4-30 Let A be the following mtix3 12 0 003 (a) Enter its characteristic equation
ID: 3114339 • Letter: 4
Question
4-30 Let A be the following mtix3 12 0 003 (a) Enter its characteristic equation below. Note you must use p as the parameter instead of and you must enter your answer as a equation, with the equals sign. (b) Enter the eigenvalues of the matrix, including any repetition. For example 16,16,24 (c) Find the eigenvectors, and then use Gram-Schmidt to find an orthonormal basis for each eigenvalue's eigenspace. Build an orthogonal matrix whose columns are these eigenvectors. Enter this matrix, as a list of row vectors, below. 1 2 3 For example the matrix4 5 6 would be entered as [1,2,3114,5,61,17.8,9] 789 (d) Enter the diagonal matrixP AP for your matrix P.Explanation / Answer
(a) The characteristic equation of the matrix A is det (A-pI3) = 0. Now, A-pI3 =
4-p
-3
0
-3
12-p
0
0
0
3-p
and det(A-pI3) = -p3+19p2 -87p+117 so that the characteristic equation of the matrix A is -p3 + 19p2 -87p+117 = 0 or, p3 - 19p2 +87p-117 = 0.
(b) The eigenvalues of A are the roots of its characteristic equation. Further, p3 - 19p2 +87p-117= (p-13)(p-3)2 so that the eigenvalues of A are 3,3,13.
(c) The eigenvectors of A are solutions to the equation (A-pI3) X = 0, where p is an eigenvalue of A. When p =3, the matrix A-3I3 can be row –reduced to its RREF as under:
Add 3 times the 1st row to the 2nd row
Then the RREF of the matrix A-3I3 is
1
-3
0
0
0
0
0
0
0
Now, if X = (x,y,z)T, then the equation (A-3I3 )X = 0 is equivalent to x-3y = 0 or, x = 3y. Then X = (3y,y,z)T = y(3,1,0)T +z(0,0,1)T. Hence the eigenvectors of A corresponding to the eigenvalue 3 are (0,0,1)T and (3,1,0)T.
Similarly, the eigenvector of A corresponding to the eigenvalue 13 is solution to the equation (A-13I3) X = 0. The RREF of A-13I3 is
1
1/3
0
0
0
1
0
0
0
Thus, the the equation (A-13I3 )X = 0 is equivalent to x+y/3 = 0 or, x = -y/3 and z = 0. Then X = (-y/3,y,0)T = 1/3(-1, 3,0)T. Hence the eigenvector of A corresponding to the eigenvalue 13 is (-1,3,0)T. The eigenspace of A corresponding to the eigenvalue 3 is span{(0,0,1)T , (3,1,0)T} and the eigenspace of A corresponding to the eigenvalue 13 is span{(-1,3,0)T}.
Let (0,0,1)T = v1 , (3,1,0)T = v2 and (-1,3,0)T =v3. We may observe that v1.v2 = 0, so that v1 and v2 are already orthogonal to each other. We only need to orthonormalise these vectors by reducing their norms to 1. Now, v is alreasdy a unit vector. Also, v2/||v2|| = (3,1,0)T/((9+1+0)) = (3/10,1/10,0)T . Then an orthonormal basis for the eigenspace of A corresponding to the eigenvalue 3 is { (0,0,1)T, (3/10,1/10,0)T}.
Further, v3/||v3|| = (-1,3,0)T/((1+9+0)) = (-1/10,3/10,0)T . Then an orthonormal basis for the eigenspace of A corresponding to the eigenvalue 3 is {(-1/10,3/10,0)T }.
The required orthogonal matrix is P =
0
3/10
-1/10
0
1/10
3/10
1
0
0
or, [0,0,1],[3/10,1/10.0],[-1/10,3/10,0].
Finally, PT AP =
3
0
0
0
3
1/10(-9+310)
0
0
1/10(13+3910)
4-p
-3
0
-3
12-p
0
0
0
3-p
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