5
$\begingroup$

What is the rational canonical form of $A$? $$A=\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2 & 3 & -1 & 4\\ 1 & 1 & -1 & 3\\ \end{bmatrix}$$

I found that the minimal polynomial $m_A(x)=(x-1)^2$ and the characteristic polynomial $c_A(x)=(x-1)^4$. Therefore the invariant factors can be $$x-1,x-1,(x-1)^2$$ or $$(x-1)^2,(x-1)^2$$ Therefore the rational canonical form may be $$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 2\\ \end{bmatrix}$$ or $$\begin{bmatrix} 0 & -1 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 2\\ \end{bmatrix}$$

How do I quickly figure out which one is the correct one?

$\endgroup$
1
$\begingroup$

The "first" $2\times 2$ principal block is clearly not an $x-1,x-1$ block as it is not the identity. Nor is the "other" $2\times 2$ principal block since it is not the identity. So we have two $(x-1)^2$ blocks.

$\endgroup$
  • $\begingroup$ Why the "first" $2\times 2$ principal block cannot be an $x-1,x-1$ block? $\endgroup$ – Kenneth.K Apr 16 '17 at 18:08
  • $\begingroup$ Surely the only matrix similar to the identity is the identity? $\endgroup$ – ancientmathematician Apr 17 '17 at 9:01
2
$\begingroup$

If the first matrix is the rational form of $A$, we should have $\dim \ker (A - I) = 3$ (because this is true for the rational form and so it should be true for $A$ as well) while if the second matrix is the rational form of $A$, we should have $\dim \ker(A - I) = 2$. Just check which of those two options holds for $A$ by computing the rank of $A - I$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.