nontrivial solution with all components nonzero

Question

I have the set of equations

$ \begin{bmatrix} 1 & -1 & -1\\ a & -b & -c\\ a^2 & -b^2 & -c^2\\ \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} =0 $

I think that, if all $x$, $y$ and $z$ are required to be non-zero, then $a = b = c$, but I don't know how to actually show this. I tried setting the determinant equal to zero, required for a non-trivial solution, but this is not sufficient as then (e.g.) $a=b$ and $z=0$ is a non-trivial solution. Is there a way to require that the all the components of a non-trivial solution be non-zero?

What I actually want to show is that, if for all $t$,

$ A \sin(\omega_1 t) - B\sin(\omega_2 t) = C \sin(\omega_3 t) $

implies that $\omega_1 = \omega_2 = \omega_3$ if $A$,$B$ and $C$ are non-zero (which I think is true). Evaluating the original equation and the first and second derivative gave the set of equations above.

Any insight into either problem would be greatly appreciated

Answer 1

Hint

Note first that if $\omega_1=\omega_2$ then $$A \sin(\omega_1 t) - B\sin(\omega_2 t) = C \sin(\omega_3 t)$$ becomes $$(A-B) \sin(\omega_1t)=C \sin(\omega_3 t)$$ from which is trivial (just derivate and combine the relations) to deduce that $\omega_3=\omega_1$.

Same way, if $\omega_1=\omega_3$ you get $$(A-C) \sin(\omega_1t)=B \sin(\omega_2 t)$$ while if $\omega_2=\omega_3$ you get $$A \sin(\omega_1t)=(B+C) \sin(\omega_2 t)$$

In all three situations you can conclude that, if two of them are equal all three are equal.

This leaves only the case when $\omega_1, \omega_2, \omega_3$ are pairwise distinct. In this case the vanDermonde determinant is non-zero, therefore your system only has the trivial solution.

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Note To make this clear, in the first problem one of the following two must happen.

• $\det\begin{bmatrix} 1 & -1 & -1\\ a & -b & -c\\ a^2 & -b^2 & -c^2\\ \end{bmatrix}\neq 0$. Then the matrix is invertible and, multiplying by the inverse you get, $x=y=z=0$.
• $\det\begin{bmatrix} 1 & -1 & -1\\ a & -b & -c\\ a^2 & -b^2 & -c^2\\ \end{bmatrix}= 0$. Then, $(a-b)(a-c)(b-c)=0$. In this case, the nly thing you can deduce is that two of you numbers must be equal.

You need to go back to the original problem, and deduce from there that the third number must be the same.

Answer 2

Let $x'=x,y'=-y,z'=-z$.Then $$\begin{bmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{bmatrix}\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.$$Let $M=\begin{bmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{bmatrix}.$ Then the determinant of $M$ is $(a-b)(b-c)(c-a)$. Multiply both sides of the matrix equation on the left by the adjoint matrix of $M.$ Thus$$(a-b)(b-c)(c-a)\begin{bmatrix}x'\\y'\\z'\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$.Thus$$(a-b)(b-c)(c-a)x'=0,(a-b)(b-c)(c-a)y'=0,(a-b)(b-c)(c-a)z'=0$$and so$$(a-b)(b-c)(c-a)x=0,(a-b)(b-c)(c-a)y=0,(a-b)(b-c)(c-a)z=0$$ Hope that helps!

Answer 3

Eliminating the minus signs from the coefficient matrix doesn’t change its determinant and turns it into a Vandermonde matrix, with determinant equal to $(a-b)(a-c)(c-b)$. The most you can say from that is that for the equation to have a nontrivial solution, at least two of the parameters must be equal, but that doesn’t get you to $a=b=c$. However, let’s examine the null space of the coefficient matrix.

Assume that $a\ne b$. Row-reducing the matrix produces $$\begin{bmatrix}1&0&{b-c\over a-b}\\ 0&1&{a-c\over a-b} \\ 0&0&(a-c)(c-b)\end{bmatrix}.$$ The matrix is rank-deficient iff $a=c$ or $b=c$. We can see from the last column of the reduced matrix that in either case null vectors of the matrix must have a zero component, so $a=b$ is a necessary condition for a solution with no zero components. By symmetry, we must also have $a=c$. The null space of the coefficient matrix is then the orthogonal complement of $(1,-1,-1)^T$, which is spanned by $(1,1,0)^T$ and $(1,0,1)^T$ and clearly has elements with no zero entries, so $a=b=c$ is also sufficient.

You could instead solve the system of equations directly. If $z$ must be nonzero, then w.l.o.g. we can set $z=1$. From the first row, we then have $y=x-1$. Substituting this into the other two equations, we obtain $$(a-b)x+(b-c)=0 \\ (a^2-b^2)x+b^2-c^2=0.$$ Substituting for $b-c$ in the last equation gives $(a-b)(a-c)x=0$, so either $a=b$ or $a=c$. Setting $a=b$ reduces the other equation to $b-c=0$. On the other hand, setting $a=c$ transforms the equation into $(b-c)(1-x)=0$, but $1-x=-y\ne0$, so once again $b-c=0$.