(1 point) It is a theorem (the \"real spectral theorem\') that any real symmetri
ID: 3116852 • Letter: #
Question
(1 point) It is a theorem (the "real spectral theorem') that any real symmetric matrix has real eigenvalues and is diagonalizable by an orthogonal matrix with 2-. In this problem you will work through this result in an example. Let A--1 24Observe that A is symmetric. -1 4 2J The eigenvalues of A are (Enter your answers as a comma separated list.) Now compute an orthonormal basis for each eigenspace, and take 2 to be the matrix whose columns are those orthonormal eigenvectors. The diagonalization of A is given by A- QDe-1, where D=Explanation / Answer
The characteristic equation of the matrix A is det(A-I3)= 0 or, 3-92+6+56 = 0 or, (-7)(-4)(+2)= 0. Hence, the eigenvalues of A are 1 =7 ,2=4 and 3 =-2. Further, the eigenvector of A associated with the eigenvalue 1 =7 is solution to the equation (A-7I3)X= 0. To solve this equation, we will reduce the matrix A-7I3 to its RREF as under:
Multiply the 1st row by -1/2
Add 1 times the 1st row to the 2nd row
Add 1 times the 1st row to the 3rd row
Multiply the 2nd row by -2/9
Add -9/2 times the 2nd row to the 3rd row
Add -1/2 times the 2nd row to the 1st row
Then the RREF of A-7I3 is
1
0
1
0
1
-1
0
0
0
Now, if X = (x,y,z)T, then the equation (A-7I3)X= 0 is equivalent to x+z = 0 or, x = -z and y-z= 0 or, y = z. Then X=(-z,z,z)T=z(-1,1,1)T.Thus, the eigenvector of A associated with the eigenvalue 1 =7 is v1 = (-1,1,1)T. Similarly, the eigenvectors of A associated with the eigenvalues 2 =4 and 3 =-2 are v2 =(2,1,1)T and v3 = (0,-1,1)T respectively.
The eigenvalues of A are 7,4,-2.
Let e1= v1/||v1||=[ 1/{ (-1)2+12+12}]v1 = (-1/3, 1/3, 1/3)T;
e2= v2/||v2||=[ 1/( 22+12+12)]v2 = (2/6, 1/6, 1/6)T; and
e3= v3/||v3||=[ 1/{ 02+(-1)2+12}]v3 = (0, -1/2, 1/2)T.
The orthonormal bases for the eigenspaces of A associated with the eigenvalues 1 =7, 2 =4 and 3 =-2 are {e1},{e2}and {e3} respectively.
Then Q=
-1/3
2/6
0
1/3
1/6
-1/2
1/3
1/6
1/2
and D =
7
0
0
0
4
0
0
0
-2
1
0
1
0
1
-1
0
0
0
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.