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I am trying to solve some sample test questions and am looking for shortcut to a problem.

Question: True or False "Every invertible matrix $A \in \mathbb{Q}^{n \times n}$ is similar to a diagonal matrix over $\mathbb{C}$"

I think the statement is true for the following reason:

Theorem: Let $D$ be a principal ideal domain and $A \in D^{n\times n}$ then $A$ is equivalent to a matrix which has the diagonal form $diag\{d_a,d_2, \ldots, d_r, 0, \ldots, 0\}$ where the $d_i \neq 0$ and $d_i | d_j$ if $i \leq j$.

The proof for this fact has a lot of steps and requires one to induct using a suitably defined notion of "length" on a non-zero element in $D$. I was wondering if there is a simple trick to this problem by using invertibility of $A$.

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Equivalent or similar? – Mark Aug 8 '11 at 8:57
Sorry for the confusion the question is stated correctly. I should not have used the theorem as a reason for why i think the statement is true. – user7980 Aug 8 '11 at 9:27
up vote 3 down vote accepted

False. Counterexample: consider $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\in\mathbb{Q}^{2\times2}$. If $A$ is similar to some diagonal matrix $D$, by inspecting its trace and determinant, one can show that $D=I_2$ and in turn $A=I_2$, which is a contradiction.

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Another argument: Since $A$ is lower triangular, its eigenvalues are its diagonal entries, i.e. $1,1$, so if it is similar to a diagonal matrix then this diagonal matrix is necessarily $I_2$. But clearly $I_2$ is only similar to itself. – Mark Aug 8 '11 at 10:29

You must distinguish between the similarity relation and the equivalence!

Two matrices $A$, $B$ are said equivalent if there exist invertible matrices $M$ and $N$ such that $B = M A N$. On a field a matrix $A$ is equivalent to the matrix $\mathrm{diag} \ \{1, \dots, 1, 0, \dots, 0 \}$, where the numbers of $1$s is the rank of $A$. On a PID it is true the weaker statement that you cited (Smith canonical form).

Two square matrices $A$, $B$ are said similar if there exists an invertible matrix $M$ such that $B = M A M^{-1}$.

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It is false. A matrix is diagonalizable if and only if the geometric multiplicity of every eigenvalue equals the algebraic multiplicity of that eigenvalue. Equivalently, a matrix is diagonalizable if and only if the sum of the geometric multiplicity equals the "size" of the matrix. A matrix is invertible if and only if all the eigenvalues are non-zero. Hence, in general invertibility of a matrix has got nothing to do with it being similar to a diagonal matrix.

For eg. $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is invertible with $A^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ but the matrix is not diagonalizable since the only eigenvalue $1$ gives rise to only one eigenvector viz $\begin{pmatrix}1 \\ 0 \end{pmatrix}$

Further, $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ is not invertible but the eigenvalues are $0$ and $1$ giving rise to the eigenvectors $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ Note that $A$ in itself is a diagonal matrix.

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