plx!hlep me analyze my question about diagonalization

tellmewhy · **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?

outermeasure · **Posted:** Sun, 11 Mar 2012 10:04:17 UTC

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?

tellmewhy · **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

Shadow · **Posted:** Sun, 11 Mar 2012 10:32:15 UTC

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 x(x^2+bx+c)

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.

tellmewhy · **Joined:** Sun, 11 Mar 2012 09:31:05 UTC **Posts:** 5

Shadow 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 x(x^2+bx+c)

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 ?

Shadow · **Posted:** Sun, 11 Mar 2012 10:39:47 UTC

tellmewhy wrote:

Shadow 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 x(x^2+bx+c)

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.

tellmewhy · **Joined:** Sun, 11 Mar 2012 09:31:05 UTC **Posts:** 5

Shadow wrote:

tellmewhy wrote:

Shadow 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 x(x^2+bx+c)

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

Shadow · **Posted:** Sun, 11 Mar 2012 11:01:19 UTC

tellmewhy wrote:

Shadow wrote:

tellmewhy wrote:

Shadow 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 x(x^2+bx+c)

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.

tellmewhy · **Joined:** Sun, 11 Mar 2012 09:31:05 UTC **Posts:** 5

Shadow wrote:

tellmewhy wrote:

Shadow wrote:

tellmewhy wrote:

Shadow 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 x(x^2+bx+c)

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"

Shadow · **Posted:** Sun, 11 Mar 2012 11:14:38 UTC

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:

$$\left|k\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\right|=\left|\begin{pmatrix} ka & kb & kc \\ kd & ke & kf \\ kg & kh & ki \end{pmatrix}\right|=k^3\left|\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\right|$ , here the matrix is 3x3, so you get that k^3

term.

Voxx · **Posted:** Mon, 12 Mar 2012 15:40:00 UTC

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

Shadow · **Posted:** Mon, 12 Mar 2012 18:52:50 UTC

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.

Voxx · **Posted:** Tue, 13 Mar 2012 00:03:55 UTC

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...

Shadow · **Posted:** Tue, 13 Mar 2012 05:18:09 UTC

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.

Voxx · **Posted:** Wed, 14 Mar 2012 08:43:39 UTC

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

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plx!hlep me analyze my question about diagonalization

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