Vtyjkyuk

We know that if $A$ is invertible, then $A^-1$ can be expressed as a polynomial of $A$, that is to say, there exists a polynomial $f(x)$ such that $$A^-1 = f(A)$$

Of course in this case, $A^*$ (the adjoint matrix of $A$) can also be expressed as a polynomial of $A$.

I wonder if $A$ is not invertible, can $A^*$ be expressed as a polynomial of $A$？

Hope for your comments.

Yes, $operatornameadj(A)$ can always be expressed as a polynomial in $A$. We can follow the same proof as for Cayley-Hamilton:

Consider $operatornameadj(A - tI)$ for a scalar $t$. We have

$$(A - tI)operatornameadj(A - tI) = det(A - tI)I = p_A(t)I$$
where $p_A(t) = (-1)^nt^n + c_n-1t^n-1 + cdots + c_1t + c_0$ is the characteristic polynomial of $A$.

Notice that $operatornameadj(A - tI)$ is also a poylnomial in $t$ of degree $le n-1$, so we can pick matrices $B_0, ldots, B_n-1$ such that $$operatornameadj(A - tI) = sum_i=0^n-1t^iB_i$$

Now we have
beginalign
p(t)I &= (A - tI)operatornameadj(A - tI) \
&= (A - tI)sum_i=0^n-1t^iB_i \
&= sum_i=0^n-1t^i AB_i - sum_i=0^n-1t^i+1B_i\
&= -t^nB_n-1 + sum_i=1^n-1t^i(AB_i - B_i-1) + AB_0
endalign

Comparing powers with $p_A(t)I = (-1)^nt^nI + c_n-1I + cdots + c_1tI + c_0I$ gives

$$B_n-1 = (-1)^n+1 I, qquad AB_i - B_i-1 = c_iI text for 1 le i le n-1, qquad AB_0 = c_0I$$

Now we can inductively express $B_i$ as poylnomials in $A$:

$$B_n-1 = (-1)^n+1 I$$
$$B_n-2 = AB_n-1 - c_n-1I = (-1)^n+1A - c_n-1I$$
$$B_n-3 = AB_n-2 - c_n-2I = (-1)^n+1A^2 - c_n-1A - c_n-2I$$
$$vdots$$
$$B_0 = AB_1 - c_1I = (-1)^n+1A^n-1 - c_n-1A^n-2 - cdots - c_2A - c_1I $$

Therefore

$$operatornameadj(A) = sum_i=0^n-1t^iB_iBigg|_t = 0 = B_0 = -Big[(-1)^nA^n-1 + c_n-1A^n-2 + cdots + c_2A + c_1IBig]$$

Yes, the adjucate matrix can always be expressed as a polynomial in $A$: Consider the characteristic polynomial of $A$, i.e. $$p(A)=A^n+c_n-1A^n-1+...+c_1A+c_0=0$$
You know that $c_0=(-1)^ndet(A)$. So you can write $$det(A)=(-1)^n+1(A^n-1+c_n-1A^n-2+...+c_1)Alabelx$$

If $det(A)neq0$ this already implies $$A^*=(-1)^n+1(A^n-1+c_n-1A^n-2+...+c_1)$$ which is a polynomial in $A$. For $det(A)=0$, you can argue from continuity. I.e.

• The expression we derived for $A^*$ is continuous in $A$


• You know that the true $A^*$ is continuous in $A$ (because it can be expressed using co-factors)


• the set of non-invertible matrices is a measure-zero subset of all matrices

UPDATED TO REMOVE UNNECESSARY TOPOLOGICAL ARGUMENTS

Let $AinmathbbM_n(K)$ and let $p_A$ be its characteristic polynomial. We have $p_A(X)=Xq(X)+(-1)^ndet(A)$ for some polynomial $qin K[X]$, so that
$$(-1)^n-1det(A)I_n=Aq(A).$$

Multiply by $textadj(A)$ on the left to get

$$(-1)^n-1det(A)textadj(A)=det(A)q(A).$$

Now consider $det(A), q(A), textadj(A)$ as polynomials in the entries of $A$, in $K[X_1,ldots,X_n^2]$. Since $det$ is a nonzero element in an integral domain, we can cancel it and get
$$textadj(A)=(-1)^n-1q(A).$$

Therefore $$textadj(A)=(-1)^n-1fracp_A(X)-(-1)^ndet(A)Xleftrvert_X=Aright.$$
is the formula we are seeking.