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I have a homework problem which asks me to find the invariant and elementary factors of $\mathbb{R}^3$ as a module over $\mathbb{R}[x]$, where the action is given by a 3 by 3 real-valued matrix $T$. That is, $f(x) \cdot v := f(T)v$.

The matrix $T$ has characteristic equation which factors into a linear and a quadratic factor over $\mathbb{R}$.

We have only proved that this decomposition exists in the class, and the proof doesn't seem applicable. I don't know the Smith Normal or Jordan forms, etc. The professor says we can solve the problem without them, but I can't see how.

From reading further my text, I think I can see that the invariant factor should just be the characteristic equation, and the the elementary factors are the factors of the characteristic equation. However, I don't see intuitively how the characteristic equation is related to the factor decomposition, and I certainly don't know how to derive this result without the machinery that the text develops.

Any guidance on how to approach this would be appreciated. Is there a more elementary method to solve the problem?

Edit: I've gotten a bit further. The matrix in question here is $T = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 5 \end{bmatrix}, $ so the space $\mathbb{R}^3$ can be seem as a direct sum of two $T$-invariant subspaces: $\mathbb{R}^3 = \operatorname{span} \{ e_1, e_2 \} \oplus \operatorname{span}\{e_3\}$. There's a fairly obvious module isomorphism from $\operatorname{span} \{ e_1, e_2 \}$ to $\mathbb{R}[x]/(x^2+1)$, and also an isomorphism from $\operatorname{span}\{e_3\}$ to $\mathbb{R}[x]/(x-5)$.

Since $x^2+1$ and $x-5$ are prime in $\mathbb{R}[x]$, the factors above are the elementary factors; and then the invariant factor can only be $\mathbb{R}[x]/((x^2+1)(x-5))$, which is the characteristic equation.

I see why this works here, but would it work in general? I.e., for general $3 \times 3$ matrices, for matrices of arbitrary dimension?

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    OK, now I see what you meant by the notation :) Thanks for clarifying!2012-11-08

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the problem is easy here because as you observed the characteristic polynomial factors into distinct irreducible factors. The same holds anytime this is true. But for more special 3 by 3 matrices, whose characteristic polynomial is say x^3, there are several cases depending on the minimal polynomial being x, or x^2, or x^3.