# Explicit linear relation for columns of adjugate matrix

Let $$A\in$$M$$_n(K)$$ with $$K$$ a field be of rank $$n-1$$. Then its adjugate matrix $$A^{adj}$$ has rank 1, as can be seen from $$A^{adj}A=\det(A)I_n=0$$, since the space of columns of $$A$$ is the kernel of $$A^{adj}$$ (some cofactor is nonzero).

This implies that any two columns of $$A^{adj}$$ are linearly dependent. Since the adjugate is polynomially computed from the original matrix, and rank $$n-1$$ implies the polynomial equation $$\det(A)=0$$, my question is: can we use this last equation (or any other) to find a explicit linear dependence relation between columns of the adjugate (e.g. by Laplace expansions or similar techniques)? I mean a relation in a generic sense, valid for all matrices of rank $$n-1$$.

Note that if this can be done, then it can be done for matrices over any commutative ring.

## 1 Answer

The last bullet point here says exactly what the adjugate is in the case of $$rank(A)=n-1$$. It is, up to a scalar, the outer product between vectors in the left and right nullspace.

• Thank you, I know this; but this is not enough: I need a polynomial way of determining those vectors, the same formula for all matrices at once (or a proof that this is impossible) – Jose Brox Jul 8 '19 at 10:00
• What do you mean by "polynomial way"? – Leo Jul 8 '19 at 12:38
• I mean that I want a linear dependence relation in which the coefficients of the columns of the adjugate are polynomial maps in the entries of the original matrix (as happens with the formula for the adjugate matrix from the cofactors of the original matrix) – Jose Brox Jul 8 '19 at 16:00
• You would need to express the (sole) eigenvectors in the kernel of arbitrary $A$ as a linear combination of its entries, which sounds too ambitious. – Leo Jul 10 '19 at 16:11