2
$\begingroup$

Let $A$ be an $n\times n$ matrix and $Z$ be the $n\times n$ matrix, whose entries are all $m$. Let $S$ be the sum of all the adjoints of $A$. Then my conjecture is

$\det(A+Z)=\det(A)+Sm$ , in particular $\det(A+Z)=\det(A)$ holds if and only if $S=0$. If the conjecture is true, how can it be proven ?

For invertible $A$, sylvester's theorem can be used. $\det(A+Z)=\det(A)\det(I+A^{-1}Z)$ The matrix $A^{-1}Z$ is the product of the row vector containing the row sums of $A^{-1}$ and the column vector $(m,...,m)$. Sylvester's theorem states that the order of the vectors can be exchanged. The scalar product of the vectors is $0$ if and only if the sum of the row sums of $A^{-1}$ is $0$ and this is the same as the sum of the adjoints of $A$ divided by the determinant of $A$. So, for invertible matrices, my conjecture should hold. But how can I manage the case when $A$ is singular ?

$\endgroup$
5
  • $\begingroup$ "sum of all the adjoints" = "sum of all entries of the adjoint"? $\endgroup$
    – darij grinberg
    Jul 26, 2014 at 19:03
  • $\begingroup$ If the adjoint (matrix) is the matrix whose entries are the adjoints, then yes. $\endgroup$
    – Peter
    Jul 26, 2014 at 19:05
  • 1
    $\begingroup$ I think you are referring to the adjugate. "Adjoint" usually refers to the adjoint with respect to a given inner product, which is usually the transpose (in $\mathbb{R}^n$ with the Euclidean inner product) or the conjugate transpose (in $\mathbb{C}^n$ with the Euclidean inner product). $\endgroup$
    – Ian
    Jul 26, 2014 at 19:08
  • $\begingroup$ I mean the determinant of the submatrix corresponding to $a_{ji}$, multiplied with $(-1)^{i+j}$, if $a_{ij}$ is the element. $\endgroup$
    – Peter
    Jul 26, 2014 at 19:11
  • $\begingroup$ If my terminology is wrong, please edit my question. $\endgroup$
    – Peter
    Jul 26, 2014 at 19:12

1 Answer 1

Reset to default
3
$\begingroup$

You can use a continuity argument. The function $$A\longmapsto \det(A+Z)-\det(A)-Sm$$ is a polynomial function from the vector space $M_n(\mathbb C)$ of all $n\times n$ complex matrices to the complex numbers, so in particular it is continuous. You claim that it is zero on the set of matrices with non-zero determinant. Since this set is dense in $M_n(\mathbb C)$, the function is in fact identically zero.

$\endgroup$
3
  • $\begingroup$ Very nice answer! $\endgroup$
    – Peter
    Jul 26, 2014 at 19:30
  • $\begingroup$ Using this, I can calculate det(X+A) in general, if both X and A are singular. $\endgroup$
    – Peter
    Jul 26, 2014 at 19:33
  • $\begingroup$ I remember that the pseudoinverse of a matrix can be calculated by using a very similar method. $\endgroup$
    – Peter
    Jul 26, 2014 at 19:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .