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In the paper Computing Rational Forms of Integer Matrices by MARK GIESBRECHT† AND ARNE STORJOHANN

It says that, When A ∈ Qn×n has all integer entries, the Frobenius form F of A has all integer entries as well.

But how about if A ∈ Znxn is diagonalizable with eigenvalues not in Z ?

Then its diagonalized form is not equal to its rcf ?

user593805
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1 Answers1

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The rational canonical form of a matrix is almost never a diagonal matrix. A typical rational canonical form is a companion matrix: $$\begin{pmatrix}0&0&0&\cdots&0&-a_0\\1&0&0&\cdots&0&-a_1\\ 0&1&0&\cdots&0&-a_2\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&-a_{n-2}\\ 0&0&0&\cdots&1&-a_{n-1}\end{pmatrix}$$ Even if the roots are all distinct and the matrix is diagonalizable, that's the form we'll choose to call the rational canonical form. When the RCF isn't of this form, it's because the minimal polynomial has lower degree than the characteristic polynomial - and then we build the RCF using companion matrix blocks. These companion matrices have integer entries when the associated polynomials have integer coefficients, which is certainly true for the minimal and characteristic polynomials of integer matrices.

In fact, the only way for the RCF to be diagonal is if the matrix is a multiple of the identity. In that case, there's no way for it to be similar to anything else.

jmerry
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  • The comment about RCF being diagonal when the matrix is a multiple of the identity is not true. Take the $2\times 2$ identity matrix: its characteristic polynomial is $(x-1)^2 = x^2-2x+1$, so its RCF is $\begin{pmatrix}0 & -1 \ 1 & 2 \end{pmatrix}$. In general, the identity matrix can be similar to matrices that are not diagonal with $1$'s along the diagonal since this essentially reflects a change of basis. – Alex Ortiz Feb 17 '19 at 04:28
  • No, $\begin{pmatrix}1&0\0&1\end{pmatrix}$ is not similar to $\begin{pmatrix}0&-1\1&2\end{pmatrix}$. They may have the same characteristic polynomial, but they don't have the same minimal polynomial. – jmerry Feb 17 '19 at 04:31
  • Ah, yes you are correct. I suppose I was thinking along the lines of when we have a diagonal matrix with distinct entries along the diagonal. In that case the reason the RCF is different from the original matrix is because there is only one invariant factor, namely the minimal polynomial which equals the characteristic polynomial. Not so for the identity, which has minimal polynomial different from characteristic polynomial. In particular, it has $x-1,x-1$ (in the case $n=2$) as its invariant factors, which give two $1\times 1$ companion matrix blocks. Thanks for the clarification! – Alex Ortiz Feb 17 '19 at 04:42
  • Thank u so much. – user593805 Feb 17 '19 at 05:00
  • I have one last question. If the minimal poly is equal to its char poly. Thus this mean its rcf is one big companion matrix ? – user593805 Feb 17 '19 at 05:02
  • Yes, the RCF is one big companion matrix if the minimal and characteristic polynomials are equal. – jmerry Feb 17 '19 at 05:16