Need help with linear algebra question.

Question

In this problem, you’ll work with some new definitions. Before attempting the problem, read the definitions carefully.

Definition 1 

Let A be an n × n matrix. 

• We define A to be happy if A^2 = A. 

• We define A to be melancholy if A^2 = 0_n.

 • We define A to be nihilistic if there is an integer k ≥ 1 such that A^k = 0_n. 

(Recall that 0_n is the n × n matrix of all zeroes.)

Questions:

(a) Find a 2 × 2 melancholy matrix, not all of the entries of which are zero.

(b) Find a matrix that is nihilistic but not melancholy. 

(c) Show that if an n × n matrix A is happy, then all of its eigenvalues are either 0 or 1. 

(d) True or false: If an n × n matrix A is nihilistic, then 0 is an eigenvalue of A. 

(e) True or false: If an n × n matrix A is happy, then both 0 and 1 are always eigenvalues of A.

Answer 1

First, let me point out that the "silly definitions" here correspond to real terms in linear algebra. Namely,

happy = idempotent,

nihilistic = nilpotent,

melancholy = nilpotent in the special case that k = 2.

Also, note that I will use 0 to denote the (n x n) 0 matrix 0_n.

Here are. your answers:

(a) Let A be the following matrix. It's easy to verify that A² = 0.

0 1

0 0

(b) Let A be the following 3 x 3 matrix:

0 1 0

0 0 1

0 0 0

Verify that A² is the following matrix:

0 0 1

0 0 0

0 0 0

Now multiply A² by A to get 0. Since A² ≠ 0, but A³ = 0, A is "nihilistic" but not "melancholy." 

(c) Let e be an eigenvalue of A corresponding the the eigenvector v. Then

(1) A v = e v ⇒ A² v = A (e v) = e (A v) = e (e v) = e² v.

On the other hand, since A² = A,

(2) A² v = A v = e v.

It follows from (1) and (2) above that

e² v = e v ⇒ e² v - e v = (e² - e) v = 0

Since v ≠ 0, we must have

e² - e = e (e - 1) = 0 ⇒ e = 0 or 1.

(d) This is TRUE.

Let A be a nilpotent (nihilistic) matrix. Then there must be a minimal value of k such that A^k = 0 (that is, A^k v = 0 v for all vectors v). The fact that k is minimal implies that there is a vector w such that A^k-1w ≠ 0, but A^kw = 0. (Otherwise, A^k-1 v = 0 v for all v, so k wouldn't be minimal.) Note that A^k w = A(A^k-1w) and consider the equation

A(A^k-1w) = 0.

Since A^k-1w ≠ 0, this implies that A^k-1w is an eigenvector of A with eigenvalue 0.

(e) This is FALSE. Consider the 2 x 2 identity matrix

1 0

0 1

The identity matrix is idempotent (happy), but only has 1 as an eigenvalue.