MATH 218 Other, Fall 2010

Brief Solutions in Math 218 Ch. 3: TF 4 is False: dim =4, so every basis of this vector space has 4 elements. But S has 3 only elts! Ch 3: TF 7 is False S is not closed under addition because, for example, det ( although each has zero determinant Ch 3: TF 18 is False: Correction: If S, T are subspaces of V, then S+T is a subspace of V. OR we justify by a counter example: Let S=X and T=Y be the x-axis & y-axis in . Then the vecors i,j are in ; while their sum ----------------------------------------------------------------------------------------------------- Ch 4: TF 8 is False: Correction: You need T to be 1-1 to preserve (arbitrary) linear independent subsets. Ch 4: TF 9 is True Both vector spaces have dimension 9, so they are isomorphic vector spaces (theorem) Ch 4: TF 22 is False Because the following rank-nullity Theorem fails dim (N(T))+ dim (R(T))=dim (domain space V) (2 +3 4) Ch 4: TF 39 is True Proof: Say . If T is 1-1, then dim(N(T))=0. Now the rank-nullty Theorem. (above) implies that dim (R(T))=5 which is impossible since R(T) is a subspace of Ch 5: TF 25 is True If the [nxn] matrix A has n distinct eigenvalues, then A has n linearly independent eigen vectors. Hence A is diagonalizable. Warning: TF 24 is False because even if we have less than n eigenvalues, the matrix A may succeed to be diagonalizable like A=I or A=5I (It is already diagonal with one eigenvalue 5). Ch 5: TF 36 is True (see the proof of Theorem 4 p. 293) Ch 5: TF 35 is True. Here we need the famous problem: A & its transpose have the same eigenvalues (5.2 problem 30). So we may assume that the sum of entries in each row is a constant c. (Note: If this constant sum is 1, A is called a probability matrix or a stochastic matrix). Now the vector (1 1 … 1) is obviously an eigenvector with eigenvalue c by inspection. Good Luck