In this note, we discuss a determinantal identity and proceed to prove it. En route, we also show that the square of a number of the form $(a^3 + b^3 + c^3 - 3 a b c)$ is also of the same form.
Show that
$\mathrm{det}\begin{pmatrix}yz - x^2&zx - y^2&xy - z^2\\
zx - y^2&xy - z^2&yz - x^2\\xy - z^2&yz - x^2&zx - y^2\end{pmatrix} =
\mathrm{det}\begin{pmatrix}r^2&u^2&u^2\\u^2&r^2&u^2\\u^2&u^2&r^2\end{pmatrix}\tag{1}$
where
$r^2 = x^2 + y^2 + z^2$ and $u^2 = xy + yz + zx$.
Note that this can surely be proven by painstakingly expanding all the determinants and multiplying all the products out; but that is not the solution we are aspiring for. We are looking for a cleaner, methodical solution that will hopefully throw some insight as to why such an identity is true.
The very first thing to appreciate upon seeing an identity of this sort is to appreciate the sheer symmetry. Such
mathematical beauty never fails to remind me of Blake's immortal lines in "The Tyger":
In order to understand the symmetry, let us consider the matrix on the LHS of the identity.
Substituting variables $a = yz - x^2, b = zx - y^2, c = xy - z^2$, the matrix on the LHS
can be seen to be:
$\mathrm{det}\begin{pmatrix}a&b&c\\b&c&a\\c&a&b\end{pmatrix}$.
Let us take a moment to admire how pretty this factoid is. The form on the RHS (sans the sign) i.e. $a^3 + b^3 + c^3 - 3 abc$ is quite common in mathematics (inequalities etc.). For instance, the AM-GM inequality for 3 quantities is equivalent to the statement that $a^3 + b^3 + c^3 - 3 abc \geqslant 0$ (for $a, b, c \geqslant 0$).
How do you prove this statement? Well - it is not too hard to compute the determinant by (almost) brute-force. One may simplify the determinant (and here, use the familiar properties of determinants, see a list here) as follows. Replace the first column by the sum of all columns of the matrix (and recall that this does not change the value of the determinant), and then take the common $(a + b + c)$ factor out.
$\mathrm{det}\begin{pmatrix}a&b&c\\b&c&a\\c&a&b\end{pmatrix} = \mathrm{det}\begin{pmatrix}a + b + c&b&c\\a + b + c&c&a\\a + b + c&a&b\end{pmatrix} = (a + b + c) \mathrm{det}\begin{pmatrix}1&b&c\\1&c&a\\1&a&b\end{pmatrix}$.So this gives us some "contextual" information about the LHS.
How about the RHS?How did we jump to such a claim? Well in this regard, one has to be familiar with the following property of determinants: $\mathrm{det}(AB) = \mathrm{det}(A)\mathrm{det}(B)$. Given that one knows this, there will be a conscious effort to mould the determinant on the LHS into something simpler (in this case, the square of something simpler).
Now, we notice that with $A = B = \begin{pmatrix}x&y&z\\z&x&y\\y&z&x\end{pmatrix}$, we have that $AB = \begin{pmatrix}x^2 + y^2 + z^2&xy + yz + zx&zx + xy + yz\\zx + xy + yz &x^2 + y^2 + z^2&yz + zx + xy \\yz + zx + xy&zx + xy + yz &x^2 + y^2 + z^2\end{pmatrix}$.In fact, the proof in Subproblem 2 gives us the following
Note that going from the matrix $\begin{pmatrix}x&y&z\\y&z&x\\z&x&y\end{pmatrix}$ to the matrix
$\begin{pmatrix}x&y&z\\z&x&y\\y&z&x\end{pmatrix}$ changes the sign of the determinant - this is
because the $2^{nd}$ and the $3^{rd}$ rows have been swapped. We can restore the sign of the
determinant by negating the second row:
$\mathrm{det}\begin{pmatrix}x&y&z\\y&z&x\\z&x&y\end{pmatrix} = \mathrm{det}\begin{pmatrix}x&y&z\\-z&-x&-y\\y&z&x\end{pmatrix}$
Taking $ A = \begin{pmatrix}x&y&z\\y&z&x\\z&x&y\end{pmatrix}$ and $B = \begin{pmatrix}x&y&z\\-z&-x&-y\\y&z&x\end{pmatrix}$ we see that $AB = \begin{pmatrix} x^2&z^2&2xz- y^2\\ 2xy - z^2&y^2&x^2\\ y^2&2yz-x^2&z^2\end{pmatrix}$
This appears somewhat promising, enough for us to try many different substitutions and tricks using the
determinants. At times in this pursuit, it felt like we were really close to the actual form desired.
However,
despite numerous methodical efforts, I was unable to prove the identity via this approach.
If someone succeeds using this approach, I would very much like to know in the comments below.
For now, it was time to rethink our strategy.
This is where complex numbers enter the stage! We will view this determinant as a function of the three variables $x, y$ and $z$; in fact this is a polynomial function (of the $3^{rd}$ degree). So if we are able to find $3$ distinct factors of this determinantal polynomial, we would be done: the product of the three factors would equal (modulo the sign) the determinant!
Well, we already know one factor of the polynomial i.e. $(x + y + z)$; see the proof above (with the necessary mapping of variables $a, b, c \rightarrow x, y, z$). Now, let $\omega$ be the cube root of unity, so that all the cube roots of $1$ are $1, \omega, \omega^2$ (where $1$ is the only real root, while $\omega, \omega^2$ are the complex roots of unity). Recall some properties of these cube roots:
Note that the determinant in (2) has $(x + y + z)$ as a factor, as we saw earlier.
So proving (2) boils down to proving that the other two expressions on the RHS
are also factors of the LHS.
Thus, for instance, we want to prove that
the complex expression
$(x + y\omega + z\omega^2)$ is also a factor of the above determinantal polynomial. To do this, multiply the
second column of the matrix by $\omega$, the third column by $\omega^2$ and (to balance it out) divide by
$\omega \times \omega^2 = \omega^3 = 1$. We get:
$
\mathrm{det}\begin{pmatrix}x&y&z\\y&z&x\\z&x&y\end{pmatrix} = \mathrm{det}\begin{pmatrix}x&y\omega&z\omega^2\\y&z\omega&x\omega^2\\z&x\omega&y\omega^2\end{pmatrix}$.
Again, using $\omega^3 = 1$ in the RHS above, and now multiplying the second row by $\omega$
and the third row by $\omega^2$, we see
$\mathrm{det}\begin{pmatrix}x&y\omega&z\omega^2\\y&z\omega&x\omega^2\\z&x\omega&y\omega^2\end{pmatrix} =
\mathrm{det}\begin{pmatrix}x&y\omega&z\omega^2\\y\omega&z\omega^2&x\\z\omega^2&x&y\omega\end{pmatrix}$.
Now sum up the columns to see that $(x + y\omega + z\omega^2)$ is also a factor of the polynomial.
Analogously, we may also see that $(x + y\omega^2 + z\omega)$ is also a factor. Also, it is easy to check that
these factors are all distinct. We can thereby conclude that the product of all three of these
factors must divide the determinant. Now, checking the signs of $x^3$ on both sides (or set $y = z = 0, x = 1$,
and evaluate the expressions on both sides), we arrive at the
conclusion that:
$
\mathrm{det}\begin{pmatrix}x&y&z\\y&z&x\\z&x&y\end{pmatrix} = -(x + y + z) (x + y\omega + z\omega^2)
(x + y\omega^2 + z\omega)$.
Now stagger the terms of the RHS above as follows: $[(x + y + z)(x + y\omega + z\omega^2)][(x + y\omega + z\omega^2)(x + y\omega^2 + z\omega)] [(x + y\omega^2 + z\omega)(x + y + z)]$.
Consider the first square bracket: $[(x + y + z)(x + y\omega + z\omega^2)]$. Expanding this out and simplifying using that $(1 + \omega + \omega^2) = 0$, we get
Thus we have that
$
{\mathrm{det}\begin{pmatrix}x&y&z\\y&z&x\\z&x&y\end{pmatrix}}^2 $
$= [(x^2 - yz) + (y^2 - zx)\omega + (z^2 - xy)\omega^2][(x^2 - yz) + (y^2 - zx)\omega^2 + (z^2 - xy)\omega]
[(x^2 - yz) + (y^2 - zx) + (z^2 - xy)]$
Using identity (4) above
$=\mathrm{det}\begin{pmatrix}yz - x^2&zx - y^2&xy - z^2\\
zx - y^2&xy - z^2&yz - x^2\\xy - z^2&yz - x^2&zx - y^2\end{pmatrix}$
thereby proving (1) or (1').
QED.
Thus, the square of a number of the form $x^3 + y^3 + z^3 - 3 xyz$ is also of the same form!