Basis and Dimension

Polynomials. $P_n$, of degree at most $n$, has basis $\{1,x,x^2,\dots ,x^n\}$ and dimension $n+1$. Note the off-by-one: $P_2$ is three-dimensional, because the constant term is a coordinate like any other.

A different basis of $P_2$ is $\{1,1+x,1+x+x^2\}$. To find the coordinates of $2+3x+4x^2$ in it, solve

Matching coefficients from the top: $c_3=4$ from $x^2$; then $c_2+c_3=3$ gives $c_2=-1$; then $c_1+c_2+c_3=2$ gives $c_1=-1$. So the coordinates are $(-1,-1,4)$.

Checking: $-1-(1+x)+4(1+x+x^2)=(-1-1+4)+(-1+4)x+4x^2=2+3x+4x^2$.

The same polynomial has coordinates $(2,3,4)$ in the standard basis and $(-1,-1,4)$ in this one.

Matrices. $M_{2\times 2}(\mathbb{R})$ has basis $\{E_{11},E_{12},E_{21},E_{22}\}$, the matrices with a single 1, and dimension 4.

The symmetric $2\times 2$ matrices form a subspace with basis

of dimension 3. The constraint $a_{12}=a_{21}$ removes exactly one degree of freedom from four. In that basis has coordinates $(3,5,-2)$, reading the diagonal entries and then the shared off-diagonal one.

Solutions of a differential equation. The real solutions of $y^{″}+y=0$ form a vector space with basis $\{\sin ,\cos \}$ and dimension 2. This is why a general solution is written $A\sin x+B\cos x$ with two arbitrary constants: the count of constants is the dimension, and a second-order linear equation has a two-dimensional solution space.

Functions, in general. The real-valued functions on $[0,1]$ have no finite basis. Neither do all polynomials, where $\{1,x,x^2,\dots \}$ is independent and infinite. Dimension as a number is a finite-dimensional notion, and most function spaces of analytical interest fall outside it.

The common thread. In every finite case the dimension counts degrees of freedom: coefficients you may choose independently. Three for a quadratic, four for a $2\times 2$ matrix, three when symmetry ties two entries together, two for the arbitrary constants of a second-order equation.

Taking the reduced columns as the basis. After reducing $A$ to find pivots in columns 1 and 3, the learner reports the reduced matrix's columns $(1,0,0)$ and $(0,1,0)$. Row operations change the columns; they preserve only which columns depend on which. The answer must be the original vectors in the pivot positions, and the reduced columns generally do not lie in the span at all.

Concluding a basis from independence alone. "These two vectors in ${\mathbb{R}}^3$ are independent, so they are a basis of ${\mathbb{R}}^3$." Independence caps the count; it does not reach it. Two independent vectors in a three-dimensional space span a plane, and the shortcut applies only when the number of vectors equals the known dimension.

Using the shortcut with an assumed dimension. "$W$ looks two-dimensional, and here are two independent vectors in it, so they are a basis." The count-plus-one-condition rule needs the dimension established beforehand. If $W$ were three-dimensional the two vectors would span only a proper subspace, and nothing in the argument would have detected it.

Treating coordinates as intrinsic. "The vector is $(3,2)$." A coordinate list means nothing without its basis: $(5,1)$ in the standard basis and $(3,2)$ in $\{(1,1),(1,-1)\}$ name the same vector. Reporting coordinates without naming the basis, or comparing coordinates computed in different bases, produces statements that cannot be checked.

What unites them. The first is a procedural slip with a conceptual cause: not knowing what row reduction preserves. The middle two mistake a necessary condition for a sufficient one. The last forgets that coordinates are a description relative to a choice, which is the same observation that makes change of basis a subject at all.

Any linearly independent set of vectors in a space is a basis for it, since independence is the condition that distinguishes a basis from an arbitrary collection.