For \(\IR^3\text{,}\) these are the vectors \(\vec e_1=\hat\imath=\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right],
\vec e_2=\hat\jmath=\left[\begin{array}{c}0 \\ 1 \\ 0\end{array}\right],\) and \(\vec e_3=\hat k=\left[\begin{array}{c}0 \\ 0 \\ 1\end{array}\right]
\text{.}\)
Observation2.6.2.
A basis may be thought of as a collection of building blocks for a vector space, since every vector in the space can be expressed as a unique linear combination of basis vectors.
For example, in many calculus courses, vectors in \(\IR^3\) are often expressed in their component form
\begin{equation*}
3\vec e_1-2\vec e_2+4\vec e_3 = 3\hat\imath-2\hat\jmath+4\hat k
.
\end{equation*}
Since every vector in \(\IR^3\) can be uniquely described as a linear combination of the vectors in \(\setList{\vec e_1,\vec e_2,\vec e_3}\text{,}\) this set is indeed a basis.
Activity2.6.3.
Label each of the sets \(A,B,C,D,E\) as
SPANS \(\IR^4\) or DOES NOT SPAN \(\IR^4\)
LINEARLY INDEPENDENT or LINEARLY DEPENDENT
BASIS FOR \(\IR^4\) or NOT A BASIS FOR \(\IR^4\)
by finding \(\RREF\) for their corresponding matrices.
If \(\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}\) is a basis for \(\IR^4\text{,}\) that means \(\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]\) doesn't have a non-pivot column, and doesn't have a row of zeros. What is \(\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]\text{?}\)
The set \(\{\vec v_1,\dots,\vec v_m\}\) is a basis for \(\IR^n\) if and only if \(m=n\) and \(\RREF[\vec v_1\,\dots\,\vec v_n]=
\left[\begin{array}{cccc}
1&0&\dots&0\\
0&1&\dots&0\\
\vdots&\vdots&\ddots&\vdots\\
0&0&\dots&1
\end{array}\right]
\text{.}\)
That is, a basis for \(\IR^n\) must have exactly \(n\) vectors and its square matrix must row-reduce to the so-called identity matrix containing all zeros except for a downward diagonal of ones. (We will learn where the identity matrix gets its name in a later module.)
Subsection2.6.1Videos
Figure2.6.6.Video: Verifying that a set of vectors is a basis of a vector space