Basis and Dimension

The linear span of (or the space spanned by) the vectors $\mathbf{x}{1}, \ldots, \mathbf{x}{\mathbf{m}}$ is defined to be the set of all linear combinations $c_{1} \mathbf{x}{\mathbf{1}}+\cdots+c{m} \mathbf{x}{\mathbf{m}}$ where $c{1}, \ldots, c_{m}$ are real numbers. The linear span is a subspace; this follows from the definition.

A set of vectors $\mathbf{x}{1}, \ldots, \mathbf{x}{\mathbf{m}}$ is said to be linearly dependent if there exist real numbers $c_{1}, \ldots, c_{m}$ such that at least one $c_{i}$ is nonzero and $c_{1} \mathbf{x}{\mathbf{1}}+\cdots+c{m} \mathbf{x}{\mathbf{m}}=\mathbf{0}$ A set is linearly independent if it is not linearly dependent. Strictly speaking, we should refer to a collection (or a multiset) of vectors rather than a set of vectors in the two preceding definitions. Thus when we talk of vectors $\mathbf{x}{1}, \ldots, \mathbf{x}_{\mathbf{m}}$ being linearly dependent or independent, we allow for the possibility of the vectors not necessarily being distinct.
The following statements are easily proved:
(i) The set consisting of the zero vector alone is linearly dependent.
(ii) If $X \subset Y$ and if $X$ is linearly dependent, then so is $Y$.
(iii) If $X \subset Y$ and if $Y$ is linearly independent, then so is $X$.
A set of vectors is said to form a basis for the vector space $S$ if it is linearly independent and its linear span equals $S$.

Let $\mathbf{e}{\mathbf{i}}$ be the $i$ th column of the $n \times n$ identity matrix. The set $\mathbf{e}{1}, \ldots, \mathbf{e}_{\mathbf{n}}$ forms a basis for $R^{n}$, called the standard basis.

If $\mathbf{x}{1}, \ldots, \mathbf{x}{\mathbf{m}}$ is a basis for $S$, then any vector $\mathbf{x}$ in $S$ admits a unique representation as a linear combination $c_{1} \mathbf{x}{\mathbf{1}}+\cdots+c{m} \mathbf{x}{\mathbf{m}}$. For if $$ \mathbf{x}=c{1} \mathbf{x}{\mathbf{1}}+\cdots+c{m} \mathbf{x}{\mathbf{m}}=d{1} \mathbf{x}{\mathbf{1}}+\cdots+d{m} \mathbf{x}{\mathbf{m}} $$ then $$ \left(c{1}-d_{1}\right) \mathbf{x}{\mathbf{1}}+\cdots+\left(c{m}-d_{m}\right) \mathbf{x}{\mathbf{m}}=\mathbf{0} $$ and since $\mathbf{x}{\mathbf{1}}, \ldots, \mathbf{x}{\mathbf{m}}$ are linearly independent, $c{i}=d_{i}$ for each $i$.
A vector space is said to be finite-dimensional if it has a basis consisting of finitely many vectors. The vector space containing only the zero vector is also finite-dimensional. We will consider only finite-dimensional vector spaces. Very often it will be implicitly assumed that the vector spaces under consideration are nontrivial, i.e., contain vectors other than the zero vector.

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