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Let $A$ be a $10 \times 10$ matrix with complex entries such that all its eigenvalues are non-negative real numbers, and at least one eigenvalue is positive.

Prove there exists a matrix $B$ such that $AB-BA=B$

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closed as off-topic by Parcly Taxel, Batominovski, B. Goddard, Mark Bennet, Scientifica Aug 21 '17 at 22:05

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    $\begingroup$ What did you try in order to solve the problem? $\endgroup$ – Itay4 Aug 21 '17 at 13:15
  • $\begingroup$ I am taking it's characteristic polynomial {det(A-XI).det(B-XI)-det(B-XI).det(A-XI)}.Then det[ {AB-(A+B)XI+X^2I^2}-det{AB-(A+B)XI+X^2I^2}].After this I am unable to achieve result. Where I is identity matrix. $\endgroup$ – Himanshu Bajpai Aug 21 '17 at 13:23
  • $\begingroup$ You need to assume more about the eigenvalues to find a non-zero matrix solution $B$. See my answer. OTOH, there is no need to assume positivity of any eigenvalues. $\endgroup$ – Jyrki Lahtonen Aug 21 '17 at 13:47
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The following can be said.

Observation. If $x$ is an eigenvector of $A$ belonging to an eigenvalue $\lambda$, then either $Bx=0$ or $Bx$ is an eigenvector belonging to eigenvalue $\lambda+1$.

Proof. $$ A(Bx)=(AB)x=(BA+B)x=BAx+Bx=B(\lambda x)+Bx=(\lambda+1)Bx. $$

So if there are no eigenvalues $\lambda_1,\lambda_2$ of $A$ such that $\lambda_1-\lambda_2=1$, then we can conclude that $B=0$. On the other hand, if there exists such pairs of eigenvalues, we can find non-trivial matrices $B$.

The basic $2\times2$ template for matrices satisfying the identity $AB-BA=B$ is $$ A=\pmatrix{\lambda+1&0\cr0&\lambda\cr},\qquad B=\pmatrix{0&1\cr0&0\cr}. $$

As an extreme example, when $A=diag(10,9,8,\ldots,1)$, the matrix $B=(b_{ij})$, where $b_{ij}=1$ when $j=i+1$ and $b_{ij}=0$ otherwise, will work. This $B$ has rank nine.


If you have friends studying quantum mechanics ask them about ladder operators.

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  • $\begingroup$ Can you please explain how $(AB)X = (BA+B)X$? $\endgroup$ – Harry May 2 at 10:18
  • $\begingroup$ @user46697 We are assuming $AB=BA+B$? $\endgroup$ – Jyrki Lahtonen May 2 at 10:41
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At least, if $A$ is a diagoalizable $n$-by-$n$ matrix over an algebraically closed field $K$ with eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$, then the eigenvalues of the map $\text{ad}_A:\mathrm{Mat}_{n\times n}(K)\to\mathrm{Mat}_{n\times n}(K)$ sending $B\mapsto AB-BA$ are the $n^2$ values given by $\lambda_i-\lambda_j$, for $i,j=1,2,\ldots,n$. Thus, in this case, there is no nonzero matrix $B$ with $\text{ad}_A(B)=B$, provided that $\lambda_i-\lambda_j\neq 1$ for any $i,j=1,2,\ldots,n$.

On the other hand, if $\lambda_i-\lambda_j=1$ for some $i,j=1,2,\ldots,n$, then we let $u_i\in K^n$ and $v_j\in K^n$ be the right eigenvector for the eigenvalue $\lambda_i$ and the left eigenvector for the eigenvalue $\lambda_j$, respectively. Therefore, $$\text{ad}_A\left(u_i\,v_j^\top\right)=\left(Au_i\right)\,v_j^\top-u_i\left(A^\top v_j\right)^\top=\left(\lambda_i-\lambda_j\right)\,\left(u_iv_j^\top\right)=u_iv_j^\top\,.$$ Although I haven't tried to think about this, the condition that $\lambda_i-\lambda_j=1$ for some $i,j=1,2,\ldots,n$ should be the necessary and sufficient condition for the existence of a nonzero $B\in\text{Mat}_{n\times n}(K)$ with $\text{ad}_A(B)=B$ (even when $A$ is not diagonalizable, and we take the $\lambda_i$'s to be the generalized eigenvalues instead).

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$B = 0$ will do. In fact, there are many examples for $A$ where this is the only solution, e.g. $A = diag(2,0,0,0,0,0,0,0,0,0)$ a diagonal matrix.

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  • $\begingroup$ Will B always zero? $\endgroup$ – Himanshu Bajpai Aug 21 '17 at 13:27
  • $\begingroup$ $B = 0$ is always one solution. There might be more in some situations, but, as I pointed out, there are matrices $A$ for which $B = 0$ is the only solution. $\endgroup$ – Dirk Aug 22 '17 at 8:25

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