Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $A$.

share|improve this question
5  
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

3 Answers 3

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.

share|improve this answer
2  
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 $B = M A M^{-1}$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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