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Let $A$ and $B$ be matrices over $\mathbb C$. Then,

  1. $AB$ and $BA$ always have the same set of eigenvalues.

  2. If $AB$ and $BA$ have the same set of eigenvalues then $AB=BA$.

  3. If $A^{-1}$ exists then $AB$ and $BA$ are similar.

  4. The rank of $AB$ is always the same as the rank of $BA$ .

  1. Suppose $AB=BA$ Let $x$ be the eigen vector of $A$ corresponding to the eigenvalue $a$. $$ABx=BAx=aBx \implies Bx$$ is the eigen vector of $A$. If the eigen space corresponding to the eigen values of $A$ is one. Then, $Bx=\lambda x \implies x$ is the eigen vector of $B$. So $AB$ and $BA$ have same set of eigen values. statement is false. Am I correct?

  2. I don't know, How to judge the statement.

  3. I don't know, How to judge the statement.

  4. Statement is false, I could obtain the counter examples.

Please check my answers. Please help me.

  • 1
    In your argument for (1), it looks like you're assuming that $AB=BA$. Do you mean to do that? – G Tony Jacobs Nov 24 '17 at 15:34
  • sorry, I had network problem yesterday. No.Actually, I was trying to show, At what condition 1. is true. @GTonyJacobs –  Nov 25 '17 at 01:19

1 Answers1

1

1.See the reference helpfully provided below.

  1. This is false. Choose any two invertible matrices which do not commute with each other. By (3), $AB$ and $BA$ are similar and therefore have the same eigenvalues.

  2. This is true since $A^{-1}(AB)A=BA.$

  3. This is false. Choose any two matrices such that $AB$ is $0$ but $BA$ is not $0$.

You might find the example of two matrices given below to be quite useful when checking other conjectures about singular matrices.

$\begin{pmatrix}0&1\\0&0\end{pmatrix}$ $\begin{pmatrix}1&0\\0&0\end{pmatrix}$=$\begin{pmatrix}0&0\\0&0\end{pmatrix}$, $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ $\begin{pmatrix}0&1\\0&0\end{pmatrix}$=$\begin{pmatrix}0&1\\0&0\end{pmatrix}$

  • What are the $2$ by $2$ matrices which disprove this conjecture in 1? – Sungjin Kim Nov 24 '17 at 16:44
  • Your matrices are not counterexample to 1, because they have the same set of eigenvalues ${0}$. – Sungjin Kim Nov 24 '17 at 16:50
  • I was giving extra help to Maneesh not answering your question which I have only just seen. Interchanging the 0 and 1 in the first column of my second matrix answers your query. –  Nov 24 '17 at 16:54
  • Still no. The statement 1 is true. In general, the characteristic polynomial of $AB$ is identical to the characteristic polynomial of $BA$. The proof is seen here: https://math.stackexchange.com/questions/311342/do-ab-and-ba-have-same-minimal-and-characteristic-polynomials – Sungjin Kim Nov 24 '17 at 16:57
  • Yes, I stand corrected. It is the eigenvectors which need not be the same. –  Nov 24 '17 at 17:01