Let $A=\begin{pmatrix} 2&-2&14\\0&3&-7\\0&0&2\end{pmatrix}$, then its rational canonical form is $R=\begin{pmatrix}2&0&0\\0&0&-6\\0&1&5\end{pmatrix}$. How can I compute a matrix $P$ such that $P^{-1}AP=R$? And in general what is the algorithm?
How to compute the change of basis matrix that conjugate a matrix to its rational canonical form
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0I think that's wrong. The characteristic polynomial of $A$ is $(x-3)(x-2)^2$, so the rational canonical form should either be diagonal, or have one $1\times 1$ block associated to $3$ and one $2\times 2$ block that is the companion matrix of $(x-2)^2$. Instead, you have one block associated to $2$ and one block that is the companion matrix of $(x-2)(x-3)$. – 2012-01-06
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0I think it's correct. The invariant factors are $(x-2),(x-2)(x-3)$. Also if you conjugate by the matrix $\begin{pmatrix}-7&-1&4\\7&1&3\\1&0&0\end{pmatrix}$ you obtain $R$, and so by uniqueness that is the rational canonical form, right? – 2012-01-06
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0What definition of Rational Canonical Form are you using? In the one I know, if the matrix is diagonalizable, the Rational Canonical Form is diagonal; and the blocks are always companion matrices of powers of irreducible factors of the characteristic polynomial. Your answer does *not* satisfy either of these conditions. Uniqueness depends on what kind of companion matrices you allow. If you change the allowable companion matrices, you change the representative of the equivalence class. – 2012-01-06
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0I would expect your book to cover how to find a basis for the Rational Canonical basis as well; assuming your blocks were correct, you find a basis for the nullspace of $2$, $\mathbf{v}_1$ and $\mathbf{v}_2$; and a basis for the nullspace of $3$, $\mathbf{v}_3$. You then take $\mathbf{v}_1$, $\mathbf{v}_2+\mathbf{v}_3$, and $A(\mathbf{v}_2+\mathbf{v}_3)$. – 2012-01-06
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0P.S. It's incorrect to talk about *the* matrix that conjugates. There are infinitely many matrices $P$ that work. – 2012-01-06
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0@ArturoMagidin: I'm probably misunderstanding what you are saying, I did some computation and I found $v_1=\begin{pmatrix} 1\\0\\0\end{pmatrix},v_2=\begin{pmatrix}7\\1\\0\end{pmatrix},v_3=\begin{pmatrix}2\\-1\\0\end{pmatrix}$. So if you ask me to take $v_1,v_2+v_3,A(v_2+v_3)$ then they are $\begin{pmatrix}1\\0\\0\end{pmatrix}$,$\begin{pmatrix}9\\0\\0\end{pmatrix}$, and $A(v_2+v_3)=2(v_2+v_3)$ . What should I do with these 3 vectors? – 2012-01-06
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0Both $v_1$ and $v_2$ should be linearly independent eigenvectors of $2$; your $v_1$ *is*, but your $v_2$ is not. $v_3$ should be an eigenvector of $3$, which your $v_3$ is. But then $A(v_2+v_3) = 2v_2 + 3v_3$, not $2(v_2+v_3)$. The basis will be $v_1$, $v_2+v_3$, $A(v_2+v_3)$; the matrix $P$ will have these vectors as its columns. – 2012-01-06
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0ok, it's right. How did you see that $v_1,v_2+v_3,A(v_2+v_3)$ is the right basis? – 2012-01-06
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0You need a vector that is annihilated by $A-2$ to go with the invariant factor $x-2$; that's just an eigenvector of $2$. Then you need a vector that is annihilated by $(A-2)(A-3)$, not by $(A-2)$, not by $(A-3)$, to go with the invariant factor $(x-2)(x-3)$; the simplest way to do that is to take an eigenvector of $2$ and add it to an eigenvector of $3$, making sure the eigenvector of $2$ is independent from the one you chose for the factor corresponding to $x-2$. Then you take the cyclic basis it generates. I'm sure this is in your textbook. – 2012-01-06
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0and in case I would have had an invariant factor of type $(x-2)^2(x-3)$ what should I have done? – 2012-01-06
1 Answers
As with the Jordan case, you need to find a Rational Canonical basis; your matrix $P$ will have the rational canonical basis as its columns.
Your computation is incorrect, though, at least under the definition I am familiar with. In the definition I am familiar with, each block in the Rational Canonical Form is the companion matrix of a polynomial of the form $\phi^k(t)$, where $\phi(t)$ is an irreducible factor of the characteristic polynomial.
The characteristic polynomial is $(x-2)^2(x-3)$. The minimal polynomial is either $(x-2)(x-3)$ or $(x-2)^2(x-3)$. The Rational Canonical form deals with the irreducible factors separately, so you will have that the Rational Canonical form of $A$ is either $$\begin{align*} \left(\begin{array}{ccc} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{array}\right) &\text{if the minimal polynomial is }(x-2)(x-3);\\ \left(\begin{array}{ccc} 0 & -4 & 0\\ 1 & 4 & 0\\ 0 & 0 & 3 \end{array}\right) & \text{if the minimal polynomial is }(x-2)^2(x-3). \end{align*}$$
It is not hard to check that $A-2I$ has rank $1$, so the nullspace is 2-dimensional; hence the eigenspace corresponding to $2$ is two dimensional, so the geometric multiplicity of $2$ equals the algebraic multiplicity. The matrix is diagonalizable, and the Rational Canonical form of $A$ is the diagonal matrix.
Find a basis of eigenvectors, that gives you the $P$.
In general, you need to find a Rational Canonical Basis, and a matrix whose columns are the elements of the Rational Canonical Basis will work as $P$.
To find them, you need to determine the size of the blocks associated to each irreducible factor of the characteristic polynomial. If you have a block of size $kd$, where $d$ Is the degree of the irreducible factor $\phi(t)$, then you need to find an element $\mathbf{v}$ of the nullspace of $\phi^k(A)$ that is not an element of the nullspace of $\phi^{k-1}(A)$. Then the basis vectors corresponding to that particular block are $\mathbf{v}$, $A\mathbf{v},A^2\mathbf{v},\ldots,A^{kd-1}\mathbf{v}$.