My Math Forum Matrices which satisfy the equation

 Linear Algebra Linear Algebra Math Forum

April 23rd, 2018, 01:08 PM   #1
Newbie

Joined: Mar 2018
From: Split, Croatia

Posts: 13
Thanks: 0

Matrices which satisfy the equation

Show that there are no real 3x3 matrices which satisfy the equation (picture below), but there are complex 3x3 matrices and real 2x2 matrices which satisfy that.
I know that this equation has no real roots, but I don't know how to apply that to matrices.
Attached Images
 20180423_230116.jpg (60.8 KB, 1 views)

 April 23rd, 2018, 01:12 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,099 Thanks: 1093 do you mean that $\lambda^2 + 1$ is the characteristic equation of the matrix? Thanks from Birgitta
April 23rd, 2018, 01:13 PM   #3
Newbie

Joined: Mar 2018
From: Split, Croatia

Posts: 13
Thanks: 0

Quote:
 Originally Posted by romsek do you mean that $\lambda^2 + 1$ is the characteristic equation of the matrix?
Yes

April 23rd, 2018, 02:51 PM   #4
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,099
Thanks: 1093

Quote:
 Originally Posted by Birgitta Yes
sorry I meant characteristic polynomial

April 23rd, 2018, 02:53 PM   #5
Newbie

Joined: Mar 2018
From: Split, Croatia

Posts: 13
Thanks: 0

Quote:
 Originally Posted by romsek sorry I meant characteristic polynomial
I assumed that. However, I've already solved the problem. Thanks

 April 23rd, 2018, 04:03 PM #6 Senior Member   Joined: Sep 2016 From: USA Posts: 443 Thanks: 254 Math Focus: Dynamical systems, analytic function theory, numerics In a nutshell, the main idea is to prove the fact that if $M$ is a matrix defined over a field, $F$, and it satisfies $\lambda^2 + 1 = 0$, then every vector in $F^3$ is a (generalized) eigenvector for either $i$ or $-i$. Note this is NOT saying that $\lambda^2 + 1$ is the characteristic polynomial for $M$ (though it is true that it must be a multiple of $\lambda^2 + 1$. EDIT: After thinking about it, I realized even this claim is not true. $\lambda^2 + 1$ need only have one linear factor in common with the characteristic polynomial. It's best to avoid thinking about the characteristic polynomial at all since it has nothing to do with the question. With this fact in hand, it should be trivial to find examples for $\mathbb{R}^2$ which work as well as examples in $\mathbb{C}^3$. I'll outline the proof below which should give you a hint. 1. Suppose $M$ is a 3x3 matrix and $M^2 + I = 0$. Then if $F = \mathbb{R}$, then $i,-i$ are eigenvalues for $M$ and moreover, they correspond to a pair of linearly independent eigenvectors. Denote these (unit vectors) by $\xi_1,\xi_2$ and note that they are conjugates (prove this and note that already this may fail if $F = \mathbb{C}$). 2. Prove that $\xi_1,\xi_2$ span a real invariant subspace with respect to $M$. Hint: prove that $\xi_1 + \xi_2$ and $\xi_1 - \xi_2$ span a 2-dimensional subspace of $\mathbb{R}^3$. Let $V = \text{span} \{\xi_1,\xi_2 \}$ denote this subspace. 3. Choose any vector $u \in \mathbb{R}^3 \setminus V$ and consider the action of $M$ on $u$ expanded in the basis $\{ \xi_1,\xi_2, u\}$ as $Mu = a_1 \xi_1 + a_2 \xi_2 + a_3 \xi_3.$ Now, $a_3 \in F$ and $a_3 \neq 0$ (why not?). Note that $a_1,a_2$ may be in $\mathbb{C}$ since I am using $\xi_1,\xi_2$ in my basis. However, this isn't a problem if you have proved (2). 4. Note that $M^2u = -u$ by assumption but also from (3), $M^2u = a_1 i \xi_1 - a_2 i \xi_2 + a_3 Mu.$ From this you can argue three things: (i) $a_1 = a_2 = 0$. (ii) $u$ is a nonzero eigenvector for $M$ with eigenvalue $a_3$. (iii) $a_3$ satisfies $a_3^2 = -1$ 5. From (4) you can conclude that such an $M$ exists only if $a_3 = \pm i$ but that requires $F = \mathbb{C}$ since you proved in (3) that $a_3 \in F$. Thanks from Birgitta Last edited by skipjack; May 2nd, 2018 at 02:07 PM.
 May 2nd, 2018, 01:14 PM #7 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,449 Thanks: 106 A 3x3 matrix has to have a $\displaystyle \lambda^{3}$ term. For a 2x2 matrix, the characteristic equation is $\displaystyle \lambda^{2} - \lambda (a_{11}+a_{22})+D = 0$ and you can easily pick real components to satisfy OP.
May 2nd, 2018, 06:23 PM   #8
Senior Member

Joined: Sep 2016
From: USA

Posts: 443
Thanks: 254

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by zylo A 3x3 matrix has to have a $\displaystyle \lambda^{3}$ term. For a 2x2 matrix, the characteristic equation is $\displaystyle \lambda^{2} - \lambda (a_{11}+a_{22})+D = 0$ and you can easily pick real components to satisfy OP.
1. This has nothing to do with the characteristic equation.

2. If what you are saying was true, the very thing the OP is supposed to prove would be false.

May 3rd, 2018, 05:33 AM   #9
Senior Member

Joined: Mar 2015
From: New Jersey

Posts: 1,449
Thanks: 106

Quote:
 Originally Posted by zylo A 3x3 matrix has to have a $\displaystyle \lambda^{3}$ term. For a 2x2 matrix, the characteristic equation is $\displaystyle \lambda^{2} - \lambda (a_{11}+a_{22})+D = 0$ and you can easily pick real components to satisfy OP.
Thought it was clear: The characteristic polynomial of a 3x3 matrix has to be of degree 3. OP is of degree 2.

May 3rd, 2018, 06:04 AM   #10
Senior Member

Joined: Oct 2009

Posts: 494
Thanks: 164

Quote:
 Originally Posted by zylo Thought it was clear: The characteristic polynomial of a 3x3 matrix has to be of degree 3. OP is of degree 2.
There is no reason why the $\lambda^2+1$ in the OP is the characteristic polynomial. In fact, it isn't for a $3\times 3$-matrix.

 Tags complex, equation, matrices, polynomial, satisfy

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post jbaldwin81 Applied Math 1 April 9th, 2017 02:47 PM toczek Calculus 3 May 6th, 2016 12:54 PM matisolla Algebra 1 May 29th, 2015 08:17 AM Jascha Linear Algebra 3 February 21st, 2012 03:00 PM MathematicallyObtuse Algebra 10 December 28th, 2010 06:18 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top