Standard and Nonstandard Bases of Rn

Coordinate Vectors and Bases

In \(\mathbb{R}^n\), the entries of a vector describe its position relative to the familiar coordinate directions. Those directions are represented by the standard coordinate vectors. They provide the simplest basis, but many other lists of vectors can also give every vector in \(\mathbb{R}^n\) through linear combinations. To check a proposed list, we will use the definition of a basis and, for lists of \(n\) vectors, row reduction.

All scalars in this tutorial are real. A list \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) in \(\mathbb{R}^n\) is a basis if it is linearly independent and spans \(\mathbb{R}^n\). Spanning means that every vector in \(\mathbb{R}^n\) is a linear combination of the list. Independence means that the equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) forces every coefficient \(c_i\) to be zero. A basis is an ordered list: changing the order leaves its span and independence unchanged, but changes the order in which its vectors are written.

For example, in \(\mathbb{R}^3\), the standard basis is \((\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\), where \(\mathbf{e}_1=(1,0,0)^{\top}\), \(\mathbf{e}_2=(0,1,0)^{\top}\), and \(\mathbf{e}_3=(0,0,1)^{\top}\). The entries of a vector are precisely the coefficients needed to combine these coordinate directions.

Proof. Let \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\in\mathbb{R}^n\). By the entries of the coordinate vectors, \(x_1\mathbf{e}_1+\cdots+x_n\mathbf{e}_n=(x_1,\dots,x_n)^{\top}=\mathbf{x}\), so the list spans \(\mathbb{R}^n\). Now suppose \(c_1\mathbf{e}_1+\cdots+c_n\mathbf{e}_n=\mathbf{0}\). The left side is the vector \((c_1,\dots,c_n)^{\top}\), so equality with the zero vector forces \(c_i=0\) for every \(i\). The list is therefore independent as well as spanning, and hence is a basis. \(\square\)

Testing a List of \(n\) Vectors

For a proposed list \((\mathbf{v}_1,\dots,\mathbf{v}_n)\) in \(\mathbb{R}^n\), put the vectors in the columns of a square matrix \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\). A coefficient vector \(\mathbf{c}=(c_1,\dots,c_n)^{\top}\) then satisfies \(\mathbf{A}\mathbf{c}=c_1\mathbf{v}_1+\cdots+c_n\mathbf{v}_n\). Thus, independence concerns the solutions of \(\mathbf{A}\mathbf{c}=\mathbf{0}\), while spanning concerns whether \(\mathbf{A}\mathbf{c}=\mathbf{b}\) can be solved for every \(\mathbf{b}\in\mathbb{R}^n\).

Proof. Suppose first that the columns form a basis. In particular, they are independent, so \(\mathbf{A}\mathbf{c}=\mathbf{0}\) has only the zero solution. Row operations preserve the solutions of a homogeneous system. If the reduced matrix had a column without a pivot, the variable in that column would be free; assigning it the value \(1\), assigning any other free variables \(0\), and solving for the pivot variables would give a nonzero solution. This contradicts independence. Therefore every column has a pivot. Conversely, suppose every column has a pivot. There are \(n\) columns and \(n\) rows, so the RREF is \(\mathbf{I}_n\). The row operations reducing \(\mathbf{A}\) to \(\mathbf{I}_n\) can also be applied to any right-hand side \(\mathbf{b}\); they produce a system \(\mathbf{I}_n\mathbf{c}=\mathbf{b}'\) for some transformed right-hand side \(\mathbf{b}'\). This system has a solution for every \(\mathbf{b}\), so the columns span \(\mathbb{R}^n\). For the homogeneous system, \(\mathbf{I}_n\mathbf{c}=\mathbf{0}\) forces \(\mathbf{c}=\mathbf{0}\), so the columns are independent. They are a basis. \(\square\)

The test gives a direct procedure: form the matrix with the proposed vectors as columns and row-reduce it. The column convention matters because the matrix-vector product then forms combinations of the proposed vectors. If the reduced matrix is the identity, the list is a basis. If it is not, the list is not a basis; examining the reduced system can also identify whether the problem is dependence, failure to span, or both.

When a List Is Not a Basis

A list can fail the basis definition in either of two ways: it can be dependent, so it contains redundancy, or it can fail to span, so some vectors cannot be built from it. For a square list in \(\mathbb{R}^n\), the pivot test detects failure of the basis condition. A concrete example makes both possible failures visible.

Common Pitfall: Entries Are Not Always Basis Coefficients

For the standard basis, a vector’s entries are its coefficients because each coordinate direction has a \(1\) in just one position. For a nonstandard basis, it is generally wrong to read the vector’s entries directly as the coefficients. In the preceding nonstandard-basis example, taking \((2,3,4)^{\top}\) itself as the coefficient vector would give the incorrect calculation

which is not \((2,3,4)^{\top}\). The correction is to solve \(\mathbf{A}\mathbf{c}=\mathbf{b}\), using the proposed basis vectors as the columns of \(\mathbf{A}\). Here that calculation gives \(\mathbf{c}=(\tfrac12,\tfrac52,\tfrac32)^{\top}\), and multiplying back verifies the answer.

When checking a proposed basis, keep the two conditions distinct. A dependence relation disproves independence; a vector outside the span disproves spanning. For a square list, the row-reduction test provides a systematic decision, while a direct multiplication or a dependence relation is a useful check on a particular conclusion.