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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?
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.
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.
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!
