Given a matrix $A$ with $\operatorname{tr} (A) = 0$, prove that there is a B such that $\forall 1\leq i\leq n :(B^{-1}AB)_{i,i}=0$

Question

I've tried using some matrices $B^{-1}$ that switch the rows, but the $B$ at the end placed the elements back in the diagonal (in different order) so I couldn't find a rule.

Answer 1

Hint: For a $2 \times 2$ matrix, consider one with $1$ and $-1$ on the diagonal. (Note that there cannot be any off-diagonal elements here, because the evalues differ). Then picking $B$ to be rotation by $\pi/4$ i.e., the matrix $$ s\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}, $$ where $s = \frac{\sqrt{2}}{2}$, produces a matrix of the desired form, and this clearly generalizes to other eigenvalues besides $1$.

How'd I find this? I looked for a vector $v$ that the original matrix sent to a vector $w$ with $v \cdot w = 0$, since that's what helps produce a 0 on the diagonal. Since the original matrix simply reflects through the $x$-axis, the vector $(1, 1)$ goes to $(1, -1)$, which is perpendicular to it. When I found two such independent vectors (a similar argument applies to $(1, -1)$, for example), I chose them as my columns for $B$ (i.e., I represented the original transform in that basis). (Along the way, I made them unit vectors so that it'd be easier to compute $B^{-1}$.)

For $3 \times 3$, there are three cases to consider: diagonal with three distinct e-values that sum to zero, diagonal with entries $a, a, -2a$, and the same thing, but with a $1$ in the superdiagonal entry above $a$ and $a$.

In each case, you need to do a similar analysis/discovery of vectors.

Answer 2

Lemma
Given any matrix $A$ of order at least $2$, there exists an invertible matrix $E$ such that $(E^{-1}AE)_{11} = 0$.

Proof: Let $A = [a_{ij}]_{n \times n}$, $n \ge 2$. If $a_{11} = 0$, the result holds with $E = I$, the identity matrix of order $n$. If $a_{12} = 0$, then let $E$ be the permutation matrix obtained by interchanging the first and second rows of $I$. Then $E^{-1}AE$ is the result of interchanging the first and second rows as well as first and second columns of $A$, and therefore has $(1,1)$ entry zero.

Suppose $a_{11}$ and $a_{12}$ are non-zero. Define the $n \times n$ matrix \begin{equation*} E = \left[\begin{array}{cc|c} a_{12} & 0 & \mathbf 0_{1 \times (n-2)}\\ -a_{11} & 1 & \mathbf 0_{1 \times (n-2)}\\ \hline \mathbf 0_{(n-2) \times 1} & \mathbf 0_{(n-2) \times 1} & I_{n-2} \end{array}\right]. \end{equation*} The inverse of $E$ is easily verified to be \begin{equation*} E^{-1} = \left[\begin{array}{cc|c} 1/a_{12} & 0 & \mathbf 0_{1 \times (n-2)}\\ -a_{11}/a_{12} & 1 & \mathbf 0_{1 \times (n-2)}\\ \hline \mathbf 0_{(n-2) \times 1} & \mathbf 0_{(n-2) \times 1} & I_{n-2} \end{array}\right]. \end{equation*}

Then the $(1,1)$ entry of $AE$ is $a_{11} a_{12} - a_{12} a_{11} = 0$. Multiplication on the left by $E^{-1}$ results in a matrix with the first row elements multiplied by $1/a_{12}$ (and other rows affected in various ways). Therefore, $E^{-1}AE$ also has $(1,1)$ entry zero. $\qquad \square$

Now, we know that similar matrices have the same trace, so $\operatorname{trace}(E^{-1}AE) = \operatorname{trace}(A)$. Thus we have the following corollary.

Corollary
Every square matrix of order at least $2$ is similar to a matrix with the same trace, and having $(1,1)$ entry zero.

Proposition
Every square matrix $A$ is similar to a matrix with the last diagonal entry equal to $\operatorname{trace}(A)$, and all preceding diagonal entries zero.

Proof: We prove by induction on $k$, $1 \le k \le n - 1$, that $A$ is similar to a matrix with the first $k$ diagonal entries $0$, and having the same trace as $A$ due to similarity. Then the result follows by letting $k = n - 1$.

From the above corollary, the result is true for $k = 1$. That is, there is a matrix similar to $A$ and having the same trace as $A$, with its first diagonal entry equal to zero. Suppose the result to be true for $k$. Thus, without loss of generality, assume that the first $k$ diagonal entries of $A$ are all zero. Let $A$ be partitioned as given below (with matrices $X$ and $Y$ of appropriate sizes). \begin{equation*} A = \left[\begin{array}{c|c} Z_{k \times k} & X\\ \hline Y & B_{(n - k) \times (n - k)} \end{array}\right] \end{equation*} Then $Z$ has a zero diagonal, and $\operatorname{trace}(B) = \operatorname{trace}(A)$.

Now, by the above lemma, we have a matrix $E$ of order $n - k$ such that $E^{-1}BE$ has $(1,1)$ entry zero. Define an $n \times n$ matrix \begin{equation*} F = \left[\begin{array}{c|c} I_k & \mathbf 0'\\ \hline \mathbf 0 & E \end{array}\right] \end{equation*} with the zero matrices $\mathbf 0$ and $\mathbf 0'$ of sizes same as those of $X$ and $Y$ respectively.

Then \begin{align*} F^{-1}AF & = \left[\begin{array}{c|c} I_k & \mathbf 0'\\ \hline \mathbf 0 & E^{-1} \end{array}\right] \left[\begin{array}{c|c} Z& X\\ \hline Y & B \end{array}\right] \left[\begin{array}{c|c} I_k & \mathbf 0'\\ \hline \mathbf 0 & E \end{array}\right] \\ & = \left[\begin{array}{c|c} I_k & \mathbf 0'\\ \hline \mathbf 0 & E^{-1} \end{array}\right] \left[\begin{array}{c|c} Z & XE\\ \hline Y & BE \end{array}\right] \\ & = \left[\begin{array}{c|c} Z & XE\\ \hline E^{-1}Y & E^{-1}BE \end{array}\right] \end{align*}

Thus, $F^{-1}AF$ is a matrix similar to $A$ (and hence with zero trace) with the first $k + 1$ diagonal entries zero (since the first diagonal entry of $E^{-1}BE$ is zero). By induction, the result is true for all $k \le n - 1$. $\qquad \square$

As a special case of the above proposition, if $\operatorname{trace}(A) = 0$, then $A$ is similar to a matrix with the last diagonal entry as well as all preceding diagonal entries zero. Thus we have the main result.

Theorem
Every matrix $A$ with zero trace is similar to a matrix with zero diagonal.