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I have this proof but did not understand the following step: $$ \det(xI - M^{-1} A M)= \det(M^{-1} xI M - M^{-1} A M)$$

The author said in the comments that it is due to "$xM$ commutes with the identity matrix $I$"

What does it mean?

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gbox
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  • I think what they mean is $xIM = MxI$. Similar to $x+y=y+x$. See commutativity-https://en.wikipedia.org/wiki/Commutative_property. – John_dydx Aug 27 '15 at 08:40

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$A$ commutes with $B$ if $AB=BA$. In this case, you have $$(xM)I = x(MI) = xM=x(IM)=I(xM)$$

so $xM$ commutes with $I$.

Because it commutes, you have $$xI=xI\cdot I = x(M^{-1}M)I =xM^{-1}(MI) = xM^{-1} IM=M^{-1}xI M$$

5xum
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  • On the last step we could bring $x$ inside the expression as it is a scalar? – gbox Aug 27 '15 at 08:50
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    @gbox Yes, you can plug in a scalar anywhere. If $a$ is a scalar and $A,B$ are matrices, then $a(AB)=(aA)B=A(aB)$ – 5xum Aug 27 '15 at 08:51
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    This looks needlessly complicated to me. Why not just $(xM)I=xM=I(xM)$? Also your second display never uses commutation of $xM$ with $I$, indeed it never even forms $xM$ to begin with. Instead it uses the (unproved here) fact that $(xA)B=A(xB)$, with $A=M^{-1}$ and $B=I$. – Marc van Leeuwen Aug 27 '15 at 09:12
  • What Marc says is also what I thought when I first saw the question. Attributing the equality $M^{-1}xIM=xI$ to the fact that $I$ commutes with every square matrix is just wrong. To prove this equality, one can use the fact that $xI$ commutes with $M$, but not the fact that $I$ commutes with $xM$. At any rate, proving the equality using commutativity is unnatural. What matters here is actually --- as Marc says --- another fact, namely, $(xA)B=A(xB)$. – user1551 Aug 27 '15 at 09:59
  • @MarcvanLeeuwen I agree, I now think my answer is bad... I hope OP accepts the better answer that you gave so I can delete mine. – 5xum Aug 27 '15 at 10:22
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I think the text should say that $xI$ (the matrix with $x$ on the diagonal and zeros elsewhere) commutes with $M$, since that is what is being used. Indeed once you know this fact you have $$ M^{-1}(xI)M=M^{-1}M(xI)=xI $$ which is what is (in the opposite direction) used. But the fact that $xI$ commutes with $M$ is obvious: computing $(xI)M$ means multiplying every row of $M$ by the indeterminate $x$, while computing $M(xI)$ means multiplying every column of $M$ by the indeterminate $x$, which obviously has the same result.

Marc van Leeuwen
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  • So generally speaking, we can say that both the Identity matrix and the scalar matrix are commutative with any matrix – gbox Aug 27 '15 at 09:48
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    @gbox: Yes that is true. Although I might pedantically add to the statement about scalar matrices "as long as scalars themselves commute". If ever you consider matrices over non-commutative rings (which is perfectly possible) then "scalar matrices" no longer always commute, because left-multiplication by a given scalar may be different from right-multiplication by it. But then scalar matrices are not a very useful notion is such a setting to begin with. – Marc van Leeuwen Aug 27 '15 at 09:56