True of False: Explain. A. Let A be an nxn matrix. Then det(5A) = 5detA. B. Let

Question

True of False: Explain. A. Let A be an nxn matrix. Then det(5A) = 5detA. B. Let A be an nxn matrix. Then det((A^T)A)greater than or equal to 0. C. Let A be an nxn invertible matrix. Then detA^-1=detA. D. Prove that if x is in Row(A), then it is orthogonal to every vector in Nul A. E. Suppose A=PRP^-1, where P is orthoginal and R is upper triangular. Show that if A is Symmetric, then R is also Symmertic. Moreover, show that R must be diagonal. D. Prove that if x is an eigenvector the matrix product AB and Bx not = to 0, then Bx is an eigenvector of BA.

Explanation / Answer

a )false - its (5^n)* (det(A)), for an n*n matrix, because each row holds a value of 5, and that number of 5's are to be taken in common, so, when the whole is considered it becomes, 5^n, for 3 by 3 matrix, its 5^3 ie 125. b) TRUE, the determinent is always supposed to be positive number, anb a positive number holds value of 0 or greater than 0. c)false - det A^(-1) = 1/det(A) d)In some sense, the row space and the nullspace of a matrix subdivide Rn into two perpendicular subspaces. For A = [1 2 5 2 4 10] , the row space has dimension 1 and basis [1 2 5] and the nullspace has dimension 2 and is the plane through the origin perpendicular to the vector [1 2 . 5] Not only is the nullspace orthogonal to the row space, their dimensions add up to the dimension of the whole space. We say that the nullspace and the row space are orthogonal complements in Rn . The nullspace contains all the vectors that are perpendicular to the row space, and vice versa. We could say that this is part two of the fundamental theorem of linear alge­ bra. Part one gives the dimensions of the four subspaces, part two says those subspaces come in orthogonal pairs, and part three will be about orthogonal bases for these subspaces. N((A^T)A) = N(A) Due to measurement error, Ax = b is often unsolvable if m > n. Our next challenge is to ?nd the best possible solution in this case. The matrix ( A^T)A plays a key role in this effort: the central equation is (A^T)Axˆ = (A^T)b. We know that A^TA is square (n × n) and symmetric. rank of (A^T)A = rank of A.

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