This MUST hold matrix multiplication satisfies the distributive property.

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Eugenia Banks
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1 The columns of AB are combinations of the columns of A. The reason is that each column of AB equals A times the corresponding column of B. But that is a linear combination of the columns of A with coefficients given by the entries in the column of B. The product is defined when n = p and then the dimensions of the product are m q. This MUST hold matrix multiplication satisfies the distributive property. This might hold. (A+B)(A-B) = A 2 + BA AB B 2 and this is A 2 B 2 if and only if AB=BA. Sometimes that is true, but not always. Cancellation does not always work, and it is quite possible for AB=AC to hold, even when B C. One way to see this is as follows: AB=AC if and only if AB-AC = 0 if and only if A(B-C) = 0. So if B-C is a nonzero matrix whose columns are all in the null space of A, then AB=AC but B doesn t have to equal C. Here is a specific example: 1 2 =

2 A square matrix A is invertible if there exists a matrix B (with the same dimensions) such that both AB and BA equal the identity matrix. In this case B is the inverse of. Given an n n matrix A, append an n n identity matrix to create [A I]. Now reduce this combined matrix to rref. If the first n columns are pivot columns, then the rref has the form [I B] and the right side, B is the inverse of A. If the first n columns of the rref are not pivot columns, then A is not invertible. An elementary matrix is one that is obtained by performing a single row operation on the identity matrix. If we do this and create the matrix E, then multiplying EA produces the same result as applying the row operation to A. We can see that every elementary row operation is invertible by constructing generic examples for each type of row operation and then using the parts of the invertible matrix theorem. For example, if we swap two rows of the identity matrix, the resulting matrix still has independent columns, and so is invertible. Similarly, if we multiply one row of the identity matrix by a nonzero scalar r, the resulting matrix is still a diagonal matrix. This allows us to compute its determinant as r and since that is nonzero, the matrix is invertible. Finally, when a multiple of one row is added to another, the identity matrix is changed into a

3 triangular matrix with all 1 s on the diagonal. Again we can compute the determinant to see that the result is invertible. Alternatively, we can argue that since every row operation is reversible by another row operation, the row operation matrices are all invertible. If A is invertible, then we can reduce it to I by applying a sequence of row operations. If we do this using elementary matrices, then we get an equation like this: E p E p-1 E 2 E 1 A = I Where E j is the jth row operation we perform on A. Multiplying both sides of this equation on the left by the inverses of the elementary matrices in reverse order, we obtain A = E 1-1 E 2-1 E p-1-1 E p -1 This shows that A is a product of elementary row operations, because the inverse of each elementary matrix is another elementary matrix. The column space of an m n matrix A is the subspace of m spanned by the columns of A. That is, it is the set of all linear combinations of the columns of A. The null space of the matrix A is the set of solutions to the equation Ax=0. The set {v 1, v 2,, v p } is linearly independent if the equation c 1 v 1 + c 2 v c p v p = 0 holds if and only if all the scalars c j are zero.

4 If V can be spanned by a finite set, then the dimension is the number of elements in any basis. That is, the dimension is n if and only if any linearly independent spanning set has n elements. If V cannot be spanned by any finite set V is infinite dimensional. The kernel of a linear transformation T consists of all the vectors v in the domain of T for which T(v) = 0. Let A be an n n invertible matrix, and let b be a vector in n. For each j between 1 and n inclusive, define a matrix B j by replacing the j th column of A by b. Then Cramer s rule states that the unique solution to the system Ax = b is the vector x=[ x 1 x 2 x n ] T where x j = det(b j )/det(a) for all j. Let A be the matrix on the left side of the equation. We can see that det(a) = (3)(4) (2)(7) = -2. Now replace the first column of A with [5 11] T to make B 1 and replace the second column with [5 11] T to make B 2 and take the determinants. We find det B 1 5 = = = 57 and 3 5 det = = = B. Therefore, according to Cramer s rule, the solution is given by x 57 = 23 2 y 2

5 This is FALSE. Independent columns indicate a nonzero determinant, but not necessarily a determinant of 1. For example, a 2x2 diagonal matrix with 2 and 3 on the diagonal has independent columns and the determinant is 6, not 1. This is TRUE it is one of the results stated in the Determinants handout. This is True. We know that det(ab) = det(a)det(b) and det(ba) = det(b)det(a). But since det(a) and det(b) are real numbers, we also know that det(a)det(b) = det(b)det(a). So we have det(ab) = det(a)det(b) = det(b)det(a) = det(ba). This is a triangular matrix, so the determinant is the product of the diagonal entries: (1)(2)(-1)(4)(5)=-40. This matrix has a column of all 0 s. That means the columns are not linearly independent, so the matrix is not invertible, so its determinant is 0. Or, if we compute the determinant by the method of minors, and choose all of our elements from the second column, then we see that all the coefficients will be 0 so the determinant is 0.

6 The invertible matrices do NOT form a subspace. In particular, the zero matrix is NOT invertible, so it is not in the set under consideration. But every subspace must include the zero element. Therefore, the set of invertible matrices is not a subspace. This H is not closed under addition. For example, if we add (3,5) + (5,3), both of which are in H, we get (8,8) which is not in H. The given information tells us that q(t) = 2(first basis element)+3(third basis element). Therefore, we have q(t) = 2 + 3(t+1) 2 = 3t 2 + 6t + 5. We have to solve the equation p(t) = a(1) + b(t+1) + c (t+1) 2. We can see by inspection that the coefficient of (t+1) 2 must be 0, because p has no quadratic term. Substituting the definition for p and expanding the right side we get 3t+5 = bt+(a+b), so we conclude that b = 3, and then a must be 2. That is, p(t) = 2(1) + 3(t+1), so [p(t)] B = [2 3 0] T.

7 A basis for the column space is given by the first, second, and fourth columns of A. We know that these are linearly independent because the corresponding columns of B are linearly independent, and because row operations do not change linear dependencies between the columns. Now we know that the column space of A is spanned by the 5 columns of A. But we can see in B that the third column is a linear combination of the first 2, and that the fifth column is a linear combination of columns 1, 2, and 4. The same must be true of the columns of A. So we see that the 3 rd and 5 th columns of A are dependent on columns 1 2 and 4, and can thus be eliminated without changing the set spanned by the columns. So we see that columns 1, 2, and 4 are linearly independent and span the column space. That makes them a basis for the column space. The nonzero rows of B form a basis for the row space. We can see that they are linearly independent, and they clearly span the row space of B. Since that is the same as the row space of A, we can see that these rows are a basis for the row space of A. The null space of A is the set of solutions to the homogeneous equation Ax = 0. Looking at B, we see that it is in rref form, so this is the rref of A. Solving for the basic variables, we find x 1 = 2x 3 x 5 x 2 = -x 3 + 2x 5 x 4 = 0x 3 x 5. This leads to the general solution of the homogeneous equation as [x 1 x 2 x 3 x 4 x 5 ] T = x 3 [ ] T + x 5 [ ] T and that shows that the two vectors on the right side of the equation are a spanning set for the null space. But they are also independent, as we can observe by looking at the entries in positions 3 and 5 neither can be a multiple of the other. Since they are independent and span the null space, they are a basis for the null space.

8 In a space of dimension n any set of more than n vectors is dependent. (That is a theorem from the book). The four vectors are independent, so they form a basis for the subspace that they span. But that is then a four dimensional subspace of V, and therefore must actually equal V. Therefore, the given vectors are a basis for V and so must be a spanning set. Either the vectors are independent or they are not. If they are independent, then they are a basis for V and so dim V = 3. If they are dependent, we can eliminate one or more to leave a spanning set that is independent. That will still be a basis for V, but now with fewer than 3 elements. In this case the dimension of V is less than 3. This shows that across all possibilities, dim V 3.

Linear Algebra: Sample Questions for Exam 2

Linear Algebra: Sample Questions for Exam 2 Instructions: This is not a comprehensive review: there are concepts you need to know that are not included. Be sure you study all the sections of the book and