I am having some doubt in the proof of Cayley Hamilton theorem. This theorem says that every matrix is a root if its characteristic polynomial.

Proof goes as follows:

Let us assume that matrix $A$ is of order $n\times n$. If $P(\lambda)$ be its characteristic polynomial, then by the definition of the characteristic polynomial

$P(\lambda) = det (A - \lambda I) = P_0 + P_1\lambda + P_2 \lambda^2 +\ldots P_n \lambda^n$.

Next, suppose that $Q(\lambda)$ be the adjoint matrix of $(A - \lambda I)$, such that

$Q(\lambda) =Q_0 + Q_1\lambda + Q_2 \lambda^2 +\ldots Q_k \lambda^k$.

I am not able to understand why the polynomial expression of $Q(\lambda)$ is of degree $k$? Can't I write $Q(\lambda)$ as follows (degree $n$ polynomial in $\lambda$)

$Q(\lambda) =Q_0 + Q_1\lambda + Q_2 \lambda^2 +\ldots Q_n \lambda^n$.

Thank you


$Q$ is in this proof the adjugate matrix of $P-\lambda I$, that is, the matrix transpose of the cofactor matrix of $P-\lambda I$. Each element of the cofactor matrix is $\pm$ the determinant of a sub-matrix of $P-\lambda I$, that is sum of products of $n-1$ elements of $P-\lambda I$, and more than one of the elements used in each of those determinant might contain a factor of $\lambda$.

So $k$, which is the greatest power of $\lambda$ appearing in $Q$, might be equal to $n$, but it also could be more or it could be less. Without proof, you can't simply say that $k=n$.

Try it for a $3\times 3$ identity matrix $A$ . The adj of $A-\lambda I$ is a diagonal matrix with entries $(1-\lambda)^2$ along the diagonal. That is of degree $2$ in $\lambda$, not degree $3$.

  • $\begingroup$ Thank you for a comprehensive answer. It was something nontrivial for me to understand. You cleared it with a complete details. Thanks again. $\endgroup$ – mathscrazy Apr 28 '17 at 23:37

If $Q(\lambda)$ is the adjoint of $P-\lambda I$, by definition $Q(\lambda)(P-\lambda I)=det(P-\lambda I) I$.

Compare the maximum degree element of each matrix of the equality. If an element $q_{i,j}$ of $Q(\lambda)$ had degree $k$, then element in each diagonal element of the matrix would have max degree element would have degree $k+1$. You can prove it writing down the product on the left. So $n=k+1$.


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