Question

1. (20 pts) For each of the following statements, please circle T (True) or F (False). You do not need to justify your answer. (a) T or F? Any eigenvector of a matrix is in the column space of the matrix. (b) T or F? The number of singular values of a matrix is also its rank. (c) T or F? If A is an m × n with m < n, then the dimension of its column space is greater than the dimension of its row space. (d) T or F? A symmetric matrix is diagonalizable. (e) T or F? The null space null(A) of a matrix A is orthogonal to the column space of AT . (f) T or F? Zero can be the eigenvalue of an elementary matrix. (g) T or F? If W is a vector space spanned by 4 vectors, them the dimension of W is 4.

Answer 1

Solution:

(a) False

Consider a matrix . Column space of = and eigenvectors of .

Clearly neither nor belongs to column space of .

(b) False

consider a matrix , clearly eigenvalues of are . so singular values of .Here we get two singular values but the rank of the matrix is one.

(c) False

We know that for any matrix A, rank(A)= dim( column space of A)=dim( row space of A)

(d) True

Every symmetric matrix is Diagonalizable.

(e) True

because for any matrix , row space of = column space of =

(f) False

Elementary matrix is a matrix which differs from the identity matrix by one single elementary row operation.so determinant of an Elementary matrix is equals to -(determinant of identity matrix= -1.since product of eigenvalues of a matrix is equal to determinant of the matrix, Hence zero is not an eigenvalue.

(g) False

consider the set .Now linear span of that is =.clearly dimension of is not equals to 4.