Basis of a Vector Space

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A basis of a vector space is a set of vectors $\mathbf{v}_1,\mathbf{v}_2, \dots$ with the properties


  • The canonical basis for $\mathbb{R}^3$ is

\begin{align} \{ \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix},\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \} \end{align}

  • There are other bases for $\mathbb{R}^3$ e.g.

\begin{align} \{ \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix},\begin{bmatrix} 2\\ 2\\ 5 \end{bmatrix},\begin{bmatrix} 3\\ 4\\ 8 \end{bmatrix} \} \end{align}

so the basis is not unique, they only have to satisfy the above conditions.

Given a space, the basis is not unique, but every basis has the same number of vectors, this number is called the dimension of the space. In $\mathbb{R}^3$ e.g. has three dimensions and there are three basis vectors. We have now a second way to look at the rank of a matrix. \begin{align} rank(\mathbf{A}) = \# \text{ of pivots}=dim C(\mathbf{A}) \end{align} The rank of the matrix $\mathbf{A}$ equals the dimension of the column space of a $C(\mathbf{A})$. The matrix \begin{align} \begin{bmatrix} 1 & 2 & 3 & 1 \\ 1& 1& 2 & 1 \\ 1& 2& 3 & 1 \end{bmatrix} \end{align} has $rank=2$. If we do Matrix Elimination, we find two pivots and the column space has dimension $dim C(\mathbf{A})=2$. The Nullspace of this matrix $\mathbf{A}$ is spanned by the two vectors. \begin{align} \begin{bmatrix} -1\\ -1\\ 1\\ 0 \end{bmatrix},\begin{bmatrix} -1\\ 0\\ 0\\ 1 \end{bmatrix} \end{align} They are a basis for the Nullspace. The dimension of the Nullspace of $\mathbf{A}$ is \begin{align} dim N(\mathbf{A})= n-r \end{align} the number of columns of $\mathbf{A}$ minus the rank of $\mathbf{A}$.

Video Lectures:

  • Gilbert Strang - Introduction to Linear Algebra Lec. 10