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Exercise: Why is the following "proof" false?

$\text{ch}_A(xI-A)=\det(xI-A)$

Substitude $A$ for $x$, and obtain $\det(A-A)=0=\text{ch}_A(A)$.

Solution: Explanation

We cannot substitude $A$ for $x$ because $A$ is a matrix and $x$ represents a scalar.

$xI$ represents the matrix $\begin{pmatrix} x\\ &x\\ &&\ddots\\ &&&x\\ \end{pmatrix}$, and we clearly cannot substitute the matrix $A$ into the entries.

user1163791
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    Is $\operatorname{ch}_A$ supposed to be the characteristic polynomial of $A$? If so, the first expression $\operatorname{ch}_A(xI-A)$ doesn't make any sense. It would make your question more readable if you could write out what false statement is being claimed with this fallacious proof. – Matthew Leingang Mar 25 '23 at 19:57
  • In this problem $x$ is an element of the field, like $\lambda$. Trying to substitute $A$ for $x$ would result in a matrix whose diagonal elements are matrices. – John Douma Mar 25 '23 at 20:01
  • The other issue is that by your logic, doesn't $ch_A(xI-A)$ also become $ch_A (0)$? – Calvin Lin Mar 25 '23 at 20:01

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