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Given $A\in M_{n \times n}(\mathbb{F})$ and $p_{A}(x)=\det(xI-A)$ why saying that $\det(AI-A)=0$ is not valid?

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marked as duplicate by user1551 linear-algebra Sep 9 '15 at 5:16

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It's not valid because in the expression $xI$, the hidden operation is scalar multiplication, i.e., multiplication of the identity matrix by the constant real value $x$. In the expression $AI$, the hidden operation is matrix multiply, which is in general different from scalar multiply.

There's a real subtlety here: the polynomial $p_A(x)$ is, as defined by the equation, a polynomial function of a single real variable. The CH theorem says that if you, after having computed the coefficients of that polynomial, now plug in the matrix $A$ for the variable $x$, and treat powers of $x$ as matrix powers of $A$, you get the zero matrix. Doing that plugging=in=a=matrix=where=a=real=number=should=be is a very weird operation, and so the result is quite surprising.

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  • $\begingroup$ And this might cause a person to wonder what a "variable" is, if it is not a real number... because what could it mean to multiply a matrix by a "variable", and do linear algebra "with a variable", and so on, leaving open the possibility of "substituting" (meaning what, then!?!) a matrix for the "variable". Hilarious! $\endgroup$ – paul garrett May 19 '15 at 16:58
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It's not valid because Cayley Hamilton says that $p_A(A)=0$ where 0 is the zero matrix, but in your argument you have a zero scalar.

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