I have to prove/disprove this:

If $\det(A+X) = \det(B + X)~ \forall X \in M_{n \times n} (\mathbb F) \rightarrow A = B$

I believe it is true but I can not think of a direct way to prove it. Any help is appreciated!

  • $\begingroup$ If $\det(A+X) = \det(B + X)~ \forall X \in M_{n \times n}$, then $\det(A-\lambda I) = \det(B -\lambda I)~ \forall \lambda \in F$. (I don't know the following!) $\endgroup$ – Arnaud May 6 '13 at 20:21

Let $M_1$ be the set of matrices such that the first row of all the matrices in $M_1$ is the negative of the first row of $A$. Then $\det(A+X)=0$ for all $X\in M_1$, so we also have $\det(B+X)=0$ for all $X\in M_1$.

I claim that the first row of $B$ must be equal to the first row of $A$. Assume contrariwise that this is not the case. Then all the matrices of the form $B+X, X\in M_1$ share the same non-zero first row (= the first row of $B-A$). But by a suitable choice of $X$ from $M_1$ we can make the remaining $n-1$ rows of $B+X$ to be anything we want. As the first row of $B+X$ was assumed to be non-zero, we can find an $X\in M_1$ such that $\det(B+X)\neq0$. This is a contradiction.

Clearly we can repeat the argument for any other row, so we can conclude that $A=B$.

  • $\begingroup$ What about the $X$ that are not in $M_1$? $\endgroup$ – TheNotMe May 9 '13 at 12:55
  • $\begingroup$ It was given that $\det(A+X)=\det(B+X)$ for all $X$. Therefore the same holds for all $X\in M_1$. Therefore (see my answer) $A$ and $B$ have the same first row. Repeat the same argument with another subset $M_2$ of matrices such that the second row of $A+X$ is all zeros for all $X\in M_2$. Rinse. Repeat. $\endgroup$ – Jyrki Lahtonen May 9 '13 at 13:01
  • $\begingroup$ Okay. What about the $X$ that are not in any $M_i$? $\endgroup$ – TheNotMe May 9 '13 at 13:23
  • $\begingroup$ Looks like I didn't need them to conclude that $A=B$! Observe that the $X$:s only appear in our assumption. We were not asked to prove anything for all $X$, so it is ok to pick and choose which $X$:s we need to make our argument work. Mind you, the other direction $A=B\implies \det(A+X)=\det(B+X)$ is true, but kinda trivial. $\endgroup$ – Jyrki Lahtonen May 9 '13 at 15:16

A characterization of the determinant is that $$ \det M=\sum_{s\in\mathfrak S_n}\varepsilon(s)\cdot\prod_{i=1}^nM_{is(i)}, $$ where $\varepsilon(s)$ is the signature of the permutation $s$ in $\mathfrak S_n$. Applying this to $M=A+X$, one gets a polynomial in $\mathbb F[X_{ij},1\leqslant i,j\leqslant n]$, namely, $$ \det(A+X)=\sum_{s\in\mathfrak S_n}\varepsilon(s)\cdot\prod_{i=1}^n(X_{is(i)}+A_{is(i)}). $$ Fix $t$ in $\mathfrak S_n$ and $1\leqslant k\leqslant n$. The coefficient of $$ \prod_{i\ne k}X_{it(i)}, $$ in $\det(A+X)$ is $\varepsilon(t)A_{kt(k)}$ hence the polynomial $\det(A+X)$ fully determines $A$.

  • $\begingroup$ +1, but I have one remark. You have shown that the matrix $A$ determines the polynomial. But we were only given that it determines the polynomial function. If the field $F$ happens to be finite, then this is not quite enough for in that case many polynomials give rise to the same function. The difference between two polynomials giving the same function is of a very restricted form, so this is not a cause for major concern. $\endgroup$ – Jyrki Lahtonen May 6 '13 at 20:44
  • $\begingroup$ And my reservation is clearly not an issue here, because all the variables $X_{ij}$ only appear with exponent one. For my issue to make a difference, the exponent would have to be at least two (in which case it might be a problem over the field of two elements). Sorry about the fuss. $\endgroup$ – Jyrki Lahtonen May 6 '13 at 20:51
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    $\begingroup$ @JyrkiLahtonen Thanks for your TWO comments, please leave them both (if you do not mind), I think that many readers will benefit from them. $\endgroup$ – Did May 6 '13 at 20:54

I just see this problem today, so this is a late answer.

We can use the same trick I used in an answer to another question. By elementary row/column reductions, we have $B-A=P(I_r\oplus0)Q$ for some invertible matrices $P$ and $Q$, with $r=\operatorname{rank}(B-A)$. Therefore, by putting $Y = P^{-1}(A+X)Q^{-1}$, the given assumption is equivalent to $\det(Y)=\det((I_r\oplus0) + Y)$ for all $Y$. Since this condition does not hold if $r\neq0$ and $Y=0\oplus I_{n-r}$, we conclude that $r$ must be zero, i.e. $B=A$.


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