Etydhv

Lets say I have a matrix A with constants x and paramater Y in the following form:

$$A=
beginbmatrix
Y & x_12 & x_13 \
x_12 & Y & x_23 \
x_13 & x_23 & Y
endbmatrix
$$

I know that if all x are the same, then $det[A] = 0$ has only one solution for Y.

My proof here is to directly compute the determinant, and then solve the suppressed cubic.

I want to also prove that if not all x are the same, that det[A]=0 has 3 distinct solutions. This is slightly more than just the converse. Directly computing is very painful, and I was able to show it only with Mathematica.

My question is how to justify these two statements more elegantly.

Your first proposition is generally incorrect. If $x_12=x_13=x_23=x$, then both $Y_1,2=x$ and $Y_3=-2x$ are solutions of $det[A]=0$. Those are different for $x neq0$. In that case, the first solution has algebraic and geometric multiplicity $2$, the second algebraic and geometric multiplicity $1$.

ADDED after more thoughts about the second proposition:
The second proposition is true. It is easy to see that

$$det[A]=f(Y)=Y^3 + 2x_12x_13x_23 - Y(x^2_12 + x^2_13 + x^2_23)$$.

In order for $f(Y)=0$ to have a root with multiplicity > 1, it has to be also a root of $f'(Y)=0$, which means

$$f'(Y)=3Y^2-(x^2_12 + x^2_13 + x^2_23)=0,$$

which implies

$$Y=pmsqrtfracx^2_12 + x^2_13 + x^2_233.$$

Putting this into $f(Y)=0$ we get

$$0 = f(Y)=Y(Y^2-(x^2_12 + x^2_13 + x^2_23)) + 2x_12x_13x_23=-frac23(x^2_12 + x^2_13 + x^2_23)Y + 2x_12x_13x_23$$

which implies

$$x_12x_13x_23 = frac13(x^2_12 + x^2_13 + x^2_23)Y.$$

Squaring both sides yields to

$$(x_12x_13x_23)^2=frac19(x^2_12 + x^2_13 + x^2_23)^2Y^2 = frac19(x^2_12 + x^2_13 + x^2_23)^2fracx^2_12 + x^2_13 + x^2_233 =frac127(x^2_12 + x^2_13 + x^2_23)^3$$

This is the same condition as Yves Daoust arrived in his answer.

Now if we assume that all the $x_ij$ are real numbers, then this condition indeed implies that all $x_ij$ must be the same. That's because, as every $x_ij$ appears squared, we can generally assume that all $x_ij ge 0$. The one can apply the theorem of the inequality of quadratic and geometric mean:

$$sqrt[3]x_12x_13x_23 le sqrtfracx^2_12 + x^2_13 + x^2_233$$

Putting this inequality to the 6th power leads to

$$(x_12x_13x_23)^2le frac127(x^2_12 + x^2_13 + x^2_23)^3$$

and we know that equality (as required for a solution for $Y$ with multiplicity > 1) only happens if $x_12=x_13=x_23$.

The polynomial is

$$y^3-(x_12^2+x_23^2+x_13^2)y+2x_12x_23x_13.$$

The multiplicity of the roots is determined by the discriminant,

$$4p^3+27q^2propto-(x_12^2+x_23^2+x_13^2)^3+27(x_12x_23x_13)^2.$$