Things to Memorize: A Partial List. January 27, 2017

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1 Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved round s long s it doesn t grow/shrink or get tilted. If you multiply vector y positive sclr, the direction stys the sme, nd the length is scled. If you multiply vector y negtive sclr, the direction ecomes opposite (ut still prllel), nd the length is scled. We cll two vectors prllel if they re sclr multiples of ech other. We geometriclly dd two vectors together y plcing the hed of one t the til of the other, nd drwing the sum vector from the til of the first to the hed of the second. This is equivlent to tking prticulr digonl of the prllelogrm formed y the two vectors. Using the ove definition, we see tht the difference of two vectors cn e drwn geometriclly y tking the two vectors with their tils together, nd drwing vector etween their heds. R n refers to the spce of vectors with n coordintes. So, R 2 is ll vectors with two coordintes tht is, ll points in the xy-plne. Similrly, R 3 is ll vectors with three coordintes tht is, ll points in xyz-spce. A unit vector is vector with length 1. In R 2, i = [1, 0] nd j = [1, 0]. So, they re the unit vectors in the directions of the positive xes. Similrly, in R 3, i = [1, 0, 0], j = [0, 1, 0], nd k = [0, 0, 1]. The length of vector = [ 1, 2,..., n ] is Dot Product = n The dot product comines two vectors (with the sme numer of coordintes) nd gives numer. (By contrst, the cross product comines two vectors in R 3 nd gives third vector in R 3.) To clculte the dot product of two vectors, you multiply together their corresponding coordintes, nd dd the products. For exmple: = [ (8)( 1) + (3)(2) + (10)(5) + (3)(6) = 66 ] 3 6 1

2 Given two vectors nd, when we tke their dot product, it tells us something out the ngle θ etween the two of them: = cos θ It follows from the previous ullet point tht two non-zero vectors re orthogonl (or perpendiculr) if nd only if their dot product is zero. This is very common use of the dot product determining orthogonlity. 2 = Projections The projection of onto is the component of tht is prllel to. The picture is this: We drw line from the tip of to tht meets t right ngle. The point where it meets is the hed of proj, nd the til is the til of. So, the lue vector is the projection. We usully show these s cute ngles, ut they work for otuse lso. We relly only cre out the line in the direction of. If the ngle is otuse, the picture is like this: We cn clculte projection s follows: proj = ( ) 2 The dot product ehves lgericlly together with sclr multipliction nd vector ddition much the wy you would like it to. For exmple, ( + c) = + c. Determinnts of Smll Mtrices The determinnt of squre mtrix is numer. So the determinnt function hs squre mtrix s its input, nd numer s its output.

3 To clculte the determinnt of 2-y-2 mtrix: [ ] det = d c c d If two non-zero vectors in R 2 re prllel (tht is, sclr multiples of ech other) then when we put them s rows in mtrix, the determinnt of tht mtrix will e 0. To clculte the determinnt of 3-y-3 mtrix: c det d e f g h i [ e f = det h i ] det [ d f g i ] + c det [ ] d e g h Tht is: first, we tke the top row, nd lternte signs +,, +. Then, to decide which su-mtrix we ll tke the determinnt of, we delete the row nd column of the entry we chose. If nd re two vectors in R 2, then we cn clculte the re of the prllelogrm spnned y them (s in the picture elow) y mking them the rows of 2-y-2 mtrix nd tking the solute vlue of the determinnt. [ Are of prllelogrm = ] det Equivlently, the solute vlue of the determinnt is sin θ, where θ is the ngle etween nd. Cross Product The cross product comines two vectors in R 3 to give third vector in R 3. Unlike dot products nd determinnts, we won t define cross product for vectors except those with three coordintes. = ( ); other properties of the cross product re given on Pge 31 of your text i j k = det gives the re of the prllelogrm spnned y nd. Tht is, = sin θ, where θ is the ngle etween nd. So, the length of the cross product is the re of the prllelogrm spnned y the two vectors.

4 Are of prllelogrm = Note: contrst this to the re of the prllelogrm formed y two vectors in R 2. The vector is orthogonl to oth nd. However, we need to e more precise thn this, ecuse there re two directions tht re orthogonl two oth vectors. Its direction (up or down) is given y the right-hnd rule. If,, nd c re three vectors in R 3, then we cn clculte the volume of the prllelepiped spnned y them y mking them the rows of 3-y-3 mtrix nd tking the solute vlue of the determinnt. c Volume of prllelepiped = det c The determinnt ove is lso the triple product of,, nd c, which cn lso e clculted s ( c) or ( ) c. Lines nd Plnes The prmetric eqution of line (in ny dimension: R 2, R 3, etc) is x = q + s Tht is, the line consists of ll points x tht cn e written s some sclr multiple of plus the constnt vector q. The vector gives the direction of the line, nd the vector q is point on the line. The component eqution of line in R 2 is x + y = c, where,, nd c re constnt sclrs. The vector [, ] is orthogonl to the line.

5 The prmetric eqution of plne (in ny dimension: R 3, R 4, etc) is x = q + s + t where nd re not prllel. Tht is, the line consists of ll points x tht cn e written s liner comintion of nd plus the constnt vector q. The vectors nd lie in the plne; tht is, if we position them to their til is point on the plne, then their hed is point on the plne s well. (We lso cll vector tht lies in the plne prllel to the plne.) The constnt vector q is point on the plne. The component eqution of plne in R 3 is x + y + cz = d, where the vector [,, c] is norml to the plne. Tht is, ny vector prllel to the plne is orthogonl to [,, c]. If d = 0, then the plne psses through the origin. The constnt d will determine the position of the plne in spce, while the coefficients,, c, determine the plne s tilt. Note tht for every point (x, y, z) in the plne, [x, y, z] [,, c] = d. The component eqution of line in R 3 is s the intersection of two plnes: 1 x + 2 y + 3 z = 4 1 z + 2 y + 3 z = 4 The vectors [ 1, 2, 3 ] nd [ 1, 2, 3 ] re orthogonl to the line. To get direction vector of the line, you cn clculte the cross product of these two vectors. Liner Systems Let 1 x + 2 y = 3 nd 1 x + 2 y = 3 e lines tht re not prllel. Then their intersection is point. So, the solution to the following system of liner equtions is single point. 1 x + 2 y = 3 1 z + 2 y = 3 Let 1 x + 2 y + 3 z = 4, 1 x + 2 y + 3 z = 4, nd c 2 x + c 2 y + c 3 z = c 4 e plnes in R 3 whose norml vectors do not ll lie on the sme plne. (Tht is, their norml vectors re linerly independent.) Then their intersection is point. So, the solution to the following system of liner equtions is single point. 1 x + 2 y + 3 z = 4 1 z + 2 y + 3 z = 4 c 1 z + c 2 y + c 3 z = c 4 If three vectors lie on the sme plne, then the prllelepiped they define is smooshed ll the wy flt, nd so hs volume zero, nd hence determinnt zero. So, the system of liner equtions in the ullet point ove hs precisely one solution if nd only if det c 1 c 2 c 3 In the cse tht the determinnt is zero, the solutions to the system of liner equtions could form line (tht is, they cn e descried using one prmeter), they could form plne (tht is, it tkes two prmeters to descrie them), they could e ll of R 3 (in the silly cse tht every eqution is 0x + 0y + 0z = 0), or there could e no solutions t ll.

6 Liner Independence A liner comintion of vectors is vector formed y dding sclr multiples of the given vectors. For exmple, the vector πc is liner comintion of the vectors,, nd c. The set of ll liner comintions of collection of vectors is clled the spn of the vectors. A collection of vectors is linerly independent if the only liner comintion of them to equl the zero vector is the liner comintion where ll sclrs re zero. Tht is: consider collection of vectors 1, s,..., n. Set up the system of liner equtions s s s n n = 0. If the only solution is s 1 = s 2 = = s n = 0, then the vectors re linerly independent. If there re other solutions, then the vectors re linerly dependent. The ove is the forml definition of liner independence. For collection of t lest two vectors, it is equivlent to the following definition: collection of vectors is linerly independent if no vector in the collection cn e written s liner comintion of the others. A collection of n linerly independent vectors from R n is clled sis of R n. Suppose you hve collection of vectors tht form sis of R n. Then ny vector in R n cn e written s liner comintion of your vectors, nd there is unique wy to do it. Chpter 3 Solving Liner Systems You should know how to solve liner system using sustitution. We cn use three row opertions on liner system without chnging its solutions: 1. Multiplying n eqution (row) y non-zero sclr 2. Adding multiple of one row to nother row 3. Swpping the verticl order of rows To sve time nd spce, we use shorthnd for systems of liner equtions. We use n ugmented mtrix, y which we men we tke mtrix (representing the coefficients of the vriles) nd ugment it y dding n extr column seprted y verticl line (the column represents the constnts in the equtions). You should know how to use Gussin Elimintion (pivoting) to solve system of liner equtions. You should recognize when mtrix is in row echelon form nd reduced row echelon form. (See the pictures on pge 79 in the text.) The rnk of mtrix (without its ugmenttion) is the numer of nonzero rows tht it hs fter eing reduced. You should know how to express the solutions to system of liner equtions using prmeters. The numer of prmeters needed to descrie the solutions to system of liner equtions with rnk r nd n vriles is n r, provided there re ny solutions t ll.

7 If mtrix hs n rows nd columns, nd rnk n, then its reduced row echelon form will hve 1s long the min digonl nd 0s everywhere else A system of liner equtions tht looks like this will hve precisely one solution, regrdless of wht its constnts re. If mtrix hs n columns, rnk n, nd t lest n rows, then its reduced row echelon form will hve 1s long the min digonl nd 0s everywhere else A system of liner equtions tht looks like this will hve either precisely one solution or no solutions t ll. Homogeneous Systems A system of liner equtions is homogeneous if the constnts re ll zero. For instnce, the system on the left is homogeneous, while the system on the right is not. 1 x + 2 y = 0 1 z + 2 y = 0 1 x + 2 y = 0 1 z + 2 y = 1 A homogeneous system lwys hs t lest one solution (where x = y = z = = 0) So, the solutions of liner eqution cn lwys e written s s s s k k (for some pproprite k, nd possily the vectors re ll 0). Any liner comintion of solutions of A 0 is itself solution of A 0. If you hve system of liner equtions, you cn form the ssocited homogeneous system y simply chnging ll the constnts to 0. This is different system, with different solutions, ut the solutions re relted in predictle mnner. Suppose A is system of liner equtions, nd A 0 is its ssocited homogeneous system. Let s s s k k e the complete set of solutions to A 0. It s possile tht A hs no solutions. If it hs t lest one solution q, then the complete set of solutions to A cn e written s q+s 1 1 +s s k k. There re lots of relted fcts: for instnce, the difference of ny two solutions of A is solution of A 0, nd the sum of ny solution of A with ny solution of A 0 is solution to A.