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Let $A$ be $3 \times 3$ real matrix with minimal polynomial $f(X)=(X-1)(X^2 +1)=X^3-X^2+X-1.$ Then By Rational Canonical Form we know that $A$ is similar to the Companion matrix of $f(X)$ which is $\begin{pmatrix}0 & 0 & 1 \\ 1 & 0 &-1 \\ 0 &1 &1\end{pmatrix}.$ Now since the ideals $(X-1)$ and $(X^2+1)$ are co-maximal in $\mathbb{R}[X]$ can we say that $A$ is similar to a block matrix whose $1 \times 1$ principal block is the jordan block of size $1$ corresponding to the eigenvalue $1$ and the remaining $2 \times 2$ principal block is the companion matrix of the polynomial $X^2 +1,$ which is actually $\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 &-1 \\ 0 &1 &0\end{pmatrix}.$

I need your confirmation. Technically I couldn't find any mistake, but if this happens then this must be a weaker version of the Jordan Canonical Form.

Thank You.

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  • $\begingroup$ I think it is Ok. Everything follows from the Structure of Modules over PID. $\endgroup$ – user185640 Dec 30 '18 at 12:57
  • $\begingroup$ You have reinvented the wheel. Your second matrix is (practically) the real Jordan normal form. $\endgroup$ – loup blanc Dec 30 '18 at 18:58
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I guess the definitions may differ depending on the (text book) source / educational background, but based on my post-grad education I have the following understanding regarding the hierarchy of canonical forms. In short, the answer to your question is yes, the form you are actually referring to is the Rational Canonical Form, and the blocks are then all called hypercompanion matrices. Just note that if you had a factor $(X^2+1)^2$ you would need a 1 on the subdiagonal between the two companion matrices for $(X^2+1)$ for it to be in rational canonical form - see this answer for more...

In terms of your specific example: Considering $A$ as a matrix over the reals the form $$A'=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 &-1 \\ 0 &1 &0\end{pmatrix}$$ (this is the second matrix you listed) is known as the rational canonical form of $A$...it generalizes the Jordan Canonical Form in the sense that a matrix over any field is similar to a matrix in rational canonical form, whereas the Jordan Canonical Form is only guaranteed for matrices over an algebraically closed field. So considering $A$ as a matrix over the reals it does not have a Jordan Canonical Form from my understanding. If a matrix has a Jordan Canonical Form then this JCF matrix would also be the Rational Canonical Form for the relevant matrix.

So if we consider $A$ as a matrix over the reals, it has rational canonical form $A'$, and it does not have a Jordan Canonical Form. If we consider $A$ as a matrix over the complex numbers, then the minimal polynomial of $A$ can be factored as $$f(X)=(X-1)(X - i)(X+i)$$ and $A$ has rational canonical form $$\begin{pmatrix}1 & 0 & 0 \\ 0 & i & 0 \\ 0 &0 &-i\end{pmatrix}$$ which is also its Jordan Canonical Form.

Note that the first matrix you show is not the Rational Canonical Form of $A$. The rational canonical form consists of hypercompanion matrices of the elementary divisors, which are powers of irreducible factors, and clearly $(X-1)(X^2 +1)$ is reducible. There is another canonical form where every matrix is similar to a block diagonal matrix where the blocks consist of companion matrices of the invariant polynomials - which is a "step" towards the Rational Canonical Form if you will - and this is where the matrix $$\begin{pmatrix}0 & 0 & 1 \\ 1 & 0 &-1 \\ 0 &1 &1\end{pmatrix}$$ fits in.

Just also to clarify further in terms of the question: what you call a Jordan Block corresponding to eigenvalue 1 would be called a hypercompanion matrix associated with $(X-1)$ in the context of the Rational Canonical Form...it is the same thing, but when referring to the canonical form involved here the second would be more correct.

If you want to see more about how the rational canonical form is defined, please see this answer.

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  • $\begingroup$ I have asked in my question if two matrices are similar over $\mathbb{R}$ or not. Is it yes or no ? $\endgroup$ – user371231 Jan 16 at 10:15
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    $\begingroup$ @user371231 it is yes...I have modified the answer $\endgroup$ – Christiaan Hattingh Jan 16 at 10:31

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