3.1: - GANG | Geometry Analysis Numerics Graphics...



Linear Algebra Midterm II Review SheetPrepared by your generous peer, Jonah Palmer, April 2017 (my comments are in red – Rob Kusner ;-)3.1:Easy 3x3 determinant equation:abcdefghi=aei-ceg+bfg-afh+cdh Computing the determinant using cofactor expansion across rows / columns:40252202-1 Cij=-1i+jdetAijdetA=a11C11+a12C12+a13C13→detA=-11+14C11+0C12+-11+32C13detA=4C11+2C13detA=4222-1+25202=4-6+210=-4Note: You should get the same answer for every row or column you use cofactor expansion on.Remember the checkerboard of ± signs -1i+j in the cofactors Cij=-1i+jdetAijHere Aij is the minor matrix resulting from removing the row i and col j from ARow operations and the effect on the determinant:abcd→7a7bcd detabcd=ad-bcdet7a7bcd=7(ad-bc)It multiplied the determinant by 7.abcd→a+kcb+kdcddetabcd=ad-bc deta+kcb+kdcd=da+kc-c(b+kd)It does not change the determinant.111-94-76-87→kkk-94-76-87 det111-94-76-87=41 detkkk-94-76-87=41kThe determinant is multiplied by k.Finding a formula relating detkA to k and det(A):A=abcd, k is a scalar detA=ad-bcdetkA=kakbkckd=k2(ad-bc)Note: Notice the example at the top of the page. When one row is filled with k’s, the determinant is multiplied by k. However, when two rows are multiplied by k’s, then the determinant is k2. 3.2:Properties of Determinants:A multiple of one row of A is added to another row to produce matrix B, then: A=BIf two rows are interchanged to produce a matrix B, then: B=-AIf one row is multiplied by k to produce a matrix B, then: B=kAIf A & B are both n×n matricies then: AB=A×BAT=ANote: A=detA, B=detBTrue / False Statements:A row replacement operation does not affect the determinant of a matrix. i.e. adding some row to another rowTRUEThe determinant of A is the product of the pivots in any echelon form U of A, multiplied by -1r, where r is the number of row interchanges made during row reduction from A to U.FALSEReduction to an echelon form may also include scaling a row by a nonzero constant, which will change the value of the determinant.If the columns of A are linearly dependent, then detA=0.TRUEIf columns of A are linearly dependent, A is not invertible.detA+B=detA+detB.FALSENot one of the properties!Showing equivalent equations using determinants:Show thatdetPAP-1=detA detPAP-1=detPdetAdetP-1→ detP-1=detP-1 detP*detP-1=1→1*detA=detA3.3:Finding Area using determinants:Find the area of the parallelogram whose vertices are listed: 0,0, 7,5,11,6,(18,11)Note: The area of the parallelogram determined by the columns of A is detA.Note: We can use any two vertices here; we’ll use 7,5 & (11,6).75, 116→A=71156 detA=-13→detA=13The area of the parallelogram is 13 square units. Note that the sign of the determinant gives info about the “right” or “left” handedness (orientation) of the parallelotope.Finding volume using determinants:Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at 3,0,-4,1,5,6, & 6,2,0.Note: If A is a 3x3 matrix, the volume of the parallelepiped determined by the columns of A is detA.30-4,156,620→A=316052-460detA=76The volume of the parallelepiped is puting the area of an image of S under the mapping x→Ax:Let S be the parallelogram determined by the vectors b1=-26 and b2=-29, and let A=4-7-27. Compute the area of the image of S under the mapping x→Ax.Area of TS=detA?area of Sdetb1b2=-2-269 =-6=6=area of S detA=4-7-27=14 Area of TS=14?6=844.1:Sets of vectors with inclusion and exclusion:Let V be the set of vectors shown below.V=xy:x≥1, y>1 It helps to draw pictures of these sets to see what’s going on, especially when looking for a counter-example to show a property fails.If u and v are in V, is u + v in V? Why?u=ab , and v=cd, then u+v=a+cb+dSince a≥1 and c≥1, then a+c≥2, which implies a+1≥1Since b>1 and d>1, then b+d>2, which implies b+d>1Since a+c≥1 & b+d>1, this means that u + v is in V.Find a specific vector u in V and a specific scalar c such that cu is not in V.Note: Any vector for u will work, make one up.u=22, then cu=2c2cNote: cu will not be in V if 2c<1. So, find a c where this is true. Ex. c=-1.If u=22, and c=-1, then cu will not be in V.A subspace of a vector space V is a subset H of V properties:The zero vector of V is in H. This is equivalent to V being nonempty.H is closed under vector addition. That is, for each u and v in H, the sum u + v is in H.H is closed under scalar multiplication. This is, for each u in H and each scalar c, the vector cu is in H.Showing H is not a subspace of R2:H=xy:16x2+9y2≤1 Find two vectors and a vector and a scalar to show that H is not a subspace of R2. Picture! (Ellipse!)Note: For two vectors to show that H is not a subspace of R2, they must show that H is not closed under addition.u=140 , v=013, u+v is not in H.Note: For a vector and a scalar to show that H is not a subspace of R2, they must show that H is not closed under scalar multiplication. u=xy is in H. The vector cu is equal to cxy=cxcy Say, c=5. Find a vector u such that cu is not in H. 16cx2+9cy2>1u=0.17-0.22Note: 160.172+9-0.222=0.898≤1cu=50.17-0.22=0.85-1.1Note: 160.852+9-1.12=22.450>1Spanning and vectors:Let H be the set of all vectors of the form 5t-t0. Find a vector v in R3 such that H=Spanv.v=5-10Note: Due to the theorem; The Span of any nonempty set is always a subspace!If v1, …, vp are in a vector space V, then Spanv1, …,vp is a subspace of V.We can say because v is in R3 and H=Spanv, H is a subspace of R3Determining subspace of vectors:v1=10-1, v2=516, v3=10212, and w=615Is w in the subspace spanned by v1,v2,v3?Note: To determine if w is in the subspace spanned by v1,v2,v3, we must determine if w is a linear combination of v1, v2 and v3.v1 v2 v3 w=1510012-1612 6151510012-1612 615~100012000 110Note: No row has the form 0…0 b where b is nonzero. Therefore, w is a linear combination of v1, v2, and v3Spanning Sets:Find a set S of vectors that spans W.W=3a-6b7b+3c6c-4a8b a b cSpanning set S=30-40,-6708, 0360Vector Space Axioms:The sum u + v is in V Some consider this to be a “given” rather than an “axiom”….u + v = v + u(u + v) + w = u + (v + w)V has a vector – such that u + 0 = uFor each u in V, there is a vector -u in V such that u + (-u) = 0The scalar multiple cu is in V Some consider this to be a “given” rather than an “axiom”….c(u + v) = cu + cv(c + d)u = cu + duc(du) = (cd)u1u = u3), 4) and 5) are the axioms of a “group” and including 2) makes it a commutative (abelian) group7) and 8) are distributive laws relating scalar “+” (one of these) with vector “+” (three of these)10) keeps things from collapsing (without it, the “trivial” scalar multiplication rule “cv=0” for all scalars c and vectors v would satisfy the other axioms)4.2:Null space of an m x n matrix A:Determine if w=241 is in NulA, where A=6-637-38-2811Also called the “kernel” kerANulA=x:x is in Rn & Ax=0Note: To test if w satisfies Aw=0, computer Aw. If the product Aw is 0, then w is in Nul(A).Aw=6-637-38-2811241=-91039 Aw≠0, so w is not in NulA. Explicit Description of Nul A by Listing Vectors that span the null space:A=16-700080Note: Need to find the general solution of Ax = 0 in terms of free variables.A 0=16-70000800~1600000100x1+6x2=0, x3=0x1=-6x2, x3=0x1x2x3x4=-6x2x20x4=x2-6100+x40001Spanning set for Nul A = -6100, 0001Show a given set is a vector space:W=pqrs :2p+q=-2s, -2p=-3s+3r2p+q+0r+2s=0 -2p+0q-3r+3s=0 Ax=2102-20-33pqrs=00 Note: this given set represents the set of all solutions to the homogeneous system of equations.Therefore; the set W=NulANote: The null space of an m x n matrix A is a subspace of Rn. Equivalently, the set of all solutions to a system Ax=0 of m homogeneous linear equations in n unknowns is a subspace of Rn.Nonzero Vectors in Nul A and Col A:A=-35-40-7-8-7-81416First find the general solution of Ax = 0 in terms of free variables.A 0=-35-400-7-80-7-8014160~1870000000000x1+87x2=0→x1=-87x2Note: To find a nonzero vector in Nul A, choose a nonzero value for the free variable, say x2=-7, and substitute into the general solution. x1=-87-7=8NulA=Span(8-7)Note: To find a nonzero vector in Col A, choose any column of A as the nonzero vector.ColA=Span(-35-7-714)Determine if a vector is in Col(A) / Nul(A):A=-10-8-6242602 and w=11-3. Determine if w is in Col(A) / Nul(A)?ColA={b : b=Ax for some x in R} Also called the “image” imANote: If the equation Ax = w is consistent, then w is in Col(A).A w=-10-8-612421602-3~1013-12011310000Note: Since the echelon form of the augmented matrix has no row of the form [0…0 b] the equation is consistent. Thus, w is in Col(A).Note: If the equation Aw = 0 is true, then w is in Nul(A). -10-8-624260211-3=000w is in Nul(A).True / False Statements:A null space is a vector space.TRUEThe column space of an m x n matrix is in RmTRUEThe column space of A, Col(A), is the set of all solutions of Ax = b.FALSEColA=b :b=Ax for some x in RnThe null space of A, Nul(A), is the kernel of the mapping x→Ax.TRUEThe kernel of a linear transformation T, from a vector space V to a vector space W, is the set of all u in V such that Tu=0. Thus the kernel of a matrix transformation Tx=Ax is the null space of A.The range of a linear transformation is a vector space. Again, most folks call this the “image”.TRUEThe range of a linear transformation T, from a vector space V to a vector space W, is a subspace of W.The set of all solutions of a homogeneous linear differential equation is the kernel of a linear transformation.TRUEThe linear transformation maps each function f to a linear combination of f and at least one of its derivatives, exactly as these appear in the homogeneous linear differential equation.4.3:Invertible Matrix Theorem:Let A be a square n x n matrix. Then the following statements are equivalent.The matrix A is an invertible matrix.The matrix A is row equivalent to the n x n identity matrix.The matrix A has n pivot positions.The equation Ax = 0 has only the trivial solution.The columns of A form a linearly independent set.The columns of A span RnDetermining if a set is a basis:Determine whether the set 110,100,111 is a basis for R3.A=111101001~100010001 Note: The columns of A form a linearly independent set and they span R3Note: The set is a basis if the set is linearly independent and they span R3Basis of a null space of a matrix: 10-3401-573-21-2 Find the basis for the null space of the matrix.A 0=10-34001-5703-21-20~10-34001-57000000x1-3x3+4x4=0x2-5x3+7x4=0x1=3x3-4x4x2=5x3-7x4x=x1x2x3x4=x33510+x4-4-701Note: The basis for the null space of the matrix will only be the basis if 3510 and -4-701 are linearly independent. They are independent however, because neither is a multiple of the other. True / False Statements:A linearly independent set in a subspace H is a basis for H.FALSEThe subspace spanned by the set must also coincide with H. One can extend to a maximal linearly independent set – that extended set will be a basis – this is how one shows any vector space has a basis.If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis for V. Dually, a minimal spanning set is a basis, but this works easily only in the finite-dimensional case.TRUESpanning Set Theorem A basis is a linearly independent set that is as large as possible. (see above)TRUEThe standard method for producing a spanning set for Nul A sometimes fails to produce a basis for Nul A.FALSEThe method always produces an independent set.If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col A.FALSEThe columns of an echelon form B of A are not necessarily in the column space of A.4.4:Finding vector x using a given basis and a coordinate vector xB:Find vector x determined by the given coordinate vector xB and the given basis B.B=-73, -56, xB=83 x=c1b1+c2b2xB=c1c2B=b1, b2x=8-73+3-56=-7142 Finding the coordinate vector xB given vector x and basis B:Find the coordinate vector xB of x relative to the given basis B=b1, b2.b1=10-7, b2=-65, x=54-41Note: We are looking for the vector xB=c1c2 here.c110-7+c2-65=54-41→b1b2 x=10-654-75-41~10301-4 x1=3, x2=-4→c1=3, c2=-4x=3b1-4b2→xB=3-4Using an inverse matrix to find xB:Use an inverse matrix to find xB for the given x and B.B=4-5, -56, x=-13Recall: PB=[b1 b2…bn]x=c1b1+c2b2+…+cnbnx=PBxBPB-1x=xBA=abcd→A-1=1detAd-b-caPB=4-5-56→PB-1=-6-5-5-4 xB=PB-1xxB=-6-5-5-4-13=-9-7Finding coordinate vectors relative to a set B:The set B=1-t2, t-t2, 1+2t-t2 is a basis for P2. Find the coordinate vector of pt=-10+t+5t2 relative to B.Note: To find the coordinate vector, split up whole numbers, t’s of the first degree, and t2’s of the second degree. c11-t2+c2t-t2+c31+2t-t2=-10+t+5t2c1-c2t2+c2t-c2t2+c3+c32t-c3t2=-10+t+5t21c1+1c3=-10 - Whole numbers here!c2+2c3=1 - t’s of the first degree here!-1c1+-1c2+-1c3=5-t2’s of the second degree here!Note: Now we need to take this system of equations and form an augmented matrix.101-100121-1-1-15~100-80105001-2c1=-8, c2=5, c3=-2pB=-85-2Finding matrix A using the inverse of PB:Let B=3-5,-47 . Since the coordinate mapping determined by B is a linear transformation from R2 into R2, this mapping must be implemented by some 2x2 matrix A. Find it.A=PB-1PB=3-4-57 A-1=1detAd-b-cadetPB=1A=PB-1=7453 ................
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