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Show that the zero linear transformation has invariant factors (and elementary divisors):$$q_1=x,q_2=x,\cdots,q_n=x$$

Here is my idea so far. If we have the zero linear transformation defined on an n-dimensional vector space then its companion matrix would be an $n\times n$ zero matrix A.

$$A=\begin{bmatrix} 0 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &0 &0 \\ 0 &0 &0 &0 \end{bmatrix}.$$ I now get the characteristic polynomial given by $|xI-A|$. Since $xI-A=\begin{bmatrix} x &0 &0 &0 \\ 0 &x &0 &0 \\ 0 &0 &x &0 \\ 0 &0 &0 &x \end{bmatrix},$$|xI-A|=x^n.$ I'm now stuck on how to show that its invariant factors are $q_1=x,q_2=x,\cdots,q_n=x$. Obviously the $q_i$'s satisfy that $q_1|q_2|\cdots|q_n$ and $q_1q_2\cdots q_n=x^n$ but how could I say that they really are the invariant factors? Couldn't I just take $q_1=x,q_2=x,q_3=x^{n-2}$? Or some other combination that will satisfy $q_1|q_2|\cdots|q_n$ and $q_1q_2\cdots q_n=x^n$? I'm really still confused with the idea of invariant factors and elementary divisors. How can I find them? How do they connect with each other. How do they connect with the charateristic polynomial and with the minimal polynomial? I'm hoping that doing some exercises will help me understand them more. I hope someone could help me with these. Thanks!

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According to Proposition 4.9 (Hungerford, Algebra, pg 363), if A be an n $\times$ n over a field K, then the matrix of polynomials $xI_n - A \in Mat_nK[x]$ is equivalent to a diagonal matrix D with nonzero diagonal entries $f_1,f_2, \dots ,f_n \in K[x]$ such that each $f_i$ is monic and $f_1|f_2| \dots |f_n$. Those polynomials $f_i$ which are not constants are the invariant factors of A.

In fact, our matrix of polynomials is a diagonal matrix D that satisfies those conditions.

Furthermore, by corollary 4.8 (Hungerford, Algebra, pg 362), the invariant factors of a linear transformation $\varphi : E \longrightarrow E$ are the invariant factors of A.

Hence, we can conclude that the invariant factors of our zero transformation are $q_1=x, q_2=x, \dots ,q_n = x$. It follows that the elementary divisors are $p_i=x$, for all $i$.

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A useful fact is that the minimal polynomial $m_A(x)$ of a matrix $A$ is the invariant factor of largest degree, (Proposition 13, Chapter 12 in Dummit and Foote if I recall correctly). Since $m_A(x)=x$ in this case, the largest invariant factor is $x$. Since the characteristic polynomial is the product of all invariant factors, it follows that the invariant factors are $x$, with multiplicity $n$.

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