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-umaryammu1512543703288.pdf. Adobe Acrobat pro File Edit View Window Help Problem3 For each matrix, find the following a) b) c) d) e) The basis for the row and column spaces. The dim of row space, the dim of column space and the rank of A. The rank ofits transpose The basis for the null space The nullity of A. Given the following matrices A-2 08 B=| 0 cos@ -sm 0 sin cos 0 1 -1 C-0-1 0 1) Are the following matrices symmetric, skew-symmetric or orthogonal, justify your answer 2) Find the eigenvalues of them and algebraic multiplicity of each 3) Find the spectrum of and the spectral radius of A. 4) Justify the relation between the eigenvalues, trace and determinant of A 5) The eigenvectors of A, geometric multiplicity and the defect of each and the basis for the eigenspaces of A for every . EN ..@.ial 4) 12/16/2017 9:48 PMExplanation / Answer
3.1 a). The RREF of A is I3. It implies that {(6,5,2),(2,0,-8),(5,4,0)} is a basis for Row(A) and {(6,5,2)T,(5,0,4)T,(2,-8,0)T} is a basis for Col(A).
b). From the RREF of A, it is apparent that dim(Row(A)) = 3, dim(Col(A))= 3 and Rank(A) = 3.
c). The rank of AT is same as Rank(A) = 3.
d). The null space of A is the set of solutions to the equation AX = 0. If X = (x,y,z)T,then, in view of the RREF of A, this equation is equivalent to x = 0,y=0 and z = 0. Hence the equation AX = 0 has only the trivial solution and the basis of Null(A) is the empty set.
e). The nullity of A is 0.
3.2.a). The RREF of B is
1
0
0
0
1
0
0
0
1
It implies that {(1,0,0),(0, cos,- sin),(0, sin, cos)} is a basis for Row(BA) and{(1,0,0)T,(0,1,0)T,(0,0,1)T} is a basis for Col(B).
b). From the RREF of B, it is apparent that dim(Row(B)) = 3, dim(Col(B))= 3 and Rank(B) = 3.
c). The rank of BT is same as Rank(B) = 3.
d). The null space of B is the set of solutions to the equation BX = 0. If X = (x,y,z)T,then, in view of the RREF of B, this equation is equivalent to x = 0,y=0 and z = 0. Hence the equation BX = 0 has only the trivial solution and the basis of Null(B) is the empty set.
e). The nullity of B is 0.
3.3.a). The RREF of C is I3. It implies that {(0,1,-1),(0,-1,0),(1,0,0)} is a basis for Row(C) and {(1,0,0)T,(0,1,0)T,(0,0,1)T} is a basis for Col(C).
b).From the RREF of C, it is apparent that dim(Row(C)) = 3, dim(Col(C))= 3 and Rank(C) = 3.
c). The rank of CT is same as Rank(C) = 3.
d). The null space of C is the set of solutions to the equation CX = 0. If X = (x,y,z)T,then, in view of the RREF of C, this equation is equivalent to x = 0,y=0 and z = 0. Hence the equation CX = 0 has only the trivial solution and the basis of Null(C) is the empty set.
e).The nullity of C is 0.
Note:
B can be reduced to its RREF as under:
1.Multiply the 2nd row by cos
2. Add sin times 3rd row to the 2nd row.
3. Add – sin times 2nd row to the 3rd row.
4. Multiply the 3rd row by 1/cos
1
0
0
0
1
0
0
0
1
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