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$\xi_A(\lambda)=det ( \lambda * I_n -A)$

Cayley-Hamilton stated that $\xi_A(A)=0_n$ Proof : $det (A-A)= det ( 0_n) =0$ Why is this proof wrong ?

First because we expect $\xi_A(A)$ to be a matrix not a scalar but I guess there is something else

  • 2
    $\lambda$ is an element of the base field and $A$ a matrix – 2017-01-23
  • 0
    Why is it ok to remplace in $ \xi_A(\lambda)$ then ? – 2017-01-23

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The matrix $X\cdot I-A$ is a matriz whose entries are polynomials. If you replace $X$ by the matrix $A$, then you end up with a matrix whose entries are matrices, not with the matrix $0$.

If $A=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}$, then $$X\cdot I-A=\begin{pmatrix}X-a_{11}&-a_{12}\\-a_{21}&X-a_{22}\end{pmatrix}$$ and the result of replacing $X$ by $A$ is the matrix in $M_2(M_2(k))$ $$\begin{pmatrix}A-a_{11}I&-a_{12}I\\-a_{21}I&A-a_{22}I\end{pmatrix},$$ which is a matrix of matrices and it clearly is not in general the zero matrix. The C-H theorem can be viewed as claiming that the determinant of this matrix is zero (this makes sense: it is a matrix with entries in the commutative ring generated by $A$ and the scalars, so determinants do make sense) is zero.