Mini-multiples: Circle all answers that are TRUE. (Sorry, no partial credit here
ID: 2984940 • Letter: M
Question
Mini-multiples: Circle all answers that are TRUE. (Sorry, no partial credit here; so THINK! and check.) (Unless stated, assume that A is an n times n matrix in the problems on this page.) A solution always exists for the equation Ax = b if b is a vector in Col A. is a dependent vector with respect to the column vectors of A. a vector in Row A. is the zero vector. none of the these If det (A) 0, then A is invertible A is invertible Col A = Row A A is diagonalizable none of the these If A is an n times n matrix and Null A = {0},then zero is NOT an eigenvalue of A. rank = n A has n linearly independent eigenvectors A has n distinct eigenvalues none of the these If a set of vectors is closed for addition and contains the zero vector, then the set of vectors is a vector space. is also closed for vector subtraction. is also closed for scalar multiplication. always has a basis. none of the these With respect to the characteristic equation of an n times n matrix A, if the algebraic multiplicity is equal to the geometric multiplicity, then matrix A has n distinct eigenvalues. matrix A has n distinct eigen spaces. matrix A has n independent eigenvectors. matrix A is diagonalizable. none of the these Mini-TRUTHS: True (under all interpretations, it is TRUE), Sometimes (depending upon conditions, it could be TRUE or FALSE), or FALSE (under all interpretations, it is FALSE). Circle your choice. TRUE, SOMEtimes, or FALSE With respect to square matrices, invertibility is equivalent to diagonalizability. TRUE, SOMEtimes, or FALSE If B = {b1, b2, b3, ..., bn} spans a vector space, V, then the dim(V) is less than or equal to n. TRUE, SOMEtimes, or FALSE Given a set of vectors B = {b1, b2, b3, ...,bn}, then span (b1, b2, b3,bn) is a vector space. TRUE, SOMEtimes, or FALSE If a basis for matrix A is B = {b1, b2, b3,..., bn} and A has n distinct eigenvalues ( lambda 1, lambda 2, lambda 3, ..., lambda n), then A is diagonalizable with A=P D P-1, where P has column vectors [b1 b2 b3 ... bn] and D is a diagonal matrix with the eigenvalues ( lambda 1, lambda 2, lambda 3, ..., lambda n). TRUE, SOMEtimes, or FALSE The solution set of the matrix equation. (A - lambda I) x = 0, is an eigenspace of A.Explanation / Answer
1) b
2) a,b
3) a,b,c,d
4)a
5) a,,b,c,d
1)sometimes
2)true
3)sometimes
4)true
5)true