Let $p_1, p_2,...,p_n$ be real polynomials of $k$ variables $x_1,...,x_k$ and assume that $$p_1^2 + \dots +p_n^2=x_1^2 + \dots + x_k^2$$

Prove that $n \geq k$.

Out of so many questions that I have attempted, this is the first question that I have absolutely no idea how to start (I am ashamed of myself). Can anyone give some hint?

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    $\begingroup$ Just a note - for the equality case, the polynomials don't have to be $p_i=x_i$. They can for example be made up of pairs like $\frac{1}{4}(x_1+x_2)$, and $\frac{1}{4}(x_1-x_2)$ (at least for $k$ even). $\endgroup$ – Peter Woolfitt Dec 15 '15 at 7:12
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    $\begingroup$ From which contest is this from ? $\endgroup$ – Ewan Delanoy Dec 15 '15 at 8:11
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    $\begingroup$ Don't be ashamed of yourself! Every single person encounters problems they don't know how to start on. We just keep learning: last year's impossible problems are this year's warm-ups. $\endgroup$ – Greg Martin Dec 18 '15 at 0:34

Let's see. To begin with, all polynomials must be linear (otherwise the square of the highest-degree term of the highest-degree polynomial has degree more than 2 and no chances to cancel out with anything).

The rest is simple: for a linear polynomial, $p(\vec x)=0$ is a hyperplane which has a normal vector. With $n<k$, $n$ vectors do not form a complete basis in a $k$-dimensional space, so there is a non-zero vector perpendicular to all of them (that is, belonging to all hyperplanes where $p_i(\vec x)=0$). So for this vector $p_1(\vec x)=\dots=p_n(\vec x)=0$, hence $p_1^2 + \dots +p_n^2=0$, but $x_1^2 + \dots + x_k^2>0$.

With $n=k$ the solutions (besides the obvious $p_i=x_i$) are plenty; just take any orthonormal basis. For example, $({2\over3}x-{1\over3}y+{2\over3}z,\; {2\over3}x+{2\over3}y-{1\over3}z,\; -{1\over3}x+{2\over3}y+{2\over3}z)$ would do in 3D.

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  • $\begingroup$ How do you prove the first assertion that a linear polynomial is a hyperplane with a normal vector? $\endgroup$ – Sandeep Silwal Dec 17 '15 at 21:58
  • $\begingroup$ Why, that's pretty well known: Wiki. $\endgroup$ – Ivan Neretin Dec 17 '15 at 22:34

As Ivan Neretin's answer points out, all polynomials must be strictly linear. That is, each polynomial could be written matricially as $p_j = {\bf a}_j' {\bf x}$, where ${\bf a}_j$ and ${\bf x}$ are $k \times 1$ matrices (polynomial coefficients and variables respectively).

Then $$p_j^2 = ({\bf a}_j' {\bf x})^2 = {\bf x}' A_j {\bf x}$$ where $A_j = {\bf a}_j{\bf a}_j'$ is a $k\times k$ (rank-one, positive definite) matrix.

Then [*], because $\sum_{i=1}^k x_i^2= {\bf x}' {\bf x}$ we want

$$ \sum_{j=1}^n {\bf x}' A_j {\bf x}= {\bf x}' A {\bf x} = {\bf x}' {\bf x} $$ where $A = \sum_{j=1}^n A_j $ (symmetric, positive definite). This implies $A=I_k$.

But we need to sum at least $k$ matrices of rank one to obtain a rank $k$ matrix. Hence $n\ge k$

[*] As said in the comments, an equivalent alternative way here is:

We must have ${\bf x}' A {\bf x}=\sum_{i=1}^k x_i^2>0$ for any ${\bf x}\ne {\bf 0}$. This implies $A {\bf x} \ne {\bf 0}$ (for ${\bf x}\ne {\bf 0}$), hence $A$ must have full rank ($k$). Then, the final paragraph above applies.

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  • $\begingroup$ The fact that ${\bf x}' A {\bf x} = {\bf x}'{\bf x}$ for all $\bf x$ does not necessarily imply that $A=I_k$. But it does imply that $A$ must have full rank, which is what you need. $\endgroup$ – Joey Zou Dec 18 '15 at 1:01
  • $\begingroup$ @JoeyZou: Why does it imply that $A$ has full rank? We choosing $e_k$ for $x$, we can get that the diagonal entries of $A$ must all be $1's$. Does it follow from this? $\endgroup$ – Sandeep Silwal Dec 18 '15 at 1:47
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    $\begingroup$ @SandeepSilwal if $A$ did not have full rank, then there exists ${\bf x}\ne 0$ such that $A{\bf x} = 0$, so ${\bf x}' A {\bf x} = 0$. But ${\bf x}'{\bf x}\ne 0$ if ${\bf x}\ne 0$. $\endgroup$ – Joey Zou Dec 18 '15 at 1:56
  • $\begingroup$ @JoeyZou : $A$ is symmetric. And ${\bf x}' A {\bf x} = {\bf x}'{\bf x} \implies {\bf x}' B {\bf x} = 0$ with $B=A-I$ symmetric. This implies $B$ is zero. $\endgroup$ – leonbloy Dec 18 '15 at 2:07
  • $\begingroup$ Ah right, $A$ is symmetric because the $A_j$'s are symmetric by construction. $\endgroup$ – Joey Zou Dec 18 '15 at 2:12

Having written everything out, I now realize that this merely a paraphrasing of leonbloy's answer, using explicit indexed sums and filling out the details, but figure I might as well post it at this point.

As has already been pointed out, each polynomial must be strictly linear. That is, for each $q$, we may write $$p_q(\mathbf{x})=\sum_{i=1}^k \alpha_q^ix_i,$$ so that $\alpha_q^i$ is the scalar of $x_i$ in the $q^{\text{th}}$ polynomial. But then $$\sum_{q=1}^n(p_q(\textbf{x}))^2=\left(\sum_{i=1}^k \left[\sum_{q=1}^n (\alpha_q^i)^2\right](x_i)^2\right)+\left(\sum_{i=1}^k \sum_{j=1}^k \left[\sum_{q=1}^n 2\alpha_q^i\alpha_q^j\right]x_ix_j\right).$$

Thus, if these polynomials $p_1,p_2,\ldots,p_n$ satisfy the property that $\sum_{q=1}^n(p_q(\textbf{x}))^2=\sum_{i=1}^k (x_i)^2$, then necessarily $$\sum_{q=1}^n (\alpha_q^i)^2=1$$ for all $i$, while $$\sum_{q=1}^n 2\alpha_q^i\alpha_q^j=0$$ for all $i\neq j$. Thinking of these $\alpha_i^q$ as indexed vectors in $n-$dimensional euclidean space, we see that this boils down to the statement: "you can't have $q$ orthogonal unit vectors in $\mathbb{R}^n$ unless $n\geq q$", which follows from the basis theorem of linear algebra. I hope that this makes sense. Please let me know if anything here is unclear at all. Great question!

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