Question

3, (a) [5 marks] what does it mean for A E Rnxn to be (i) symmetric? (ii) orthogonal? (ii) diagonalisable? (iv) orthogonally diagonalisable? (b) [4 marks] Suppose that A ERn is orthogonally diagonalisable. Prove that A is symmetric. (c) [11 marks] Let A be the matrix 6 -2 Show that the eigenvalues are 7 and -6. Show that any corresponding eigenvectors vi and v2 are orthogonal with respect to the Euclidean inner product (d) [5 marks] Hence prove that the matrix A in (c) is orthogonally diagonalisable.

Answer 1

3(a).

    (i). If a nxn real matrix equals its transpose,i.e. if A = A^T, then A is said to be a symmetric matrix.

  (ii). A real orthogonal matrix is a square matrix with real entries whose columns and rows are orthogonal unit vectors.

    (iii). A real nxn matrix A is said to be diagonalizable, if there exists a nxn diagonal matrix D and a nxn invertible matrix P such that A = PDP^-1.

    (iv). A real nxn matrix A is said to be orthogonally diagonalizable if there is an orthogonal matrix S such that S^-1AS is a nxn diagonal matrix.

(b). Let A be a real nxn matrix. Then, as per the Spectral theorem, A is orthogonally diagonalizable if and only if A is symmetric.

(c ). The the solutions to its characteristic equation det(A-?I₂) = 0 or, ?²-?-42 = 0 or, (?-7)(?+6) = 0. Thus, the eigenvalues of A are 7 and -6. The eigenvector of A associated with its eigenvalue 7 is solution to the equation(A-7I₃)X = 0. The RREF of A-7I₃ is

1	-3/2
0	0

Now, if X = (x,y)^T, then the equation (A-7I₃)X = 0 is equivalent to x-3y/2 = 0 or, x = 3y/2. Then X = (3y/2,y)^T = (y/2)(3,2)^T. Therefore,the eigenvector of A associated with its eigenvalue 7 is v₁ = (3,2)^T. Similarly, the eigenvector of A associated with its eigenvalue -6 is v₂ = (-2,3)^T.Further, v₁.v₂ =(3,2)^T . (-2,3)^T = -6+6 = 0. Hence the eigenvectors v₁ and v₂ of A associated with its eigenvalues 7 and -6 are orthogonal.

(d). It may be observed that A^T = A i.e.tric. Hence, by the Spectral theorem, A is orthogonally diagonalizable.