Linear Algebra – Why Can’t We Substitute A for ? in Cayley-Hamilton Theorem?

linear algebramatrices

Cayley Hamilton Theorem states that if $A$ is an $n \times n$ matrix over the field $F$ then $p(A) = 0$.

We note that $p(\lambda) = \det(\lambda I – A)$. Hence, why can't we just substitute $\lambda$ with $A$ and directly prove Cayley-Hamilton Theorem by saying that $p(A) = p(\lambda) = \det(\lambda I – A) = \det(AI – A) = 0$?

Best Answer

There is another way to see that the proof must be flawed: by finding the interesting consequences this proof technique has. If the proof would be valid, then we would also have the following generalisation:

Faulty Lemma. Suppose that $A$ and $B$ are $n\times n$ matrices. Let $p_A$ be the characteristic polynomial for $A$. If $B - A$ is singular, then $B$ must be a zero of $p_A$.

Faulty proof: We have $p_A(B) = \det(BI - A) = \det(B - A) = 0$.$$\tag*{$\Box$}$$

This has the following amazing consequence:

Faulty Corollary. Every singular matrix is nilpotent.

Faulty proof: Let $B$ be a singular matrix and let $A$ be the zero matrix. Now we have $p_A(\lambda) = \lambda^n$. Furthermore, by the above we have $p_A(B) = 0$, because $B - A$ is singular. Thus, we have $B^n = 0$ and we see that $B$ is nilpotent.$$\tag*{$\Box$}$$

In particular, this proves that we have $$ \pmatrix{1 & 0 \\ 0 & 0}^2 = \pmatrix{0 & 0 \\ 0 & 0}. $$ This comes to show just how wrong the proof is!

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