# Prove that if $v$ is an eigenvector for A, than $v$ is also and eigenvector for $adj(A)$.

Let $$A$$ be a complex square matrix and $$\operatorname{adj}(A)$$ its adjugate (so the entries of $$\operatorname{adj}A$$ are minors of $$A$$, up to sign). Prove that if $$v$$ is an eigenvector for $$A$$, than $$v$$ is also an eigenvector for $$\operatorname{adj}(A)$$.

• Can you provide your defintion of the adjugate matrix? – John Aug 6 at 9:43
• $A^{-1} =\frac{1}{detA} adj(A)$ – nene123 Aug 6 at 9:45
• I can't do it but this should be put on hold. – limitsandlogs224 Aug 11 at 0:04

An argument that works over any commutative ring:

It is well-known that $$\operatorname{adj} A$$ can be written as a polynomial in $$A$$. In other words, there exists a polynomial $$p$$ (in one variable) over the ground ring such that $$\operatorname{adj} A = p\left(A\right)$$. (This $$p$$ can be described explicitly: We have \begin{align} p\left(t\right) = \left(-1\right)^{n-1} \sum_{i=0}^{n-1} c_{n-1-i} t^i , \end{align} where $$c_j$$ denotes the coefficient of $$t^{n-j}$$ in the characteristic polynomial $$\det\left(tI_n-A\right)$$ of $$A$$. See Theorem 5.14 in my The trace Cayley-Hamilton theorem for a proof. Alternatively, you can find proofs at https://mathoverflow.net/questions/32133/expressing-adja-as-a-polynomial-in-a .)

Now, let $$v$$ be an eigenvector of $$A$$, and let $$\lambda$$ be the corresponding eigenvalue. Thus, $$A v = \lambda v$$. Consider the polynomial $$p$$ defined above, which satisfies $$\operatorname{adj} A = p\left(A\right)$$. From $$A v = \lambda v$$, we can easily obtain (by induction) that $$A^i v = \lambda^i v$$ for each nonnegative integer $$i$$. Hence, $$p\left(A\right) v = p\left(\lambda\right) v$$ (since the polynomial $$p\left(t\right)$$ is a linear combination of powers $$t^i$$). Thus, $$v$$ is an eigenvector of the matrix $$p\left(A\right)$$. In other words, $$v$$ is an eigenvector of the matrix $$\operatorname{adj} A$$ (since $$\operatorname{adj} A = p\left(A\right)$$).

This is an interesting question, in fact at first I thought it wasn't true when the eigenvalue is $$0$$, but it turns out that it is !

We have the well-known formula $$adj(A)A = \det(A)I_n$$. From this it follows at once that if $$v$$ is an eigenvector of $$A$$ with nonzero eigenvalue $$\lambda$$, $$v$$ is an eigenvector of $$adj(A)$$ with eigenvalue $$\det(A)/\lambda$$.

If $$Av =0$$ however, this does not work so we have to use a trick : perturb $$A$$ by $$\epsilon>0$$ to get $$A_\epsilon = A +\epsilon I_n$$. Then $$v$$ is an eigenvector of $$A_\epsilon$$ with eigenvalue $$\epsilon$$ which is nonzero. It follows that $$adj(A_\epsilon)v = \frac{\chi_A(-\epsilon)}{\epsilon}$$.

Note that since $$A$$ has $$0$$ as an eigenvalue, its determinant is $$0$$ so $$\chi_A(0) =0$$. Therefore, letting $$\epsilon \to 0$$ in the previous equality gives the derivative of $$\chi_A$$ : $$adj$$ is continuous so that it yields $$adj(A)v = -\chi_A'(0)v$$

(We even have an explicit formula for the eigenvalue !)

One might wonder whether this generalizes to other fields, since I used the topology of $$\mathbb C$$ (or $$\mathbb R$$) to let $$\epsilon$$ tend to $$0$$; the only problem is that you have to be careful in the last step : you want to first work in $$K(X)$$ if $$K$$ is your base field, then notice that the fraction $$\det(A+XI_n)/X$$ is actually a polynomial (because $$\det(A)=0$$), so we may work in $$K[X]$$, and then let $$X=0$$.

(Morally you can work in the ring if dual numbers $$K[\epsilon]$$ but the technicalities seem more annoying to write)

The easiest (and the most natural) solution is to note that $$\operatorname{adj}(A)$$ is a polynomial in $$A$$ (as indicated in darij grinberg's answer). For a more elementary approach, you may consider the rank of $$A$$. Suppose $$A$$ is $$n\times n$$. There are three possibilities:

1. $$\operatorname{rank}(A)=n$$. This case is easy, as $$\operatorname{adj}(A)=\det(A)A^{-1}$$.
2. $$\operatorname{rank}(A). This is case is also easy, as $$\operatorname{adj}(A)=0$$.
3. $$\operatorname{rank}(A)=n-1$$. Then the left or right null spaces of $$A$$ are one-dimensional and $$\operatorname{adj}(A)=uv^T$$ for some eigenvector $$u$$ in the right null space of $$A$$ and some eigenvector $$v$$ in the left null space of $$A$$. Now, suppose $$x$$ is an eigenvector of $$A$$.
• If $$Ax=0$$, since the null space of $$A$$ is one-dimensional, we must have $$x=cu$$ for some scalar $$c$$. Thus $$\operatorname{adj}(A)x=uv^T(cu)=(v^Tu)x$$.
• If $$x$$ is an eigenvector of $$A$$ corresponding to some nonzero eigenvalue $$\lambda$$, then $$0=(v^TA)x=v^T(Ax)=\lambda v^Tx$$. Hence $$v^Tx=0$$ and $$\operatorname{adj}(A)x=uv^Tx=0$$.
• I wonder whether using your method could be proved more general statement i.e. if $AB=BA=0$ and $rank(A)+rank(B)=n$ ( I'm not sure whether this condition is really needed) then A,B have the same eigenvectors ( ... if you regard necessary I can ask a new question). – Widawensen Aug 7 at 7:24
• @Widawensen I don't think so. E.g. when $A=0\oplus X$ and $B=Y\oplus 0$ are two block matrices with conformable sizes and $X,Y$ are square sub-blocks, $e_1$ is not necessarily an eigenvector of $B$ and $e_n$ is not necessarily an eigenvector of $A$. – user1551 Aug 7 at 7:50
• Yes, it seems to be a good counterexample. – Widawensen Aug 7 at 7:55

It is because if A, B are similar square matrices, adj(A) and adj(B) are also similar (§). If v is an eigenvector of A, then you pick a basis e_1...,e_n with v=e_1, the result will be obvious in that basis.

In order to prove (§), define K:=the field of fractions of Z[X_1,...,X_{2n^2}] you may consider for every nxn invertible matrices F,G with coefficients in K, the identity (GFG^{-1})^{-1} = 1/det(F) adj(GFG^{-1}) = 1/det(F)G*adj(F)G^{-1}=GF^{-1}G^{61} which yields to adj(GF*adj(G)) = det(G)^{n-2}Gadj(F)*adj(G). This holds for every invertibles F,G and so it holds for any F,G with coefficients in Z[X_1,...,X_{2n^2}] and so this polynomial identity holds in any commutative ring. The result can be deduced from this (by dividing again by the determinant of the base change matrix)