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Finding a Matrix that is Similar to A=[1 1;0 1] 
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#1
Nov912, 06:32 PM

P: 84

Is there one?
I know A=[1 1;0 1] and A^{1}=[1 1;0 1] So I know that A and A^{1} have the same eigenvalues, I know that this is not sufficient to say that A and A^{1} are similar (or maybe) but the dimension of the Eigenspace with eigen value 1 is 1. In other words the geometric multiplicity does not equal the algebraic multiplicity. So does this ultimately mean that A and A^{1} are not similar? 


#2
Nov912, 07:06 PM

P: 606

First, let us see if what I understand is what you meant: $$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\;\;\;,\;\;\;A^{1}=\begin{pmatrix}1&\!\!1\\0&1\end{pmatrix}$$ When we talking of square matrices [itex]\,n\times n\;\;,\;n\leq 3\,[/itex], similarity is determined by the characteristic and the minimal polynomials. Since $$p_A(x)=p_{A^{1}}(x)=(x1)^2=m_A(x)=m_{A^{1}}(x)$$ both matrices are similar. I'll let you to find out what is the general form of the matrix [itex]\,P\,[/itex] that fulfills $$P^{1}AP=A^{1}$$ DonAntonio 


#3
Nov912, 07:11 PM

P: 84

thank you very much DonAntonio that was quite helpful!



#4
Nov1012, 12:59 AM

P: 84

Finding a Matrix that is Similar to A=[1 1;0 1]
you know what, I just read this again, sorry for jumping the gun
I'm not assuming that A and A^{1} are similar, and when I took a good look at this problem again there is only 1 eigenvector for A, so there aren't 2 linearly independent eigenvectors so far i understand that if two matrices are similar then their characteristic polynomials are the same and as a consequence their eigenvalues. in this case specifically we can say that (x1)^{2} is the minimum polynomial. Is that enough to say that the two matricies are similar? i guess the main concern of my question comes from the eigenvectors that make up the similarity matrix. in this problem i only have 1 eigenvector, so how can I find a P s.t. P is invertible? 


#5
Nov1012, 01:11 PM

P: 606

In this very particular case it is enough that both the characteristic and the minimal polynomials of two matrices are equal, just as I mentioned in my first post, since then they both have the very same Jordan Canonical Form (in fact, the matrix A is already in JCF) , since we're talking of square matrices of order less than 4. If these were matrices of order 4 or more then the above would not suffice. And it is unimportant about the eigenvalues, though if there were two difrerent eigenvalues then the matrix would be diagonal, which in this case is impossible. About P: you don't need P to be constructed out of eigevectors of the matrix to show some matrix is similar to another one. DonAntonio 


#6
Nov1012, 07:57 PM

P: 84




#7
Nov1012, 09:56 PM

P: 606

Well, the ultimative test: they both must have the very same and exact JCF, which means the very same eigenvalues with the same algebraic and geometric multiplicities each one...and THEN one must also check the corresponding Jordan blocks are a match. DonAntonio 


#8
Nov1412, 04:57 PM

P: 84

Hmm I just thought of a question. Suppose A is nxn with eigenvalue 1 with algebraic multiplicity n, then A^{1} would have eigenvalue 1 with the same algebraic multiplicity.
Can we always say that A and A^{1} are similar? We can't make any assumptions on the minimal polynomial, but if they have the same minimal polynomial then A and A^{1} would be similar to the same jordan matrix. So where can one go from here? 


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