Ejercicios

Show that if A is a symmetric non-singular matrix, then A-1 is also symmetric. Proof: Since A is symmetric, At = A. Since A is nonsingular, A-1 exists. Now, since inverting and transposing performed in either orther yield (A-1)t = (At)-1, the same result = A−1, since At=A. Thus A−1 is symmetric, since it is equal to its own transpose . Q.E.D. (Note: Q.E.D. is an abbreviation for the Latin phrase“Quod erat demonstrandum” which means “That which was to be demonstrated”, that is, the end of the proof.) Let A be an n × n matrix and let B = A + At and C = A − At (a) Show that B is symmetric and C is skew-symmetric. Proof: To show B is symmetric, we must show that it equals its own transpose. Bt = (A + At )t , by assumption by a property of transpose = At + (At )t , = At + A, by another propertyof transpose since matrix addition is commutative =B Thus B = Bt and so B is symmetric. To show C is skew-symmetric, we must show that C = −Ct . by assumption Ct = (A − At )t, = At − (At )t, by a property of transpose t = A − A, by another property of transpose factoring out the scalar −1 = −(A − At ), = −C, by definition of C t Thus C = −C which implies C = − Ct and we see that C isskew-symmetric. Q.E.D. (b) Show that every n × n matrix can be represented as a sum of a symmetric matrix and a skewsymmetric matrix. Proof: Let A be an n×n matrix. Then At exists and is also an n×n matrix. By part (a), A + At is symmetric and A − At is skew-symmetric. Now we notice that (A + At ) + (A − At ) = 2A since matrix addition is associative and commutative. This is close to what we want, but notexactly what we want. We have a symmetric matrix and a skew-symmetric matrix that add to give 2A, the matrix A times the scalar 2. We fix the problem by multiplying both sides by 1/2. ½ [(A + At ) + (A − At )] = ½ (2A) ½ (A + At ) + ½ (A − At ) = A since scalar multiplication distributes over matrix addition. Finally, we note that multiplying a symmetric matrix by a scalar yields a symmetric matrix(the reader should verify this). Similarly, a scalar times a skew-symmetric matrix yields a skewsymmetric matrix (verify). Thus ½ (A + At ) and ½ (A − At ) are symmetric and skew-symmetric respectively and we have expressed A as the sum of a symmetric matrix and a skew-symmetric matrix. Q.E.D.

If A and B are n x n matrices, then (A − B)2 = A2 − 2AB + B2 Solution: Let us begin by analyzing (A −B)2 using facts we know about matrix operations. Note that (A − B)2 = (A − B) (A − B), by definition of (A − B)2 = A(A − B) − B(A − B) since matrix multiplication distributes over matrix addition, = A2 − AB − BA + B2 for the same reason. Now we see clearly that (A − B)2 will be equal to A2 − 2AB +B if and only if AB = BA. Recall, however, that this is not in general true for matrix multiplication.This fact suggests that the given statement is not true in general and indicates how to build a counterexample, i.e., by choosing 2x2 matrices A and B that do not commute with each other.

⎛ 0 1⎞ ⎛1 1 ⎞ ⎛ 1 0⎞ ⎛1 2⎞ 2 2 Thus we let A = ⎜ ⎜ 1 1⎟ and B = ⎜1 0 ⎟ . Then AB = ⎜ 2 1 ⎟ , BA = ⎜ 0 1 ⎟ and (A − B) = I = ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 2 1⎞ ⎛ 1 2⎞ ⎛1 1 ⎞ 2 2 2 I. On the other hand, A2 =⎜ ⎜ ⎜ ⎟ ⎟ ⎜1 2 ⎟ and B = ⎜ 1 1⎟ so that A − 2AB + B = ⎜ − 2 1 ⎟ . Since this ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ matrix is not the identity, we have shown that in general (A − B)2 ≠ A2 − 2AB + B2
If AB = AC and A ≠ 0 (the zero matrix), then B = C.

Solution: We show this to be a false statement by constructing a counterexample. ⎛0 0⎞ ⎛1 2⎞ ⎛1 2⎞ Let A = ⎜ ⎜ 1 0 ⎟ , B = ⎜ 3 4 ⎟ and C = ⎜ 5 6 ⎟ . A calculation showsthat AB = AC = ⎜ ⎜ ⎟ ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Now A ≠ 0 and AB = AC, but B ≠ C.

⎛0 0⎞ ⎜ ⎜1 2⎟ . ⎟ ⎝ ⎠

Let A and B be n x n matrices and let C = AB. Use determinants to show that if either A or B is singular, then C is singular.

Solution:
by assumption; C = AB ⇒ det(C) = det(AB) taking the determinant of each side; ⇒ det(C) = det(A) det(B) by a property of det Recall that the determinant of a...