New York University Linear Algebra Questions Response

The homework include College Linear Algebra Questions. I will also provide the notes from the lecture.

Linear Algebra (MATH140/MA-UY3044)
Fall 2019
Homework 9: sections 6.2, 6.4, 6.5 (lectures 18, 19)
(Due Sunday 11/17/2019 before midnight)
1. (10 points) Diagonalize the following symmetric matrices into QΛQT , where Q is an
orthogonal matrix.


7 −4 4
5 0 , with eigenvalues λ = 13, 7, 1.
(a) A =  −4
4
0 9

2
 0
(b) B = 
 0
0
0
1
0
1
0
0
2
0

0
1 
, with eigenvalues λ = 0, 2
0 
1


0.3
0
0
2. (16 points) Let A =  0.2 0.1 0.4 .
0.1
0 0.4
(a) (4 points) Find the eigenvalues of A.
(b) (6 points) Find the eigenvectors associated with each eigenvalue of A.
(c) (5 points) Diagonalize A, and use it to compute lim Ak .
k→∞
(d) (1 points) Suppose A is an n × n matrix that is diagonalizable (so it has n linearly
independent eigenvectors). What must be true for lim Ak to exist? What is
k→∞
needed for Ak → 0? Justify your answer.
3. (3 points) Show that if A and B are similar, then det(A)=det(B).
Linear Algebra (MATH140/MA-UY3044)
Fall 2019
4. (3 points) The algebraic multiplicity of an eigenvalue λ, i.e., the number of times λ is
repeated, is always greater than or equal to its geometric multiplicity, i.e., the number
of linearly independent eigenvectors associated with λ . Find h in the matrix A below,
such that λ = 4 has a geometric multiplicity of 2, where

4
 0
A=
 0
0
2
2
0
0

3 3
h 3 
.
4 14 
0 2
5. (8 points) Label the following statements as True or False. To receive full credit,
explain why a given statement is either true or false.
(a) Two distinct eigenvectors corresponding to the same eigenvalue are always linearly
dependent.
(b) If λ is an eigenvalue of an n × n matrix A, then λ is also an eigenvalue of AT .
(c) An n×n matrix A is diagonalizable if the algebraic multiplicity of each eigenvalue
equals its geometric multiplicity.




2 1
1 0
(d) The matrices
and
are similar.
0 1
0 2
6. (10 points) Consider the sequence xk+2 = 3xk+1 − 2xk for k ≥ 0. Starting with
an initial condition x0 = 0, x1 = 1, compute x63 by finding a general formula for xk in
terms of the initial conditions.