:03,539 we can do the translation

401 00:11:03,549 –> 00:11:05,000 from the matrix (basis) B

402 00:11:05,099 –> 00:11:06,609 into the matrix (basis) C.

403 00:11:07,309 –> 00:11:08,650 Moreover, you should see

404 00:11:08,659 –> 00:11:10,210 this whole procedure I’ve

405 00:11:10,219 –> 00:11:11,369 demonstrated here.

406 00:11:11,390 –> 00:11:12,690 Would also work.

407 00:11:12,700 –> 00:11:14,570 For example, for R five,

408 00:11:15,289 –> 00:11:16,820 the only thing that changes

409 00:11:16,830 –> 00:11:17,859 is that you have a little

410 00:11:17,869 –> 00:11:19,659 bit more work in solving

411 00:11:19,669 –> 00:11:20,599 this system here.

412 00:11:21,260 –> 00:11:22,919 However, the whole idea is

413 00:11:22,929 –> 00:11:24,150 exactly the same.

414 00:11:24,210 –> 00:11:25,929 So now you can apply it to

415 00:11:25,940 –> 00:11:27,400 all other examples.

416 00:11:28,299 –> 00:11:28,760 OK.

417 00:11:28,770 –> 00:11:30,049 So now you know how to

418 00:11:30,059 –> 00:11:32,020 calculate a change of basis

419 00:11:32,030 –> 00:11:32,679 matrix.

420 00:11:33,460 –> 00:11:35,080 So with the next video, we

421 00:11:35,090 –> 00:11:36,799 will go more abstract again.

422 00:11:36,919 –> 00:11:38,239 Therefore, I really hope

423 00:11:38,250 –> 00:11:39,390 we meet again and have a

424 00:11:39,400 –> 00:11:40,070 nice day.

425 00:11:40,130 –> 00:11:40,869 Bye bye.

Q1: Let $V = \mathcal{P}1$ be the vector space of polynomials with degree less or equal than $1$ and $\mathcal{E} = (m_0, m_1)$ be the standard monomial basis. Now consider the basis $\mathcal{B} = (m_0 + 2m_1, 3 m_0 + 4 m_1) $ and $\mathcal{C} = ( m_0 , 2 m_0 + 2 m_1) $. What is the system of linear equations we have to solve in order to get $T{\mathcal{C} \leftarrow \mathcal{B}}$?

A1: $$ \left(\begin{array}{cc|cc} 1 & 2 & 1 & 3 \ 0 & 2 & 2 & 4\ \end{array}\right) $$

A2: $$ \left(\begin{array}{cc|cc} 1 & 2 & 3 & 1 \ 4 & 5 & 6 & 1\ \end{array}\right) $$

A3: $$ \left(\begin{array}{cc|cc} 1 & 0 & 1 & 3 \ 2 & 2 & 2 & 4\ \end{array}\right) $$

A3: $$ \left(\begin{array}{cc|cc} 1 & 0 & 0 & 0 \ 2 & 2 & 0 & 0\ \end{array}\right) $$