Linear algebra final exam questions part2 4, (12 points) For the matrix A = | 1-
ID: 3168935 • Letter: L
Question
Linear algebra final exam questions part2 4, (12 points) For the matrix A = | 1-3-11 , find the eigenvalue(s) (in- cluding the complex ones, if any) and determine bases for the eigenspaces and generalized eigenspaces. For each eigenvalue, determine the algebraic and geometric multiplicity. 5. (12 points) Let A- Is there a matrix P with real entries and a 0 -1 diagonal matrix D such that D = P-1 AP? If yes, find them, if not, explain why 6, (4+2+2=8 points) (a) Write down the definitions of "self-adjoint matrix" and "unitary matrix" (b) which values, if any, of a and b make the m atrix C self adjoint? l. (c)-For which real number pairs (a, b), if any, is the matrix C = unitary? Extra Credit Problem. (5 points) Let U and V be two n × n real unitary matrices. Show that UV is also unitaryExplanation / Answer
4. The characteristic equation of the matrix A is det(A-I3)= 0 or, 3+62+12+8 = 0 or, (+2)3 =0. Thus, the only eigenvalue of A is -2 ( of algebraic multiplicity 3). The eigenvectors associated with this eigenvalue are solutions to the equation (A+2I3)X = 0. To solve this equation, we will reduce A+2I3 to its RREF as under:
Interchange the 1st row and the 2nd row
Add 1 times the 1st row to the 3rd row
Then the RREF of A+2I3 is
1
-1
-1
0
0
0
0
0
0
Now, if X = (x,y,z)T, the equation (A+2I3)X = 0 is equivalent to x-y-z = 0 or, x = y+z. Then nX = (y+z,y,z)T = y(1,1,0)T+z(1,0,1)T. Thus, the eigenvectors of A are (1,1,0)Tand(1,0,1)T. Therefore, the geometric multiplicity of the eigenvalue -2 of A is 2 ( no. of associted linearly independent eigenvectors). The eigenbasis for A is {(1,1,0)T,(1,0,1)T }.
5. The characteristic equation of the matrix A is det(A-I2)= 0 or, 2--2 = 0 or,(-2)(+1)=0. Therefore, the eigenvalues of A are 1 = 2 and 2 =-1. The eigenvector associated with the eigenvalue 2 is solution to the equation (A-2I2)X= 0. The RREF of A-2I2 is
0
1
0
0
Now, if X = (x,y)T, then the equation (A-2I2)X= 0 is equivalent to y = 0. Then X = (x,0)T = x(1,0)T. Hence the eigenvector associated with the eigenvalue 2 is (1,0)T. Similarly, the eigenvector associated with the eigenvalue -1 is (-1,3)T. Further, let D =
2
0
0
-1
and P =
1
-1
0
3
Then D = P-1AP.
Note:
The entries on the leading diagonal of D are the eigenvalues of A and the columns of P are the corresponding eigenvectors of A.
Please post the remaining question again,
1
-1
-1
0
0
0
0
0
0
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