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If $A,B$ are square matrices and $A^2=A,B^2=B,AB=BA$, then calculate $\det (A-B)$.

My solution: consider $(A-B)^3=A^3-3A^2B+3AB^2-B^3=A^3-B^3=A-B$, then $\det(A-B)=0\vee 1\vee -1$

The result of the book is the same as mine, but their solution is different. They begin: "Since $A^2=A$ and $B^2=B$, $A$ and $B$ are diagonalizable; moreover, since $AB=BA$, then there exists a invertible matrix $P$ such that $P^{-1}AP$ and $P^{-1}BP$ are diagonal matrices."

The statement has two parts, and I don't understand both. So I really need a specific explanation for those two parts.

Thanks in advance.

Tien Kha Pham
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2 Answers2

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For any square matrix $\;A\;$ :

$\;A^2=A\implies A(A-I)=0\implies\;$ the minimal polynomial of $\;A\;$ divides $\;x(x-1)\;$ and it is thus a product of different linear factors $\;\iff A\;$ is diagonalizable.

Also, if two matrices are diagonalizable, then they are simultaneosly diagonalizable iff they commute.

Timbuc
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  • Could you give a proof for "if $A,B$ commute, then they are simultaneously diagonalizable"? – Tien Kha Pham Feb 17 '15 at 17:45
  • @TienKhaPham The proof is tricky for general $A,B$ that commute. See my answer for the case when all eigenvalues of $A$ are distinct. The general problem actually appeared as a preliminary exam problem for graduate students in mathematics at UC Berkeley (a top 5 math program). – user2566092 Feb 17 '15 at 17:48
  • @TienKhaPham Try here http://math.stackexchange.com/questions/236212/simultaneously-diagonalizable-proof – Timbuc Feb 17 '15 at 17:53
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If $AB = BA$, and $A$ and $B$ are diagonalizable, then you can find a basis matrix $P$ that simultaneously diagonalizes $A$ and $B$. To see this, consider an eigenvector $v$ of $A$. Then $A(Bv) = BAv = B \lambda v$, so $Bv$ is an eigenvector of $A$ too, with the same eigenvalue. But if $A$ is diagonalizable with distinct eigenvalues this can only happen if $v$ is an eigenvector of $B$. The argument for non-distinct eigenvalues of $A$ is a little trickier but it can be done, along the same lines.

user2566092
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