University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2024
Views:

350

Pages:

2
Author:

Beckett Gray

Question 1 It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace corresponding to lambda. Find ℎ h in the matrix 𝐴 A below such that the eigenspace for 𝜆 = 5 λ=5 is two-dimensional: 𝐴 = [ 5 − 2 6 − 1 0 3 − 6 0 0 0 ℎ 4 0 0 0 1 ] . A= 5 0 0 0 −2 3 0 0 6 −6 h 0 −1 0 4 1 . Answer Row reduce the augmented matrix for the equation ( 𝐴 − 5 𝐼 ) 𝑥 = 0 (A−5I)x=0: [ 0 − 2 6 − 1 0 − 2 − 6 0 0 0 ℎ − 5 4 0 0 0 − 4 ] → [ 0 − 2 6 − 1 0 0 0 3 0 0 ℎ − 5 4 0 0 0 0 ] . 0 0 0 0 −2 −2 0 0 6 −6 h−5 0 −1 0 4 −4 → 0 0 0 0 −2 0 0 0 6 0 h−5 0 −1 3 4 0 . For a two-dimensional eigenspace, the system above needs two free variables. This happens if and only if ℎ = 6 h=6: [ 0 − 1 − 3 0 0 0 0 4 0 0 1 0 0 0 0 0 ] . 0 0 0 0 −1 0 0 0 −3 0 1 0 0 4 0 0 . Question 2 Suppose 𝐴 A is a 6 × 9 6×9 matrix. If Nullity ( 𝐴 𝑇 𝐴 𝑇 ) = 2 (A T A T )=2, then Nullity ( 𝐴 ) = 2 (A)=2. Answer It is given that 𝐴 𝐴 AA is a 6 × 9 6×9 matrix and Nullity ( 𝐴 𝑇 ) = 2 (A T )=2. Let rank ( 𝐴 ) = 𝑘 rank(A)=k. Then we know that rank ( 𝐴 𝑇 ) = 𝑘 . rank(A T )=k. By Rank-nullity theorem we have: rank ( 𝐴 ) + Nullity ( 𝐴 ) = 9 and rank ( 𝐴 𝑇 ) + Nullity ( 𝐴 𝑇 ) = 6. rank(A)+Nullity(A)=9andrank(A T )+Nullity(A T )=6. It follows that: rank ( 𝐴 𝑇 ) = 6 − 2 = 4 ⟹ 𝑘 = 4. rank(A T )=6−2=4⟹k=4. Therefore, we get: Nullity ( 𝐴 ) = 9 − 𝑘 = 9 − 4 = 5. Nullity(A)=9−k=9−4=5. Hence, the given statement is False. Question 3 Show that if 𝐴 A and 𝐵 B are 𝑛 × 𝑛 n×n matrices such that 𝐴 𝐵 = 0 AB=0 (where 𝑂 O is the zero matrix), then 𝐴 A is not invertible or 𝐵 B is not invertible. Answer If 𝐴 A is not invertible, we are done. If 𝐴 A is invertible, we can multiply 𝐴 𝐵 = 0 AB=0 by 𝐴 − 1 A −1 from the left: 𝐴 − 1 ( 𝐴 𝐵 ) = 𝐴 − 1 𝑂 ⟹ 𝐵 = 0. A −1 (AB)=A −1 O⟹B=0. So 𝐵 = 0 B=0, meaning that 𝐵 B is not invertible. Question 4 𝑇 T must be a linear transformation, we assume. Can you find the 𝑇 T standard matrix? 𝑇 : 𝑅 2 → 𝑅 4 , 𝑇 ( 𝑒 1 ) = ( 3 , 1 , 3 , 1 ) and 𝑇 ( 𝑒 2 ) = ( − 5 , 2 , 0 , 0 ) , T:R 2 →R 4 ,T(e 1 )=(3,1,3,1)andT(e 2 )=(−5,2,0,0), where 𝑒 1 = ( 1 , 0 ) e 1 =(1,0) and 𝑒 2 = ( 0 , 1 ) e 2 =(0,1). Answer To find the standard matrix for the linear transformation 𝑇 : 𝑅 2 → 𝑅 4 T:R 2 →R 4 , we need to determine how 𝑇 T maps the standard basis vectors 𝑒 1 e 1 and 𝑒 2 e 2 of 𝑅 2 R 2 to 𝑅 4 R 4 . Given: 𝑇 ( 𝑒 1 ) = ( 3 , 1 , 3 , 1 ) , 𝑇 ( 𝑒 2 ) = ( − 5 , 2 , 0 , 0 ) . T(e 1 )=(3,1,3,1),T(e 2 )=(−5,2,0,0). The standard matrix for 𝑇 T can be formed by arranging the images of the basis vectors as columns: 𝐴 = [ 𝑇 ( 𝑒 1 ) , 𝑇 ( 𝑒 2 ) ] . A=[T(e 1 ),T(e 2 )]. Substituting the given values, we have: 𝐴 = [ 3 − 5 1 2 3 0 1 0 ] . A= 3 1 3 1 −5 2 0 0 . Therefore, the standard matrix for the linear transformation 𝑇 T is: 𝐴 = [ 3 − 5 1 2 3 0 1 0 ] . A= 3 1 3 1 −5 2 0 0 .