Linear Algebra Homework: Matrix and Polynomial Problems Solved

This document provides a comprehensive solution to a linear algebra homework assignment. The solution addresses several problems, including demonstrating non-similar matrices with identical eigenvalues and eigenvectors, determining matrix congruence, proving a greatest common divisor property for polynomials, proving the linearity of a transformation based on Euclidean division, and determining the diagonalizability of a matrix in GF(5). The solutions employ rigorous mathematical reasoning and detailed calculations to arrive at the correct answers. The assignment covers fundamental concepts in linear algebra, such as eigenvalues, eigenvectors, matrix similarity, congruence, polynomial factorization, and linear transformations. The solutions provide clear explanations and steps, making it a valuable resource for students studying linear algebra and seeking to understand and solve similar problems.

Running head: LINEAR ALGEBRA
LINEAR ALGEBRA
Name of the Student
Name of the University
Author Note

1. Let the two matrixes A and B are
A =
[0 1 0 0
0 0 0 0
0 0 0 1
0 0 0 0 ], B =
[0 1 0 0
0 0 1 0
0 0 0 0
0 0 0 0 ]
Now, the eigenvalues of A are 0 and eigenvectors are
[1
0
0
0 ],
[−1
0
0
0 ],
[0
0
1
0 ],
[ 0
0
−1
0 ]
The eigenvalues of B are also 0 and so as the eigenvector.
As the eigenvalues and the eigenvectors are same hence the minimum polynomial equation,
Eigen space dimensions are also same.
Now, A2 = 0, B2 ≠ 0 and however, let B = P−1∗A∗P.
Then B2=P−1∗A2∗P
This becomes contradictory as if A2 is null matrix then P−1∗A2∗P needs to be null matrix.
Hence no matrix P exists such that B = P−1∗A∗P.
Hence, A and B matrixes are not similar but satisfies the same eigenvalues, eigenvector
conditions.
2. if two matrixes A and B are congruent over M4 then
B=M T∗A∗M
Then, det ( B )=det ( MT )∗det ( A )∗det ( M )
 det (B)/det( A)=de t2 (M )(As det ( M T ) = det ( M ))
Hence, det (B)/det( A) must be a perfect square if the matrixes are congruent in M4.

Now, A =
[0 2 1 1
2 0 1 −1
1 1 0 0
1 −1 0 −1 ], B =
[−1 0 −1 −1
0 0 0 −1
−1 0 −1 −2
−1 −1 −2 −2 ]
Now, det(B)/det(A) = 0/0 form which is not a perfect square as the square term is undefined.
Hence, A and B are not congruent in M4.
3. It is given that P ∈ [𝑋] which satisfies P(1) = 0 ∧ 𝑃(2) = 5
Now, X2 + X – 2 = X^2 + 2X –X – 2 = (X-1)(X+2).
Now, if gcd(P, X2 + X – 2¿ = (X-1) then (X-1) must be in P.
Now, P(1) = 0 means (X-1) is a factor of P.
Now, P(2) = 5 means (X-2) is not factor of P as if (X-2) is a factor then P(2) = 0.
Hence, the maximum factor that can be taken out from P and X2 + X – 2 is (X-1) or
gcd(P, X2 + X – 2¿ = X-1 (proved).
4. Given, 𝔽 and S ∈𝔽[X] is a polynomial. Here, T is defined as a function which matches
every polynomial P with the remainder of Euclid division of P with S.
Hence, by Euclid lemma P = q*S + 𝔽[X]. (where q is any integer)
Now, by properties of linear transformation:
If T is a linear transformation then
T(cP) = c*T(P) (where c is any constant)
Now, T(cP) = T(c(q*S + 𝔽[X])) = c(T(q*S + 𝔽[X])) (As T is defined for every polynomial P
hence c can be taken common from all the polynomial)
Thus T(cP) = c*T(P)

Hence, The transformation T: [𝑋] → 𝔽[𝑋] is a linear transformation (proved)
5. Given, A = [3 0 0
1 1 3
2 1 4 ] is a matrix in GF(5). A is not a diagonal matrix as its all off-
diagonal matrix elements are not zero.
However, A can be diagonizable if all the eigenvalues of the matrix are different from each
other. The Eigen-values of the matrix can be computed by the following method.
det ( [3 0 0
1 1 3
2 1 4 ]−λ [1 0 0
0 1 0
0 0 1 ]) = 0
 det ( [3−λ 0 0
1 1− λ 3
2 1 4−λ ]) = 0
 λ = 4.7913, 0.2087 and 3
Hence, as all the eigenvalues are unique hence the matrix is diagonizable.