# S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
 It is currently Fri, 24 May 2013 03:33:47 UTC

 All times are UTC [ DST ]

 Page 1 of 2 [ 16 posts ] Go to page 1, 2  Next
 Print view Previous topic | Next topic
Author Message
 Post subject: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 09:38:12 UTC

Joined: Sun, 11 Mar 2012 09:31:05 UTC
Posts: 5
Q1
given:M is a real 3×3 matrix , M and −M are similar
show
(a) 0 is an eigenvalue of M
(b) If some eigenvalue of M is non-zero, then M is diagonalizable

Q2
given:
Suppose that a 3×3 matrix A with real entries satisfies
the equation M^3 + M^2 − M + 2I = 0.
(a) Find the eigenvalues of M.
(b) Is A diagonalizable? Why?

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 10:04:17 UTC
 Moderator

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6008
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
tellmewhy wrote:
Q1
given:M is a real 3×3 matrix , M and −M are similar
show
(a) 0 is an eigenvalue of M
(b) If some eigenvalue of M is non-zero, then M is diagonalizable

Q2
given:
Suppose that a 3×3 matrix A with real entries satisfies
the equation M^3 + M^2 − M + 2I = 0.
(a) Find the eigenvalues of M.
(b) Is A diagonalizable? Why?

Show your work! What have you tried? Where are you stuck?

_________________

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 10:23:39 UTC

Joined: Sun, 11 Mar 2012 09:31:05 UTC
Posts: 5
outermeasure wrote:
tellmewhy wrote:
Q1
given:M is a real 3×3 matrix , M and −M are similar
show
(a) 0 is an eigenvalue of M
(b) If some eigenvalue of M is non-zero, then M is diagonalizable
(a)as M is similar to -M>>>M=PM(P^-1)>>>M is diagonalizable>>M is the form of {a 0 0;0 b 0; 0 0 c}
a, b and c is the eigenvalues of M
detM=0>>>abc=0>>>a=0,b=0 or c=0 is it right?
actually, i hv not idea about b part

Q2
given:
Suppose that a 3×3 matrix A with real entries satisfies
the equation M^3 + M^2 − M + 2I = 0.
(a) Find the eigenvalues of M.
(b) Is A diagonalizable? Why?

(a)M^3 + M^2 - M +2I = 0>>>(M+2I)(M-Z iI)(M-K iI)=0 Z,K are complex numbers
i think -2, Z and K would be the eigenvalues, but finally i found it is not probably true
as same as Q1, i have not idea about b part yet

Show your work! What have you tried? Where are you stuck?

plx help me to analyze this last 2 question for my assignment

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 10:32:15 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
No, the order of eigenvalues is not important. The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M. Then the characteristic polynomial of M is for some . If some eigenvalue is nonzero then one of b or c must be non-zero.

For the second one I almost agree with your factorization of the characteristic polynomial, except as-written they're purely imaginary. Now use the fact that it has distinct roots.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 10:39:02 UTC

Joined: Sun, 11 Mar 2012 09:31:05 UTC
Posts: 5
No, the order of eigenvalues is not important. The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M. Then the characteristic polynomial of M is for some . If some eigenvalue is nonzero then one of b or c must be non-zero.

For the second one I almost agree with your factorization of the characteristic polynomial, except as-written they're purely imaginary. Now use the fact that it has distinct roots.

i dont know about The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M
does it have any explanation ?

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 10:39:47 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
tellmewhy wrote:
No, the order of eigenvalues is not important. The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M. Then the characteristic polynomial of M is for some . If some eigenvalue is nonzero then one of b or c must be non-zero.

For the second one I almost agree with your factorization of the characteristic polynomial, except as-written they're purely imaginary. Now use the fact that it has distinct roots.

i dont know about The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M
does it have any explanation ?

Multiply the entries of an nxn matrix by a constant and you multiply the determinant by that constant to the n.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 10:44:07 UTC

Joined: Sun, 11 Mar 2012 09:31:05 UTC
Posts: 5
tellmewhy wrote:
No, the order of eigenvalues is not important. The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M. Then the characteristic polynomial of M is for some . If some eigenvalue is nonzero then one of b or c must be non-zero.

For the second one I almost agree with your factorization of the characteristic polynomial, except as-written they're purely imaginary. Now use the fact that it has distinct roots.

i dont know about The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M
does it have any explanation ?

Multiply the entries of an nxn matrix by a constant and you multiply the determinant by that constant to the n.

i am not a smart guy, i dont understand ur words, can u give me an example plx orz

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 11:01:19 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
tellmewhy wrote:
tellmewhy wrote:
No, the order of eigenvalues is not important. The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M. Then the characteristic polynomial of M is for some . If some eigenvalue is nonzero then one of b or c must be non-zero.

For the second one I almost agree with your factorization of the characteristic polynomial, except as-written they're purely imaginary. Now use the fact that it has distinct roots.

i dont know about The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M
does it have any explanation ?

Multiply the entries of an nxn matrix by a constant and you multiply the determinant by that constant to the n.

i am not a smart guy, i dont understand ur words, can u give me an example plx orz

I too am not smart, I cannot understand what you're saying. Please use common American English.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 11:10:13 UTC

Joined: Sun, 11 Mar 2012 09:31:05 UTC
Posts: 5
tellmewhy wrote:
tellmewhy wrote:
No, the order of eigenvalues is not important. The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M. Then the characteristic polynomial of M is for some . If some eigenvalue is nonzero then one of b or c must be non-zero.

For the second one I almost agree with your factorization of the characteristic polynomial, except as-written they're purely imaginary. Now use the fact that it has distinct roots.

i dont know about The fact that M and -M are similar DOES tell you 0 is an eigenvalue of M
does it have any explanation ?

Multiply the entries of an nxn matrix by a constant and you multiply the determinant by that constant to the n.

i am not a smart guy, i dont understand ur words, can u give me an example plx orz

I too am not smart, I cannot understand what you're saying. Please use common American English.

sorry, i am not a american even my english is not good
and what i say is that can you give me an example for "Multiply the entries of an nxn matrix by a constant and you multiply the determinant by that constant to the n"

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Sun, 11 Mar 2012 11:14:38 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
tellmewhy wrote:
sorry, i am not a american even my english is not good
and what i say is that can you give me an example for "Multiply the entries of an nxn matrix by a constant and you multiply the determinant by that constant to the n"

It means, if you have a constant then:

, here the matrix is 3x3, so you get that term.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Mon, 12 Mar 2012 15:40:00 UTC
 Member

Joined: Tue, 20 Oct 2009 18:18:45 UTC
Posts: 33
Location: Lisbon, Portugal
if M and -M are similar:
PM = (-M)P <-> M = [P^(-1)](-M)P -> det(M) = det ([P^(-1)](-M)P) <-> det(M) = det(P^(-1))*det(-M)*det(P)

determinants are real numbers so you can use the commutative property:

det(M) = det (-M)*det(P^(-1))*det(P)
det(M) = det(-M)*1
det(M) = (-1)^3*det(M)
det(M) = -det(M)
det(M) + det(M) = 0
2det(M) = 0
det(M) = 0 -> 0 is eigenvalue of M

0 is eigenvalue if (M - 0*I)X=0 <-> MX=0
we want that this system has multiple solutions (the eigenvetors) so M can not be invertible...
hope this help a little bit

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Mon, 12 Mar 2012 18:52:50 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
Voxx wrote:
if M and -M are similar:
PM = (-M)P <-> M = [P^(-1)](-M)P -> det(M) = det ([P^(-1)](-M)P) <-> det(M) = det(P^(-1))*det(-M)*det(P)

determinants are real numbers so you can use the commutative property:

det(M) = det (-M)*det(P^(-1))*det(P)
det(M) = det(-M)*1
det(M) = (-1)^3*det(M)
det(M) = -det(M)
det(M) + det(M) = 0
2det(M) = 0
det(M) = 0 -> 0 is eigenvalue of M

0 is eigenvalue if (M - 0*I)X=0 <-> MX=0
we want that this system has multiple solutions (the eigenvetors) so M can not be invertible...
hope this help a little bit

How do you know that determinant 0 implies a zero eigenvalue? The trick here is not about invertibility, it's about the characteristic polynomial.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Tue, 13 Mar 2012 00:03:55 UTC
 Member

Joined: Tue, 20 Oct 2009 18:18:45 UTC
Posts: 33
Location: Lisbon, Portugal
if det(M)=0 <-> the system MX=0 has multiple solutions

if 0 is a eigenvalue of M <-> (M -0*I)X = 0 is undetermined (multiple solutions)
But (M - 0*I)X=0 <-> MX = 0
so det(M) = 0 and 0 is the eigenvalue

another aproach: if A = P^(-1)*D*P (A admits a diagonalization)
det(A) = det(P^(-1)*D*P) = det (D)
so det (A) = product of eigenvalues (the components of the diagonal matrix)
i think this result is correct...

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Tue, 13 Mar 2012 05:18:09 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
Voxx wrote:
if det(M)=0 <-> the system MX=0 has multiple solutions

if 0 is a eigenvalue of M <-> (M -0*I)X = 0 is undetermined (multiple solutions)
But (M - 0*I)X=0 <-> MX = 0
so det(M) = 0 and 0 is the eigenvalue

another aproach: if A = P^(-1)*D*P (A admits a diagonalization)
det(A) = det(P^(-1)*D*P) = det (D)
so det (A) = product of eigenvalues (the components of the diagonal matrix)
i think this result is correct...

It is, but not for the reasons you're saying. I mean, you already had to assume A is diagonalizable to prove the result.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination

Top

 Post subject: Re: plx!hlep me analyze my question about diagonalizationPosted: Wed, 14 Mar 2012 08:43:39 UTC
 Member

Joined: Tue, 20 Oct 2009 18:18:45 UTC
Posts: 33
Location: Lisbon, Portugal
i'm quite sure that if A is not invertible, A admits 0 as a eigenvalue and the eigenvectors are the kernel of A. This is independent of A being diagonalizable

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 2 [ 16 posts ] Go to page 1, 2  Next

 All times are UTC [ DST ]

#### Who is online

Users browsing this forum: No registered users

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forum

Search for:
 Jump to:  Select a forum ------------------ High School and College Mathematics    Algebra    Geometry and Trigonometry    Calculus    Matrix Algebra    Differential Equations    Probability and Statistics    Proposed Problems Applications    Physics, Chemistry, Engineering, etc.    Computer Science    Math for Business and Economics Advanced Mathematics    Foundations    Algebra and Number Theory    Analysis and Topology    Applied Mathematics    Other Topics in Advanced Mathematics Other Topics    Administrator Announcements    Comments and Suggestions for S.O.S. Math    Posting Math Formulas with LaTeX    Miscellaneous