# $x$ and $x^2$ linearly dependent?

Consider the functions $$x$$ and $$x^2$$ on $$\mathbb{R}$$. Clearly, they are linearly independent.
But consider the following argument.

Consider the matrix $$A = \begin{bmatrix} x & x^2\\ 0 & 0\\ \end{bmatrix}$$

Clearly, the determinant is zero. This implies the existence of a nonzero matrix, $$B =\begin{bmatrix} a\\ b\\ \end{bmatrix}$$ such that $$AB=0$$.

This implies that $$ax+bx^2=0$$ for some nonzero $$a$$ or some nonzero $$b$$. But this implies that $$x$$ and $$x^2$$ are linearly dependent.

Clearly, false.

Where’s the flaw?

• To explain it in basic terms (without needing ring theory): You have proven that $$\forall x \in \mathbb{R} : \exists a, b \in \mathbb{R}_0 : ax + bx^2 = 0$$ What you actually need to prove to conclude linear dependence is $$\exists a, b \in \mathbb{R}_0 : \forall x \in \mathbb{R} : ax + bx^2 = 0$$ – Zeno Aug 10 '19 at 9:57

They are linearly independent as functions of variable $$x$$. Yes, you are right that given a specific value of $$x$$ you can find non-zero $$a,b\in\mathbb{R}$$ such that $$ax+bx^2=0$$. But you will never find non-zero $$a,b$$ which will work for all $$x\in\mathbb{R}$$.
• But can you please show how we can not conclude anything for the functions $x$ and $x^2$ from this? – Atom Aug 10 '19 at 9:23
• Look, your argument would be correct if $A$ was a matrix of real numbers. In that case if the determinant was zero then indeed the system $Ax=0$ would have a non-trivial solution over $\mathbb{R}$. But it's not true that if the determinant of a matrix of functions is zero then its columns are linearly dependent over $\mathbb{R}$. Real functions belong to a different ring. – Mark Aug 10 '19 at 9:35
When discussing linear independence of the columns of a matrix, you must allow the coefficients in the linear combinations to come from the same space as the entries in the matrix (otherwise linear algebra doesn't in any way work the way you're used to). And clearly, if we allow $$a$$ and $$b$$ to be polynomials, then $$ax+bx^2=0$$ has nontrivial solutions.
$$x$$ and $$x^2$$ are considered to be elements of the vectorial space $$\mathbb{R}[X]$$ or $$\mathbb{R}[X]_2$$ endowed with the basis $$1,x,x^2$$. Then \begin{align*} x=&0\times 1 + 1\times x+ 0\times x^2\\ x=&0\times 1 + 0\times x+ 1\times x^2 \end{align*} ,hence $$\begin{pmatrix} 0 \\ 1\\0\end{pmatrix}$$, $$\begin{pmatrix} 0 \\ 0\\1\end{pmatrix}$$ are coordinates of $$x$$ and $$x^2$$ respectively. Clearly, this two vectors are linearly independents.