The matrix with entries $\pm 1$ is surjective?

Question

Consider matrix $A$ of size $n\times 2^n$ such that each column is a vector in $\{-1, +1\}^n$. It is clear that there are exactly $2^n$ distinct columns.

I have the impression that the linear map corresponding to $A$ is surjective? Indeed, it is clear in 2D, 3D. It is correct in higher dimension?

To prove the surjection, I know that we can use: definition of surjective function, rank–nullity theorem, or computing rank. But I could not find any solution.

Can you show that $(1, 0, 0, \dotsc)$ lies in the image of $A$? — Izaak van Dongen, Mar 25 '22 at 10:03
In order to do that, you simply need to find $n$ linearly independent columns in $A$. For example, you can take vectors whose first $k$ entries are $+1$ and last $n-k$ entries are $-1$ (here $k=\overline{1,n}$). (Why are they linearly independent?) — richrow, Mar 25 '22 at 10:05
You can take $(1, v)-(-1, v)=(2, 0_{2^n-1})$ is a $2$-scaling of your vector. Here $v\in {\pm 1}^{2^n-1} $. — Leonard Neon, Mar 25 '22 at 10:07
@richrow, Indeed, your sub-matrix defines a bijective map. I can prove it by using Gaussian elimination. Is there a simpler method? — Leonard Neon, Mar 25 '22 at 10:13
Just to clarify : Does the matrix have all possible $2^n$ column-vectors that can be made of the elements $-1$ and $1$ ? — Peter, Mar 25 '22 at 10:19
@Wuestenfux, I think that Hadamard matrix is square, here $A$ is rectangular matrix. — Leonard Neon, Mar 25 '22 at 10:25

Ben Grossmann · Accepted Answer · 2022-03-25T13:23:40.067

Two quick solutions:

Solution 1: Let $e_1,\dots,e_n$ denote the standard basis of $\Bbb R^n$ (i.e. the columns of the identity matrix). Let $v$ denote the column vector $v = (1,1,\dots,1)$. Each vector $e_i$ can be expressed as a linear combination of the columns of $A$ in the following way: $$ e_i = \frac 12 v - \frac 12 (v - 2e_i). $$ Thus, each vector $e_i$ lies in the column space of $A$. Since the vectors $e_1,\dots,e_n$ form a basis, we can conclude that the associated linear map is surjective.*

Solution 2: The case of $n=2$ can be handled separately**. For $n \geq 2$, note that $A$ contains as a submatrix $$ M = \pmatrix{-1&1&\cdots & 1\\ 1&-1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & -1}. $$ With the techniques outlined here, it's easy to see that the determinant of $M$ will be $$ \det(M) = (-2)^{n-1}\cdot(n-2) \neq 0. $$ Thus, $M$ in invertible, which means it has linearly independent columns. So, the columns of $A$ (which contain the columns of $M$) span $\Bbb R^n$.

* (Note 1): Concretely, the $n=2$ case can be handled by considering the invertible submatrix $M = \pmatrix{1&1\\1&-1}$.

** (Note 2): Solution 1 can be reframed as finding a $2^n \times n$ matrix $B$ such that $AB = I_{n \times n}$, i.e. a "right-inverse" to $A$.

Thank you so much for your help! It is indeed elegant! – Leonard Neon Mar 25 '22 at 18:47 — Leonard Neon, Mar 25 '22 at 18:47

The matrix with entries $\pm 1$ is surjective?

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