Physics of Semiconductor Devices Vol.10

Transcription

1 10-1 Vector Spce Physics of Semicoductor Devices Vol.10 Lier Algebr for Vector Alysis To prove Crmer s rule which ws used without proof, we expli the vector lgebr tht ws explied ituitively i Vol. 9, by logicl method gi [1] Vector Additio d Sclr Multiplictio We employ the followig ottio for the lie, the ple, d three-dimesiol spce: (i) The rel umber lie is deoted R 1 or R. (ii) The set of ordered pirs (x, y) of rel umbers is deoted R. (iii) The set of ll ordered triples (x, y, z) of rel umbers is deoted R 3. However, whe we deote the ottio s R, i this cse my tke y umbers. The opertio of dditio c be exteded from R to R d R 3. For R 3, this is doe s follows. Give the two triples ( 1,, 3 ) d (b 1, b, b 3 ), we defie the sum to be ( 1,, 3 ) + (b 1, b, b 3 ) = ( 1 + b 1, + b, 3 + b 3 ). The elemet (0, 0, 0) is clled the zero elemet (or just zero) of R 3. The elemet ( 1,, 3 ) is the dditive iverse (or egtive) of ( 1,, 3 ), d we will write ( 1,, 3 ) (b 1, b, b 3 ) for ( 1,, 3 ) + ( b 1, b, b 3 ). The dditive iverse, whe dded to the vector itself, of course produces zero: ( 1,, 3 ) + ( 1,, 3 ) = (0,0,0). There re severl importt product opertios tht will defie o R 3. Oe of these, clled the ier product, ssigs rel umber to ech pir of elemets of R 3. We shll discuss it i detil i ext Sectio. Aother product opertio for R 3 is clled sclr multiplictio (the word sclr is syoym for rel umber ). This product combies sclrs (rel umbers) d elemets of R 3 (ordered triples) to yield elemets of R 3 s follows: Give sclr α d triple ( 1,, 3 ), we defie the sclr multiple by α( 1,, 3 ) = (α 1, α, α 3 ). Additio d sclr multiplictio of triples stisfy the followig properties: (i) (αβ)( 1,, 3 ) = α[β( 1,, 3 )] (ssocitivity) (ii) (α + β)( 1,, 3 ) = α( 1,, 3 ) + β( 1,, 3 ) (distributivity) (iii) (iv) (v) (vi) α[( 1,, 3 ) + (b 1, b, b 3 )] = α( 1,, 3 ) + α(b 1, b, b 3 ) α(0,0,0) = (0,0,0) 0( 1,, 3 ) = (0,0,0) 1( 1,, 3 ) = ( 1,, 3 ) (distributivity) (property of zero) (property of zero) (property of the uit elmet) The idetities re prove directly from the defiitios of dditiol d sclr multiplictio. For istce, (α + β)( 1,, 3 ) = ((α + β) 1, (α + β), (α + β) 3 ) = (α 1 + β 1, α + β, α 3 + β 3 ) = α( 1,, 3 ) + β( 1,, 3 ). 1 / 17

2 For R, dditio d sclr multiplictio re defied just s i R 3, with the third compoet of ech vector dropped off. All the properties (i) to (vi) still hold The Ier Product, Legth, d Distce The me ier product is sme meig s the sclr product used i vol. 9. Suppose we hve two vectors d b i R 3 (Fig. 10-1) d we wish to determie the gle betwee them, tht is, the smller gle subteded by d b i the ple tht they sp. The ier product ebles us to do this. Let us first develop the cocept formlly d the prove tht this product does wht we clim. Let = 1 i + j + 3 k d b = b 1 i + b j + b 3 k. We defie the ier product of d b, writte b, to be the rel umber b = 1 b 1 + b + 3 b 3. Note tht the ier product of two vectors is sclr qutity. Sometimes the ier product is deoted <, b>; thus <, b> d b me exctly the sme thigs. Certi properties of the ier product follow from the defiitio. If, b, d c re vector i R 3 d α d β re rel umbers, the (i) 0; = 0 if d oly if = 0. (ii) α b = α( b) d βb = β( b). (iii) (b + c) = b + c d ( + b) c = c + b c. (iv) b = b. It follows the Pythgore theorem tht the legth of the vector = 1 i + j + 3 k is The legth of the vector is deoted by. This qutity is ofte clled the orm of. Becuse = , it follows tht Uit Vectors = ( ) 1 =. Vectors with orm 1 re clled uit vectors. For exmple, the vectors i, j, k re uit vectors. Observe tht for y ozero vector, / is uit vector; whe we divide by, we sy tht we hve ormlized. I the ple defie the vector i θ = (cosθ)i + (siθ)j, which is the uit vector mkig gle θ with the x xis. Becuse i θ = (cos θ + si θ) 1/ = Distce If d b re vectors, we hve see tht the vector b is prllel to d hs the sme mgitude s the directed lie segmet from the edpoit of to the edpoit of b. It follows tht the distce from the edpoit of to the edpoit of b is b (see Fig. 10-1) The Agle Betwee Two Vectors Let us ow show tht the ier product does ideed mesure the gle betwee two vectors. / 17 x z Figure 10-1 Q b θ b P y

3 The Let d b be two vectors i R 3 d let θ, where 0 θ π, be the gle betwee them (Fig. 10-1). b = b cos θ. It follows from the equtio b = b cos θ tht if d b re ozero, we my express the gle betwee them s θ = cos 1 b ( b ). To prove this theorem, we should remember the lw of cosies s followig, c = + b b cos θ Cosider trigle with sides of legth, b, c, where θ is the mesuremet of the gle opposite the side of legth c. This trigle c be plced o the Crtesi coordite system by plottig the followig poits, s show i Fig By the distce formul c = ( b cos θ) + b si θ = + b b cos θ. If we pply the lw of cosies from trigoometry to the trigle with oe vertex t the origi d djcet sides determied by the vector d b (s see i the Fig. 10-1), it follows tht b = + b b cos θ. Becuse b = (b ) (b ), =, b = b b, we c rewrite the precedig equtio s (b ) (b ) = + b b b cos θ. We c lso expd (b ) (b ) s follows Thus, Tht is, (b ) (b ) = (b ) b (b ) = b b b b + = + b b b. + b b b = + b b b cos θ. b = b cos θ The Cuchy-Schwrz Iequlity Theorem 1 shows tht the ier product of two vectors is the product of their legths times the cosie of the gle betwee them. This reltioship is ofte of vlue i problems of geometric ture. A importt cosequece of Theorem 1 is: Cuchy-Schwrz Iequlity: For y two vectors d b, we hve b b with equlity if d oly if is sclr multiple of b, or oe of them is 0. 3 / 17 θ (b cos θ, b si θ) b Figure 10- QED c b cos θ

4 If is ot sclr multiple of b, the θ, the gle betwee them, is ot zero or π, d so cos θ < 1, d thus the iequlity holds; i fct, if d b re both ozero, strict iequlity holds i this cse. Whe is sclr multiple of b, the θ = 0 or π d cos θ = 1, so equlity holds i this cse. QED If d b re ozero vectors i R 3 d θ is the gle betwee them, we see tht b = 0 if d oly if cos θ = 0. Thus, the ier product of two ozero vectors is zero if d oly if the vectors re perpediculr. Hece, the ier product provides us with coveiet method for determiig whether two vectors re perpediculr. Ofte we sy tht perpediculr vectors re orthogol. The stdrd bsis i, j, d k re mutully orthogol d legth 1; y such system is clled orthoorml. We shll dopt the covetio tht zero vector is orthogol to ll vectors. p = The orthogol projectio of v o is the vector v The Trigle Iequlity A useful cosequece of the Cuchy-Schwrz iequlity, which is clled the trigle iequlity, reltes the legth of vector d b d of their sum + b. Geometriclly, the trigle iequlity sys tht the legth of y sides of trigle is o greter th the sum of the legth of the two other sides (see Fig. 10-3). + b + b. For vectors d b i spce, While this result my be cler geometriclly, it is useful to give proof usig the Cuchy-Schwrz iequlity, s it will geerlize to -dimesiol vectors. We cosider the squre of the left-hd side: By the Cuchy-Schwrz iequlity, we hve Thus, + b = ( + b) ( + b) = + b + b. + b + b + b + b = ( + b ) + b ( + b ) ; tkig squre roots proves the result. QED b + Figure Mtrices d Determit Mtrices We defie mtrix to be rry [ ], where 11, 1, 1, d re four sclrs. The determimt of such mtrix is the rel umber defied by the equtio = / 17

5 Mtrices A 3 3 mtrix is rry [ ], 33 where, gi, ech ij is sclr; ij deotes the etry i the rry tht is i the i th row d the j th colum. We defie the determit of 3 3 mtrix by the rule = Without some memoic device, formul 10- would be difficult to memorize. The rule to ler is tht you move log the first row, multiplyig 1j by the determit of the mtrix obtied by ccelig out the first row d the j th colum, d the you dd these up, rememberig to put mius i frot of the 1 term. For exmple, the determit multiplyig by the middle term of formul 10-, mely, , is obtied by crossig out the first row d the secod colum of the give 3 3 mtrix: Properties of Determits A importt properties of determits is tht iterchgig two rows or colums results i chge of sig. For determits, this is cosequece of the defiitio s follows: For colums, we hve d for rows = = ( ) = = = ( ) = I the cse of 3 3, results of iterchgig two colums become three wys , , Here, we prove the left-sided determit oly. However, other determits c be proved similrly = = = 1 3. QED Now, the results of two rows iterchgig will be , , / 17

6 Agi, we prove left-sided determit oly = = = 11 ( ) + 1 ( ) 13 ( ) = = 1 3 QED A secod it to you to verify this property of determits is tht we c fctor sclr out of y row or colum. For determits, this mes α 11 1 α 1 = 11 α 1 1 α = α = α 11 α 1 1 = 11 1 α 1 α. Similrly, for 3 3 determits we hve α 11 α 1 α α = α 1 3 = 1 α 3, α 3 33 d so o. These results follow from the defiitios. I prticulr, if y row or colum cosists of zeros, the the vlue of the determit is zero. A third fudmetl fct bout determits is the followig: If we chge row (or colum) by ddig other row (or, respectively, colum) to it, the vlue of the determit remis the sme. For the cse, this mes tht 1 b 1 b = 1 + b 1 + b 1 = b 1 b b b + = 1 + = b 1 + b b b 1 b 1 + b For the 3 3 cse, this mes b 1 + b 3 + b b 1 b b 3 = b 1 b b 3 = b 1 + b b b 3, c 1 c c 3 c 1 c c 3 c 1 + c c c 3 d so o. Agi, this property c be proved usig the defiitio of the determit. Closely relted to these properties is the fct tht we c expd 3 3 determit log y row or colum usig the sigs i the followig checkerbord ptter: For istce, we c expd by miors log the middle row: = Cross Products Now tht we hve estblished the ecessry properties of determits, we re redy to proceed with the cross product of vectors. 6 / 17

7 The Cross Product Suppose tht = 1 i + j + 3 k d b = b 1 i + b j + b 3 k re vector i R 3. The cross product of d b, deoted b, is defied to be the vector or, symboliclly, b = 3 b b i 3 3 b b j b k, 3 i j k b = 1 3. b 1 b b 3 b Eve though we oly defied determits for rrys of rel umbers, this forml expressio ivolvig vectors i useful memory id for the cross product. Certi lgebric properties of the cross product follow from the defiitio. If, b, d c re vectors d α, β, d γ re sclrs, the (i) b = b (ii) (βb + γc) = β( b) + γ( c) d (α + βb) c = α( c) + β(b c). Note tht = ( ), by property (i). Thus = 0. I prticulr, i i = 0, j j = 0, k k = 0. i j k i j k i j k i j = = k j k = = i k i = = j which c be remembered by cyclicly permutig i, j, k like this To give geometric iterprettio of the cross product, we first itroduce the triple product. Give three vectors, b, d c, the rel umber ( b) c i k j is clled the triple product of, b, d c (i tht order). To obti formul for it, let = 1 i + j + 3 k b = b 1 i + b j + b 3 k, d c = c 1 i + c j + c 3 k. The ( b) c = ( 3 b b i b 1 b j b k) (c 1 i + c j + c 3 k) = 3 b b 3 c b 1 b 3 c + 1 b 1 b c 3. This is the expsio by miors of the third row of the determit, so 1 3 ( b) c = b 1 c 1 b c b 3. c 3 If c 1 = α 1 + βb 1 c 1 = α + βb c 3 = α 3 + βb 3 The b 1 7 / 17

8 ( b) c = 3 b b 3 (α 1 + βb 1 ) 1 3 b 1 b 3 (α + βb ) + 1 b 1 b (α 3 + βb 3 ) = α b 1 b b 3 + β b 1 b b 3 = b b b 3 This result mes tht if c is vector i the ple sped by the vectors d b, the ( b) c = 0. I other words, the vector b is orthogol to y vector i the ple sped by d b, i prticulr to both d b. Next, we clculte the legth of b. Note tht b = 3 b b b 1 b b 1 b = ( b 3 3 b ) + ( 1 b 3 3 b 1 ) + ( 1 b b 1 ) = b b + 1 b b b + b 1 b 3 3 b 1 b 3 3 b 1 1 b b 1 = 1 b 1 + b + 3 b 3 + b b + 1 b b b + b 1 which equls 1 b 1 b 3 b 3 b 3 b 3 3 b 3 1 b 1 1 b 1 b = ( )(b 1 + b + b 3 ) ( 1 b 1 + b + 3 b 3 ), b ( b) = b b cos θ = b si θ where θ is the gle betwee d b, 0 θ π. Tkig squre root d usig k = k, we fid tht b = b si θ Combiig our results, we coclude tht b is vector perpediculr to the ple P sped by d b with legth b si θ. We see from Fig tht this legth is lso the re of the prllelogrm (with bse d height b si θ) sped by d b. There re still two possible vectors tht stisfy these coditios becuse there re two choices of directio tht re perpediculr (or orml) to P. This is cler from Fig. 10-4, which shows the two choices 1 d 1 perpediculr to P, with 1 = 1 = b si θ. Which vector represets b, 1 or 1? Accordig to cross product results of i, j, d k, the followig right-hd rule determies the directio of b i geerl. Tke your right hd d plce it so your figers curl from towrd b through the cute gle θ, s i Fig The your thumb poits i the directio of b Geometry of Determits Usig the cross product, we my obti bsic geometric iterprettio of d 3 3 determits. Let = 1 i + j d b = b 1 i + b j be two vectors i the ple. If θ is the gle betwee d b, we hve see tht b = b si θ is the re of the prllelogrm with djcet sides d b. The cross product s determit is 0 1 θ b b si θ 1 = Figure 10-4 b Figure 10-5 θ b P 8 / 17

9 i j k b = 1 0 = 1 b 1 b k. b 1 b 0 Thus, the re b is the bsolute vlue of the determit 1 b 1 b = 1 b b 1. The bsolute vlue of the determit 1 b 1 b is the re of the prllelogrm whose djcet sides re the vectors = 1 i + j d b = b 1 i + b j. The sig of the determit is + whe, rottig i the couterclockwise directio, the gle from to b is less th π. There is iterprettio of determits of 3 3 mtrices s volumes tht is logous to the iterprettio of determits of mtrices s res. 1 3 D = b 1 c 1 b c b 3 c 3 The bsolute vlue of the determit is the volume of the prllelepiped whose djcet sides re the vectors = 1 i + j + 3 k, b = b 1 i + b j + b 3 k d c = c 1 i + c j + c 3 k. This reltio hs bee proved i volume Equtios of Ples Let P be ple i spce, P 0 = (x 0, y 0, z 0 ) poit o tht ple, d suppose tht = A i + B j + C k is vector orml to tht ple ( see Fig. 10-7). Let P = (x, y, z) be poit i R 3. The P lies o the ple P if d oly if the vector P 0 P = (x x 0 )i +(y y 0 )j + (z z 0 )k is perpediculr to. Tht is, P 0 P = 0, or, equivletly, Thus, + C k is (Ai + Bj + Ck) [(x x 0 )i + (y y 0 )j + (z z 0 )k] = 0. A(x x 0 ) + B(y y 0 ) + C(z z 0 ) = 0. The equtio of the ple P through (x 0, y 0, z 0 ) tht hs orml vector = A i + B j A(x x 0 ) + B(y y 0 ) + C(z z 0 ) = 0; tht is, (x, y, z) P if d oly if Ax + By + Cz + D = 0, where D = A x 0 B y 0 C z 0. The four umbers A, B, C, d D re ot determied uiquely by the ple P. To see this, ote tht (x, y z) stisfies the equtio A x + B y + C z + D = 0 if d oly if it lso stisfies the reltio (λa)x + (λb)y + (λc)z + (λd) = 0 for y costt λ 0. Furthermore, if A, B, C, D d A, B, C, D determie the sme ple P, the A = λ A, B = λ B, C = λ C, D = λ D for sclr λ. Cosequetly, A, B, C, D re determied by P up to sclr multiple. 9 / 17

10 z z E = (x 1, y 1, z 1 ) P P Q v y P 0 y R = (x 0, y 0, z 0 ) P y (x, y) r θ x x Figure 10-6 x Figure 10-7 Figure 10-8 Two ples re clled prllel whe their orml vectors re prllel. Thus, the ples A 1 x + B 1 y + C 1 z + D 1 = 0 d A x + B y + C z + D = 0 re prllel whe 1 = A 1 i + B 1 j + C 1 k d = A i + B j + C k re prllel; tht is, 1 = σ for costt σ Distce: Poit to Ple Let us ow determie the distce from poit E = (x 1, y 1, z 1 ) to the ple P described by the equtio A(x x 0 ) + B(y y 0 ) + C(z z 0 ) = A x +B y + C z + D = 0. To do so, cosider the uit orml vector = Ai + Bj + Ck A + B + C, which is uit vector orml to the ple. Drop perpediculr from E to the ple d costruct the trigle REQ show i Fig The distce d = EQ is the legth of the projectio of v = RE (the vector from R to E) oto ; thus Distce = v = [(x x 0)i + (y y 0 )j + (z z 0 )k] A + B + C If the ple is give i the form A x +B y + C z + D = 0, the for y poit (x 0, y 0, z 0 ) o it, D = (A x 0 + B y 0 + C z 0 ). Substitutio ito the previous formul gives the followig: Distce = Ax 1 + By 1 + Cz 1 + D A + B + C 10-3 Cylidricl d Sphericl Coordites The distce from (x 1, y 1, z 1 ) to the ple A x +B y + C z + D = 0 is A stdrd wy to represets poit i the ple R is by mes of rectgulr coordites (x, y). However, i physics, polr coordites i the ple c be extremely useful. As portryed i Fig. 10-8, the coordites (r, θ) re relted to (x, y) by the formul x = r cos θ d y = r si θ r 0 d 0 θ π Cylidricl Coordites The cylidricl coordites (r, θ, z) of poit (x, y, z) re defied by (see Fig. 10-9) x = r cos θ, y = r si θ, z = z / 17

11 z z (x, y, z θ r y y x Figure 10-9 x Figure To express r, θ, d z i terms of x, y, d z, d to esure tht θ lies betwee 0 d π, we c write t 1 y x r = x + y θ = { π + t 1 y x π + t 1 y x if x > 0 d y 0 if x < 0 if x > 0 d y < 0 z = z, where t 1 (y/x) is tke to lie betwee π / d π /. The requiremet tht 0 θ π uiquely determies θ d r 0 for give x d y. If x = 0, the θ = π / for y > 0 d 3 π / for y < 0. If x = y = 0, θ is udefied. I other words, for y poit (x, y, z), we represet the first d secod coordites i terms of polr coordites d leve the third coordite uchged. Formul 10-3 shows tht, give (r, θ, z), the triple (x, y, z) is completely determied, d vice vers, if we restrict θ to the itervl [0, π) (sometimes the rge ( π, π] is coveiet) d require tht r > 0. To see why we use the term cylidricl coordites, ote tht if the coditios 0 θ π, < z < hold d if r = is some positive costt, the the locus of these poits is cylider of rdius (see Fig ) Sphericl Coordites Cylidricl coordites re ot the oly possible geerliztio of polr coordites to three dimesios. Recll tht i two dimesios the mgitude of the vector x i + y j (tht is, x + y ) is the r i the polr coordite system. For cylidricl coordites, the legth of the vector x i + y j + z k, mely, ρ = x + y + z, is ot oe of the coordites of tht system isted, we used the mgitude r = x + y, the gle θ, d the height z. We ow modify this by itroducig the sphericl coordite system, which does use ρ s coordite. Sphericl coordites re ofte useful for problem tht possess sphericl symmetry (symmetry bout poit), wheres cylidricl coordites c be pplied whe cylidricl symmetry (symmetry bout lie) is ivolved. Give poit (x, y, z) R 3, let 11 / 17 x z θ ϕ ρ Figure r (x, y, z) z y

12 ρ = x + y + z, d represet x d y by polr coordites i the x y ple: x = r cos θ y = r si θ 10 4 where r = x + y d θ is determied by formul The coordite z is give by z = ρ cos φ, where ϕ is the gle (chose to lie betwee 0 d π, iclusive) tht the rdius vector v = x i + y j + z k mkes with the positive z xis, i the ple cotiig the vector v d the z xis (see Fig ). Usig the dot product, we c express ϕ s follows: v k cos φ = v, tht is, φ = v k cos 1 ( v ). We tke s our coordites the qutities ρ, θ, ϕ. Becuse r = ρ si φ, We c use formul 10-4 to fid x, y, z i terms of the sphericl coordites ρ, θ, ϕ. where The sphericl coordites of poits (x, y, z) i spce re the triples (ρ, θ, ϕ), defied s follows: x = ρ si φ cos θ, y = ρ si φ si θ, z = ρ cos φ, 10 5 ρ 0, 0 θ π, 0 φ π Dimesiol Euclide Spce Vector i -spce I Previous sectios we studied the spces R = R 1, R, d R 3 d gve geometric iterprettios to them. For exmple, poit (x, y, z) i R 3 c be thought of s geometric object, mely, the directed lie segmet or vector emtig from the origi d edig t the poit (x, y, z). We c therefore thik of R 3 i either of two wys: (i) (ii) Algebriclly, s set of triples (x, y, z), where x, y, d z re rel umbers Geometriclly, s set of directed lie segmets These two wys of lookig t R 3 re equivlet. For geerliztio it is esier to use defiitio (i). Specificlly, we c defie R, where is positive iteger (possibly greter th 3), to be the set of ll ordered -tuples (x 1, x,, x ), where the x i re rel umbers. For istce, (1, 5,, 3) R 4. The set R so defied is kow s Euclide -spce, or -vectors. By settig = 1,, or 3, we recover the lie, the ple, d three-dimesiol spce, respectively. We luch our study of Euclide -spce by itroducig severl lgebric opertios. There re logous to those itroduced i sectio 10-1 for R d R 3. The first two, dditio d sclr multiplictio, re defied s follows: d for y rel umber α. The vectors (i) (x 1, x,, x ) + (y 1, y,, y ) = (x 1 + y 1, x + y,, x + y ); (ii) α(x 1, x,, x ) = (αx 1, αx,, αx ). e 1 = (1,0,0,,0), e = (0,1,0,,0),, e = (0,0,,0,1) 1 / 17

13 re clled the stdrd bsis vectors of R, d they geerlize the three mutully orthogol uit vectors i, j, k of R 3. The vector x = (x 1, x,, x ) c the be writte s x = x 1 e 1 + x e + +x e. For two vectors x = (x 1, x, x 3 ) d y = (y 1, y, y 3 ) i R 3, we defie the dot or ier product x y to be the rel umber x y = x 1 y 1 + x y + x 3 y 3. This defiitio esily exteds to R ; specificlly, for x = (x 1, x,, x ), y = (y 1, y,, y ), we defie the ier product of x d y to be x y = x 1 y 1 + x y + +x y. I R, the ottio <x, y> is ofte used i plce of x y for the ier product. Cotiuig the logy with R 3, we re led to defie the otio of the legth or orm of vector x by the formul Legth of x = x = x x = x 1 + x + + x. If x d y re two vector i the ple (R ) or i spce (R 3 ), the we kow tht the gle θ betwee them is give by the formul cos θ = x y x y. The right side of this equtio c be defied i R s well s i R or R 3. It still represets the cosie of the gle betwee x d y lie i two-dimesiol subspce of R (the ple determied by x d y) d our usul geometry ides pply to such ples. It will be useful to hve vilble some lgebric properties of the ier product. There re summrized i ext theorem. write For x, y, z R d α, β, rel umbers, we hve (i) (ii) (iii) (iv) (αx + βy) z = α(x z) + β(y z). x y = y x. x x 0. x x = 0 d oly if x = 0. Ech of the four ssertios c be proved by simple computtio. For exmple, to prove property (i) we (αx + βy) z = (αx 1 + βy 1, αx + βy,, αx + βy ) (z 1, z,, z ) The other proofs re similr. QED The x y x y. = (αx 1 + βy 1 )z 1 + (αx + βy )z + + (αx + βy )z = αx 1 z 1 + βy 1 z 1 + αx z + βy z + + αx z + βy z = α(x z) + β(y z). Let x, y be vectors i R. Let = y y d b = x y. If = 0, the theorem is clerly vlid, becuse the y = 0 d both sides of the iequlity reduce to 0. Thus, we my suppose 0. By the theorem 3 we hve 0 (x + by) (x + by) = x x + bx y + b y y = (y y) x x (y y)(x y). Dividig by y y gives 0 (y y)(x x) (x y), tht is, (x y) (x x)(y y) = x y. Tkig squre roots o both sides of this iequlity yields the desired result. QED 13 / 17

14 There is useful cosequece of the Cuchy-Schwrz iequlity i terms of legths. The trigle iequlity is geometriclly cler i R 3 d ws discussed i sectio The lytic proof of the trigle iequlity tht ws give i sectio 10-1 works exctly the sme i R d proves the followig: x + y x + y. Let x, y be vectors i R. The If Theorem 4 d its corollry re writte out lgebriclly, they become the followig useful iequlities: x i y i ( x i ) i=1 i=1 ( (x i + y i ) ) i= Geerl Mtrices 1 1 ( y i ) i=1 ( x i ) i=1 1 1 ; + ( y i ) Geerlizig d 3 3 mtrices, we c cosider m mtrices, which re rrys of m umbers: A = [ 1 m1 m ]. m We shll lso write A s [ i j]. We defie dditio d multiplictio by sclr compoetwise, just s we did for vectors. Give two m mtrices A d B, we c dd them to obti ew m mtrix C = A + B, whose ij th etry is the sum of ij d b ij. It is cler tht A + B = B + A. Give sclr λ d m mtrix A, we c multiply A by λ to obti ew m mtrix λ A = C, whose ij th etry c ij is the product λ ij. Next we tur to mtrix multiplictio. If A = [ ij ], B = [b ij ] re mtrices, the the product AB = C hs etries by c ij = ik b kj, k=1 which is the dot product of the i th row of A d j th colum of B: i=1 1 i th row j th colum c ij = 11 1 i1 i 1 b 11 b 1j b 1 b 1 b j b. Similrly, we c multiply m mtrix (m rows, colums) by p mtrix ( rows, p colums) to obti m p mtrix (m rows, p colums) by the sme rule. Note tht for AB to be defied, the umber of colums of A must equl the umber of rows of B. 14 / 17

15 Ay m mtrix A determies mppig of R to R m defied s follows: Let x = (x 1,, x ) R ; cosider the 1 colum mtrix ssocited with x, which we shll temporrily deote x T x 1 x T = [ ], x d multiply A by x T (cosidered to be 1 mtrix) to get ew m 1 mtrix: 11 1 x 1 y 1 Ax T = [ ] [ ] = [ ] = y T, m1 m x y m correspodig to the vector y = (y 1,..., y ). Thus, lthough it my cuse some cofusio, we will write x = (x 1,, x ) d y = (y 1,..., y m ) s colum mtrices x 1 y 1 x = [ ], y = [ ] x y m whe delig with mtrix multiplictio; tht is, we will idetify these two forms of writig vectors. Thus, we delete the T o x T d view x T d x s the sme. Thus, Ax = y will relly me the followig: Write x s colum mtrix, multiply it by A, d let y be the vector whose compoets re those of the resultig colum mtrix. The rule x Ax therefore defies mppig of R to R m. This mppig is lier; tht is, it stisfies A(x + y) = Ax + Ay A(αx) = α(ax), α sclr, s my be esily verified. Coversely, y lier trsformtio of R to R m is represetble i this wy by m mtrix. If A = [ ij ] is m mtrix d e j is the j th stdrd bsis vector of R, the i th compoet of A e j is ij. I symbols, (A e j ) i = ij Properties of Mtrices d Crmer s rule where Mtrix multiplictio is ot, i geerl, commuttive: If A d B re mtrices, the geerlly AB BA. A mtrix is sid to be ivertible if there is mtrix B such s tht AB = BA = I, I = is the idetity mtrix: I hs the property tht I C = C I = C for y mtrix C. We deote B by A 1 d cll A 1 the iverse of A. The iverse, whe it exists, is uique. If A is ivertible, the equtio Ax = y c be solved for the vector x by multiplyig both side by A 1 to obti x = A 1 y. I sectio10-, we defied the determit of 3 3 mtrix. This c be geerlized by iductio to determits. 15 / 17

16 If A is squre mtrix, the the mior of etry ij is deoted by M ij is defied to the determit of the submtrix tht remis fter the i th row d j th colum re deleted from A. The umber ( 1) i + j M ij is deoted by C ij is clled the cofctor of etity ij. Note tht mior M ij d its correspodig cofctor C ij re either sme or egtives of ech other d tht the reltig sig ( 1) i + j is either + 1 or 1 i ccordce with the ptter with i the checkbord rry which ws lredy show i sectio 10-. Thus, it is ever relly ecessry to clculte ( 1) i + j to clculte C ij []. <Exmple cofctor expsio of 4 4 determit> = If A is mtrix, the the umber obtied by multiplyig the etries i y row or colum of A by the correspodig cofctors d ddig the resultig products is clled the determit of A. d the sums themselves re clled cofctor expsios of A. Tht is, A = 1j C 1j + j C j + + j C j [cofctor expsio log the j th colum], A = i1 C i1 + i C i + + i C i [cofctor expsio log the i th row]. IF A is y mtrix d C ij is the cofctor of ij, the the mtrix C 11 C 1 C 1 C [ 1 C C ] C 1 C C is clled the mtrix of cofctors from A. The trspose of this mtrix is clled the djoit of A d is deoted by C 11 C 1 C 1 C [A] dj = [ 1 C C ] C 1 C C T C 11 C 1 C 1 C = [ 1 C C ]. C 1 C C There re severl techiques to get the iverse of mtrix A. We itroduce oe of those techiques without provig. If A is ivertible mtrix, the A 1 = 1 A [A] dj. I previous volumes, we hve used Crmer s rule without provig to get the solutio of lier system. Here, we prove Crmer s rule. 16 / 17

17 the the system hs uique solutio. This solutio is x 1 = A 1 A If Ax = b is system of lier equtios i ukows such tht A is ivertible, x = A A x = A A where A j is the mtrix obtied by replcig the etries i the j th colum of A by the etries i the mtrix b 1 b b = [ ]. b If A is ivertible, the Multiplyig the mtrices out gives The etry i the j th row of x is therefore C 11 C 1 C 1 b 1 x = A 1 b = 1 A [A] djb = 1 A [ C 1 C C b ] [ ] C 1 C C b b 1 C 11 + b C b C 1 x = 1 A [ b 1 C 1 + b C + + b C ] b 1 C 1 + b C + + b C x j = b 1C 1j + b C j + + b C j A 10 6 Now let j 1 b 1 1j A j = j 1 b j+1 1 j 1 b j+1 Sice A j differs from A oly i the j th colum, it follows tht the cofctors of etries b 1, b,..., b i A j re the sme s the cofctors of the correspodig etries i the j th colum of A. The cofctor expsio of A j log the j th colum is therefore Substitutig this result i 10-6 gives A j = b 1 C 1j + b C j + + b C j x j = A j A QED Refereces [1] Jerrold E. Mrsde, Athoy Tromb, Vector Clculus 6 th Editio, W. H. Freem d Compy, p19-66 pp53 (01) [] Howrd Ato, Elemetry Lier Algebr 10 th Editio, WILEY, p pp558 (010) 17 / 17