Linear Algebra Final Exam

Summarize 3 main differences between matrix multiplication and the multiplication of real numbers. In each case give an example. : 1. AB need not equal BA A = 1 2 B = 2 -1 AB = 4 5 BA = 0 0 2 4 1 3 8 10 7 142. AB may be zero matrix with A and B not equal to 0A = 1 2 B = 4 -6 AB = 0 0 2 4 -2 3 0 03. AB may equal AC with B not equal to CA = 1 2 B = 2 1 C = -2 7 AB=AC= 8 5 But B not =C 2 4 3 2 5 -1 16 10

Prove Theorem 1.7 - If A is a nonsingular matrix, then A^-1 is nonsingular and (A^-1)^-1 = A. : If A is a nonsingular matrix, then A^-1 is nonsingular and (A^-1)^-1 = A. Proof: We have A^-1A=In=AA^-1. Since inverses are unique we conclude (A^-1)^-1 = A.

Let A be a 4X4 matrix with det(A) = -59. : 1. If A is transformed to RREF B, what is B?- I42. Does A^-1 exist? Justify-Yes, since det(A) not = 0, A is invertible.3. How many solutions does Ax=0 have?-Only one, just the trivial solution.

Let A be an mXn matrix and Ax=0, where x is vector in R^n. Show that the solution space of the homogeneous system of the null space of the matrix A is a subspace of R^n. : Let x and y be solutions for the homogeneous system:So that Ax=0 and Ay=0A(x+y) = Ax + Ay = 0 + 0 = 0, so x + y is a solutionIf there is a scalar c:A(cx) = c(Ax) - c0 = 0, so cx is a solution-Thus closed under addition and multiplication, also know that it contains the zero vector.

Definition: A basis for a subspace S of R^n is a set of vectors in S that spans s and is linearly independent. What does each of these ensure? : Linear independence ensures that there are not too many vectors in a basis.Spanning ensure that there are not too few.

Suppose an nXn matrix A is invertible, what can you say about ColA? NulA? : Since A is invertible, the equation Ax=0 has only the trivial solution, therefor NulA={0}.By definition, the columns of any matrix always span the column space, so in this case, ColA is all of R^n, so ColA = R^n.

Prove Theorem 4.20-If a is an nXn matrix, then rankA = n if and only if A is row equivalent to In. : If RankA = n, then A~BrankB = n, B = In, A~InIf A~In, rankA = rankIn = n.

What can you say about the application of rank to the solutions of the linear system Ax=b, where b is an arbitrary nX1 matrix? : Theorem 4.21The linear system Ax=b has a solution if and only ifrankA - rank[A|b] that is, if and only if the rank of the coefficient and augmented matrices are equal.Ex, consider: 1 2 x1 = 4 -2 3 x2 -1Since rankA = rank[A|b] = 2, the linear system has a solution.

Prove Theorem 6.4:Let L:V->W is a linear transformation of a vector space V into a vector space W, then:A)KerL is a subspace of V.B)L is one-to-one if and only if KerL = {0}. : A) we show that if v and w are in kerL, then so are v+w and cv for any real numbers of c. If v and w are in kerL, then:L(v) = 0w, and L(w) = 0w, Since L is a linear transformation:L(v+w) = L(v) + L(w) = 0w + 0w = 0w, so v+w is in kerLL(cv) = cL(v) = c0w = 0, so cv is also in kerL, so kerL is a subspace of VB)If we let L be one-to-one, we show that kerL={0v}. Let v be in kerL, then L(v) = 0w, also we know L(0v) = 0w. Then:L(v) = L(0v). Since L is one-to-one, we conclude that v = 0v.So kerL = {0v}.Conversely, if kerL = {0v}, we need to show L is one-to-one, Let L(v1) = L(v2) for v1 and v2 in V.Then L(v1) - L(v2) = 0w.So that L(v1-v2) = 0w. Meaning v1-v2 is in kerL, so v1-v2=0v.Hence v1=v2, and L is one-to-one.

Show that every matrix transformation is a linear transformation. : Let A be an mXn matrix. A matrix transformation as a function L:R^n->R^m defined by L(u) = Au.We now show every matrix transformation is a linear transformation.If u and v are vectors in R^n then:L(u+v) = A(u+v) = Au + Av = L(u) = L(v)If c is a scalar then:L(cu) = A(cu) = c(Au) = cL(u).

Show that every eigenspace is a subspace of R^n. : Let V be the set of all eigenvalues corresponding to lambda, together with the zero vector. To show v is a subspace, we show that it is closed under vector addition and scalar multiplication.Let x1 and x2 be vectors in V and c be a scalar then:Ax1 = lambdax1 and Ax2 = lambdax2So, Ax1 + Ax2 = lambdax1 + lambdax2A(x1 + x2) = lambda(x1 + x2)Thus x1 + x2 is an eigenvector corresponding to lambda(closed under addition).Since Ax1 = lambdax1cAx1 = clambdax1A(cx1) = lambda(cx1)So it is closed under scalar multiplication.

Subspace : A) Closed under additionB) Closed under scalar multiplicationC) Contains the zero vector

Dimension : The dimension of a nonzero vector space V is the number of vectors in a basis for v.

Rank : The rows of A, considered as vectors in Rn, span a subspace of Rn called the row space of A.The columns of A, considered as vectors in R^m, span a subspace of R^m called the column space of A.