Question

Question 4. The spectral decomposition (or the orthogonal eigenvalue decomposi- tion) of a matrix A whose determinant is zero is given by A = (2) [11* • -*] +/- +] + (-1). tao ta + (e)- vv V2 for some v € Ry, and a real number c ER. (a) (5 points) Find the eigenvalues of A and the value of c. You must justify your answer. (b) (5 points) Find v. (c) (5 points) The matrix A can expressed as A=U DUT for some 3 x 3 orthogonal matrix U (that is for some 3 x 3 matrix U with orthonormal columns) and a 3 x 3 diagonal matrix D. :-1 0 0 If D 0c0 then find U. 0 0 2 } 1 (d) (5 points) Given a, b € R, find a vector x € R3 in terms of a,b such that x"Ax = 02-62 (Hint: A properly performed change of variable x = Py yields x" Ax = cy} + czyż + c3y for some constants C1,C2,C3 € R.)

Answer 1

a) From the spectral decomposition of A, the eigenvalues of A are 2, -1,c. As det(A)=0, so the product of the eigenvalues of A is 0. Hence c=0 and the eigenvalues are 2,-1,0.

b) From spectral decomposition,            
A=.5[1 0 -3; 0 0 0; -3 0 1]. As v is eigenvector corresponding to eigenvalue 0, v=[0 1 0]^t.

c) Let w=[1/√2 0 -1/√2]^t, u=[1/√2 0 1/√2]^t and D=diag (-1,0,2). As A=UDU^t,

U=[u v w]=[1/√2 0 1/√2; 0 1 0; 1/√2 0 -1/√2].

d) for x in R^3, let Y=U^t x in x^t Ax=a^2-b^2, we get    Y^t DY=a^2-b^2, or 2(y3)^2-(y1)^2=a^2-b^2.

So y3=a/√2, y1=b.   [ Where Y=(y1 y2 y3)^t].

Now Y=[(x1+x3)/√2 x2   (x1-x3)/√2]^t.

Hence x1=(a+b√2)/√2, x3=(b√2-a)/√2, take x2=0.

x=[(a+b√2)/√2 0 (b√2-a)/√2]^t.