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Let $A$ be a complex $n×n$ matrix with $tr(A) = 0$. Show that there exists an invertible $n×n$ matrix $B$ such that all diagonal entries of $BAB^{-1}$ are zeros.

$\bf Edit:$ Is the fact that $\forall A\in M_{m\times n}(D)$, $D$ is $PID$, then $A$ is equivalent to a matrix which has the diagonal form $diag\{d_1,...,d_2,0,..,0\}$ where $d_i|d_j, i\leq j$and the other entries are zeros, applicable in this question?

Ronald
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    What have you done so far (have you tried induction ?)? – Bertrand R Aug 21 '13 at 18:04
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    Hint: The trace is invariant under similarity transforms. – AlexR Aug 21 '13 at 18:05
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    What can I do further $tr(A)=0=tr(BAB^{-1})$? I do not know how to use induction here either – Ronald Aug 21 '13 at 18:12
  • I edited the question – Ronald Aug 21 '13 at 19:07
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    No don't use that. Why not trying something that can introduce a zero in the first entry by changing the first basis vector? Then, after you can do that, can you see room to apply induction to some submatrix or something? For that matter, have you tried the base case? Ok, the $2 \times 2$ case, rather. – Evan Aug 21 '13 at 19:55
  • @Evan I will try it. But do u think the theorem above is not useful even to prove that $A$ and $A^T$ are similar? do u know good reference with these examples? – Ronald Aug 21 '13 at 20:04

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Note that all the diagonal entries of $A$ are already $0$ if they are all equal, otherwise there are at least 2 diagonal elements of $A$ that are non-zero and different from each other. Assume without loss of generality that it's the first two diagonal elements that are non-zero and different from each other. Then all you need to do is show that you can find an invertible $2 \times 2$ matrix $X$ that produces $X A_2 X^{-1}$ such that the first diagonal element of the product is equal to zero, where $A_2$ is the upper-left $2 \times 2$ block of $A$. Then you define $B$ to have $X$ as its upper-left $2 \times 2$ block, and fill in the rest of the diagonal of $B$ with ones (and zeros everywhere else). When you compute $BAB^{-1}$, you will get one more diagonal entry equal to $0$ than you had in $A$. Then by induction you can keep choosing two diagonal elements that are non-zero and not equal, and repeat this process, to build up a product of matrices $B = B_n B_{n-1} \cdots B_1$ that will give all zeros on the diagonal of $BAB^{-1}$.

For the $2 \times 2$ case finding a suitable $X$ that gives first diagonal element of $XA_2X^{-1}$ equal to 0, just write down the symbolic expression for the first diagonal entry in terms of the entries of $A_2$ and $X$ (and you can skip dividing by the determinant of $X$ since you want to set the expression to $0$) and just find suitable entries in $X$ that make $X$ invertible and that set the diagonal entry expression to 0. A hint: First consider the case that $A_2$ has at least one non-zero off-diagonal element, and you should be able to find a suitable $X$ which has one $0$ in its first row. Then assume $A_2$ is diagonal and recall that the diagonal elements are not equal. Then you should be able to find a suitable $X$ that has its first row set to $(1,1)$.

user2566092
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