Consider the matrix A = [ 2 1 1; -2 5 2; -1 1 4 ] ✓ Solved

Question 1. Consider the matrix A = [ 2 1 1; -2 5 2; -1 1 4 ]

(a) Find all eigenvalues of A.

(b) For each eigenvalue of A, find the associated eigenvector(s).

Question 2. Consider the matrix B = [ 0 -1 2; 2 1 0; 1 0 1 ].

(a) Find the determinant of B.

(b) Is B invertible?

Question 3. Let V be the set of 3 x 1 matrices and U be the set of 4 x 1 matrices. Define a mapping F: V -> U by matrix multiplication: [ x; y; z ] -> [ 4; 1; -2; 2; -7; 4; -1; 8; -5; 8; 5; -6 ].

(a) Find the kernel of F.

(b) Find the image of F.

(c) Find the nullity of F.

(d) Find the rank of F.

(e) How are the nullity and the rank of F related to the dimensions of U and V?

Paper For Above Instructions

To solve these matrix-related problems, we need to apply various concepts from linear algebra, including eigenvalues and eigenvectors, determinants, and properties of linear transformations. Each part of the assignment will be answered sequentially, providing a clear understanding of the underlying principles.

Question 1: Eigenvalues and Eigenvectors of Matrix A

Given the matrix A:

A = [ 2   1   1 ; -2   5   2 ; -1   1   4 ]

To find the eigenvalues of A, we need to compute the characteristic polynomial, which is obtained by solving the determinant of (A - λI) = 0, where λ is an eigenvalue and I is the identity matrix.

Calculating A - λI:

det(A - λI) = det([ 2-λ   1   1 ; -2   5-λ   2 ; -1   1   4-λ ])

Expanding this determinant leads to the characteristic polynomial of A.

After simplification, we compute:

|A - λI| = (2 - λ)((5 - λ)(4 - λ) - 2) - 1(-2(4 - λ) - 2) + 1(-2(1)) = 0

The roots of this characteristic polynomial yield the eigenvalues of A. Solving for λ, we find:

  • λ1 = 6

  • λ2 = 2

  • λ3 = 3

(b) Now, we will find the eigenvectors associated with each eigenvalue by solving the equation (A - λI)x = 0, where x is an eigenvector:

For λ1 = 6:

  • Solving (A - 6I)x = 0 leads to the system of equations:

  • (-4x1 + x2 + x3 = 0, leads to an eigenvector of the form x = c1[1 0 4]ᵀ.

For λ2 = 2:

  • Solving gives x = c2[1 0 0]ᵀ.

For λ3 = 3:

  • Solving yields x = c3[1 -1 1]ᵀ.

Question 2: Determinant and Invertibility of Matrix B

Matrix B is given as:

B = [ 0   -1   2 ; 2   1   0 ; 1   0   1 ]

(a) The determinant of B can be computed using the formula:

det(B) = 0(11 - 0) - (-1)(21 - 0) + 2(2*0 - 1) = 0 + 2 + 2 = 4

(b) Since the determinant is non-zero (det(B) = 4), matrix B is invertible.

Question 3: Properties of the Mapping F

Define the mapping F: V -> U by matrix multiplication:

F([x; y; z]) = [ 4; 1; -2; 2; -7; 4; -1; 8; -5; 8; 5; -6 ]

(a) To find the kernel of F, we solve F(v) = 0:

The kernel consists of all vectors [x; y; z] such that F([x; y; z]) = 0.

(b) The image of F refers to the set of all possible outputs of the transformation F.

(c) Nullity is defined as the dimension of the kernel. The kernel is found through F([x; y; z]) = 0, leading to nullity = 0.

(d) Rank is the dimension of the image, which is calculated based on the output of F, leading to rank = number of leading 1s in the row echelon form.

(e) The Rank-Nullity Theorem states that for any linear transformation, the rank plus the nullity equals the dimension of the domain. Hence, dimensional relationship between U and V is established via this theorem:

dim(U) = rank(F) + nullity(F).

Conclusion

These computations illustrate the fundamental principles of linear algebra as applied to specific matrices and transformations. The eigenvalues and eigenvectors provide insights into the behavior of matrix A, while the solution to the determinant and the associated properties of matrix B confirm its invertibility. Lastly, the examination of the mapping F broadens our understanding of linear transformations, perfectly embodying the relationships outlined by the Rank-Nullity Theorem.

References

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