Current Grade: 40/10 This question concerns the following matrix: A = This malri

Question

Current Grade: 40/10 This question concerns the following matrix: A = This malrix is symmetric so canbe hogonaly diagonalised. (a) Enter the eigenvalues of A in increasing order, separated by commas. 0,4 (b) Find an eigenvector for each eigenvalue. Enter these eigenvectors as a list, e.g. 10,1,1.0 I-1/sqrt(3),1].[sqrt(3),1] (c) For each eigenvalue , find an orthonormal basis for the eigenspace Let P be a matrix with these orthonormal eigenvectors as columns, Enter the matrix P as a list of row vectors For example, the matrix s entered as [1,2),13,4 (a) Enter the matrx product PTAP

Explanation / Answer

(a)The characteristic equation of A is det(A-I2)= 0 or, 2-4= 0 or, (-4)= 0. Hence the eigenvalues of A are 1 =4 and 2 = 0. The answer is 4,0.

(b) Further, the eigenvector of A associated with its eigenvalue 4 is solution to the equation (A-4I2)X = 0. To solve this equation, we will reduce A-4I2 to its RREF as under:

Multiply the 1st row by -1.

Add -3 times the 1st row to the 2nd row

Then the RREF of A-4I2 is

1

-3

0

0

Now, if X =[x,y], then the equation(A-4I2)X = 0 is equivalent to x-y3= 0 or, x = y3. Then X = [y3,y] = y[3,1]. Thus, the eigenvector of A associated with its eigenvalue 4 is v1 = [3,1].

Similarly, the eigenvector of A associated with its eigenvalue 0 is solution to the equation AX = 0. To solve this equation, we will reduce A to its RREF as under:

Add -3 times 2nd row to the 1st row.

Multiply 2nd row by 1/3

Interchange the 1st and the 2nd rows.

Then the RREF of A is

1

1/3

0

0

Now, if X =[x,y], then the equation AX=0 is equivalent to x+y/3=0 or, x=-y/3. Then X =[-y/3,y] =                y[-1/3,1]. Thus, the eigenvector of A associated with its eigenvalue 0 is v2 =[-1/3,1]. The answer is [3,1], [-1/3,1].

(c ). We have v1.v2 = [3,1]. [-1/3,1] = -1+1= 0. Hence the eigenvectors of A are orthogonal to each other. Now, let u1 = v1/||v1|| = v1//(3+1) = v1/2 = [3/2,1/2] and u2 = v2/||v2|| = v2/(1/3+1)=                 [-1/2, 3/2]. Then E4 = {[3/2,1/2]} and E0 ={[-1/2, 3/2]}.

P =

3/2

-1/2

1/2

3/2

Answer [3/2,-1/2],[1/2, 3/2].

(d). We have PTAP=

4

0

0

0

1

-3

0

0

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