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Newsgroups: sci.math
From:
G Patel <gaya.pa... @gmail.com>
Date: Wed, 4 Nov 2009 18:56:01 -0800 (PST)
Local: Thurs, Nov 5 2009 2:56 am
Subject: eigenvalues of A
true or false.... eigenvalues of A are same as eigenvalues of kA (because if v is eigenvector for A, then kv is eigenvector for kA)
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Newsgroups: sci.math
From:
Arturo Magidin <magi... @member.ams.org>
Date: Wed, 4 Nov 2009 19:06:40 -0800 (PST)
Local: Thurs, Nov 5 2009 3:06 am
Subject: Re: eigenvalues of A
On Nov 4, 8:56 pm, G Patel <gaya.pa... @gmail.com> wrote:
> true or false.... eigenvalues of A are same as eigenvalues of kA > (because if v is eigenvector for A, then kv is eigenvector for kA)
What is k, a scalar? If so, then:
(i) What are the eigenvalues of a diagonal matrix? (ii) If A is a diagonal matrix, what is kA? (iii) What are the eigenvalues of kA, then?
If k is not a scalar, then what is it?
-- Arturo Magidin
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Newsgroups: sci.math
From:
Gerry Myerson <ge... @maths.mq.edi.ai.i2u4email>
Date: Thu, 05 Nov 2009 14:09:20 +1100
Local: Thurs, Nov 5 2009 3:09 am
Subject: Re: eigenvalues of A
In article <5da73f76-99b2-455e-8788-0034cc6d9... @t2g2000yqn.googlegroups.com>, G Patel <gaya.pa... @gmail.com> wrote:
> true or false.... eigenvalues of A are same as eigenvalues of kA
true if k = 1 and in some other fairly trivial cases, false most of the time. If I is the identity matrix, are the eigenvalues of I the same as the eigenvalues of 2I?
> (because if v is eigenvector for A, then kv is eigenvector for kA)
I'm not familiar with this use of the word, "because." -- Gerry Myerson (ge... @maths.mq.edi.ai) (i -> u for email)
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Newsgroups: sci.math
From:
G Patel <gaya.pa... @gmail.com>
Date: Wed, 4 Nov 2009 19:48:07 -0800 (PST)
Local: Thurs, Nov 5 2009 3:48 am
Subject: Re: eigenvalues of A
On Nov 4, 9:56 pm, G Patel <gaya.pa... @gmail.com> wrote:
> true or false.... eigenvalues of A are same as eigenvalues of kA > (because if v is eigenvector for A, then kv is eigenvector for kA)
if c is eigenvalue of A, then Ax = cx for nonzero x, multiply both sides by nonzero scalar k: (kA)x = c(kx) , thus c is also eigenvalue of kA?
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Newsgroups: sci.math
From:
Arturo Magidin <magi... @member.ams.org>
Date: Wed, 4 Nov 2009 20:25:25 -0800 (PST)
Local: Thurs, Nov 5 2009 4:25 am
Subject: Re: eigenvalues of A
On Nov 4, 9:48 pm, G Patel <gaya.pa... @gmail.com> wrote:
> On Nov 4, 9:56 pm, G Patel <gaya.pa
... @gmail.com> wrote:
> > true or false.... eigenvalues of A are same as eigenvalues of kA > > (because if v is eigenvector for A, then kv is eigenvector for kA)
> if c is eigenvalue of A, then Ax = cx for nonzero x, multiply both > sides by nonzero scalar k: (kA)x = c(kx) , thus c is also eigenvalue > of kA?
An eigenvalue of A is a scalar c such that there exists a nonzero vector x for which Ax=cx. Note that the vector on the left side of "Ax=cx" is the *same* as the vector on the right side of "Ax=cx".
Question: is the vector on the left side of your "(kA)x = c(kx)" the same as the vector on your right hand side?
And to put it even more plainly, *again*:
What are the eigenvalues of, say, the identity matrix?
What are the eigenvalues of k times the identity?
PS: You are utterly confused; go talk to your professor because you are just not getting this.
-- Arturo Magidin
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