2.8: Bases as Coordinate Systems
- Page ID
- 77007
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Learn to view a basis as a coordinate system on a subspace.
- Recipes: compute the \(\mathcal{B}\)-coordinates of a vector, compute the usual coordinates of a vector from its \(\mathcal{B}\)-coordinates.
- Picture: the \(\mathcal{B}\)-coordinates of a vector using its location on a nonstandard coordinate grid.
- Vocabulary word: \(\mathcal{B}\)-coordinates.
In this section, we interpret a basis of a subspace \(V\) as a coordinate system on \(V\), and we learn how to write a vector in \(V\) in that coordinate system.
If \(\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}\) is a basis for a subspace \(V\), then any vector \(x\) in \(V\) can be written as a linear combination
\[x=c_1v_1+c_2v_2+\cdots +c_mv_m\nonumber\]
in exactly one way.
- Proof
-
Recall that to say \(\mathcal{B}\) is a basis for \(V\) means that \(\mathcal{B}\) spans \(V\) and \(\mathcal{B}\) is linearly independent. Since \(\mathcal{B}\) spans \(V\), we can write any \(x\) in \(V\) as a linear combination of \(v_1,\: v_2,\cdots ,v_m\). For uniqueness, suppose that we had two such expressions:
\[\begin{aligned}x&=c_1v_1+c_2v_2+\cdots +c_mv_m \\ x&=c_1'v_1+c_2'v_2+\cdots +c_m'v_m.\end{aligned}\]
Subtracting the first equation from the second yields
\[0=x-x=(c_1-c_1')v_1 +(c_2-c_2')v_2+\cdots +(c_m-c_m')v_m.\nonumber\]
Since \(\mathcal{B}\) is linearly independent, the only solution to the above equation is the trivial solution: all the coefficients must be zero. It follows that \(c_i−c_i'\) for all \(i\), which proves that \(c_1=c_1',\: c_2=c_2',\cdots ,c_m=c_m'\).
Consider the standard bases of \(\mathbb{R}^{3}\) from Example 2.7.3 in Section 2.7.
\[e_1=\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad e_2=\left(\begin{array}{c}0\\1\\0\end{array}\right),\quad e_3=\left(\begin{array}{c}0\\0\\1\end{array}\right).\nonumber\]
According to the above fact, Fact \(\PageIndex{1}\), every vector in \(\mathbb{R}^{3}\) can be written as a linear combination of \(e_1\), \(e_2\), \(e_3\), with unique coefficients. For example,
\[v=\left(\begin{array}{c}\color{Red}{3}\\ \color{blue}{5} \\ \color{Green}{-2}\end{array}\right)\color{black}{=}\color{red}{3}\color{black}{\left(\begin{array}{c}1\\0\\0\end{array}\right)+}\color{blue}{5}\color{black}{\left(\begin{array}{c}0\\1\\0\end{array}\right)}\color{Green}{-2}\color{black}{\left(\begin{array}{c}0\\0\\1\end{array}\right)=}\color{Red}{3}\color{black}{e_1+}\color{blue}{5}\color{black}{e_2}\color{Green}{-2}\color{black}{e_3.}\nonumber\]
In this case, the coordinates of \(v\) are exactly the coefficients of \(e_1\), \(e_2\), \(e_3\).
What exactly are coordinates, anyway? One way to think of coordinates is that they give directions for how to get to a certain point from the origin. In the above example, the linear combination \(3e_1+5e_2−2e_3\) can be thought of as the following list of instructions: start at the origin, travel \(3\) units north, then travel \(5\) units east, then \(2\) units down.
Let \(\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}\) be a basis of a subspace \(V\), and let
\[x=c_1v_1+c_2v_2+\cdots+c_mv_m\nonumber\]
be a vector in \(V\). The coefficients \(c_1,\:c_2,\cdots,c_m\) are the coordinates of \(x\) with respect to \(\mathcal{B}\). The \(\mathcal{B}\)-coordinate vector of \(x\) is the vector
\[[x]_{\mathcal{B}}=\left(\begin{array}{c}c_1 \\c_2\\ \vdots \\ c_m\end{array}\right)\quad\text{ in }\mathbb{R}^{m}.\nonumber\]
If we change the basis, then we can still give instructions for how to get to the point \((3,\:5,\:−2)\), but the instructions will be different. Say for example we take the basis
\[v_1=e_1+e_2=\left(\begin{array}{c}1\\1\\0\end{array}\right),\quad v_2=e_2=\left(\begin{array}{c}0\\1\\0\end{array}\right),\quad v_3=e_3=\left(\begin{array}{c}0\\0\\1\end{array}\right).\nonumber\]
We can write \((3,\:5,\:−2)\) in this basis as \(3v_1+2v_2−2v_3\). In other words: start at the origin, travel northeast \(3\) times as far as \(v_1\), then \(2\) units east, then \(2\) units down. In this situation, we can say that “\(3\) is the \(v_1\)-coordinate of \((3,\:5,\:−2)\), \(2\) is the \(v_2\)-coordinate of \((3,\:5,\:−2)\), and \(−2\) is the \(v_3\)-coordinate of \((3,\:5,\:−2)\).”
The above definition, Definition \(\PageIndex{1}\), gives a way of using \(\mathbb{R}_m\) to label the points of a subspace of dimension \(m\): a point is simply labeled by its \(\mathcal{B}\)-coordinate vector. For instance, if we choose a basis for a plane, we can label the points of that plane with the points of \(\mathbb{R}^2\).
Define
\[v_1=\left(\begin{array}{c}1\\1\end{array}\right),\quad v_2=\left(\begin{array}{c}1\\-1\end{array}\right),\quad\mathcal{B}=\{v_1,\:v_2\}.\nonumber\]
- Verify that \(\mathcal{B}\) is a basis for \(\mathbb{R}^2\).
- If \([w]_{\mathcal{B}}=\left(\begin{array}{c}1\\2\end{array}\right)\), then what is \(w\)?
- Find the \(\mathcal{B}\)-coordinates of \(v=\left(\begin{array}{c}5\\3\end{array}\right)\).
Solution
- By the basis theorem in Section 2.7, Theorem 2.7.3, any two linearly independent vectors form a basis for \(\mathbb{R}^2\). Clearly \(v_1,\:v_2\) are not multiples of each other, so they are linearly independent.
- To say \([w]_{\mathcal{B}}=\left(\begin{array}{c}1\\2\end{array}\right)\) means that \(1\) is the \(v_1\)-coordinate of \(w\), and that \(2\) is the \(v_2\)-coordinate:
\[w=v_1+2v_2=\left(\begin{array}{c}1\\1\end{array}\right)+2\left(\begin{array}{c}1\\-1\end{array}\right)=\left(\begin{array}{c}3\\-1\end{array}\right).\nonumber\] - We have to solve the vector equation \(v=c_1v_1+c_2v_2\) in the unknowns \(c_1,\: c_2\). We form an augmented matrix and row reduce:
\[\left(\begin{array}{cc|c} 1&1&5 \\ 1&-1&3\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad\left(\begin{array}{cc|c} 1&0&4 \\ 0&1&1\end{array}\right).\nonumber\]
We have \(c_1=4\) and \(c_2=1\), so \(v=4v_1+v_2\) and \([v]_{\mathcal{B}}=\left(\begin{array}{c}4\\1\end{array}\right)\).
In the following picture, we indicate the coordinate system defined by \(\mathcal{B}\) by drawing lines parallel to the “\(v_1\)-axis” and “\(v_2\)-axis”. Using this grid it is easy to see that the \(\mathcal{B}\)-coordinates of \(v\) are \(\left(\begin{array}{c}5\\1\end{array}\right)\), and that the \(\mathcal{B}\)-coordinates of \(w\) are \(\left(\begin{array}{c}1\\2\end{array}\right)\).
Figure \(\PageIndex{1}\)
This picture could be the grid of streets in Palo Alto, California. Residents of Palo Alto refer to northwest as “north” and to northeast as “east”. There is a reason for this: the old road to San Francisco is called El Camino Real, and that road runs from the southeast to the northwest in Palo Alto. So when a Palo Alto resident says “go south two blocks and east one block”, they are giving directions from the origin to the Whole Foods at \(w\).
Figure \(\PageIndex{2}\): A picture of the basis \(\mathcal{B}=\{v_1,\: v_2\}\) of \(\mathbb{R}^2\). The grid indicates the coordinate system defined by the basis \(\mathcal{B}\); one set of lines measures the \(v_1\)-coordinate, and the other set measures the \(v_2\)-coordinate. Use the sliders to find the \(\mathcal{B}\)-coordinates of \(w\).
Let
\[v_1=\left(\begin{array}{c}2\\-1\\1\end{array}\right)\quad v_2=\left(\begin{array}{c}1\\0\\-1\end{array}\right).\nonumber\]
These form a basis \(\mathcal{B}\) for a plane \(V=\text{Span}\{v_1,\: v_2\}\) in \(\mathbb{R}^3\). We indicate the coordinate system defined by \(\mathcal{B}\) by drawing lines parallel to the “\(v_1\)-axis” and “\(v_2\)-axis”:
Figure \(\PageIndex{3}\)
We can see from the picture that the \(v_1\)-coordinate of \(\color{red}{u_1}\) is equal to \(1\), as is the \(v_2\)-coordinate, so \([\color{red}{u_1}\color{black}{]_{\mathcal{B}}=\left(\begin{array}{c}1\\1\end{array}\right)}\). Similarly, we have
\[ [\color{blue}{u_2}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}-1 \\ \frac{1}{2}\end{array}\right)\quad [\color{Green}{u_3}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}\frac{3}{2} \\ -\frac{1}{2}\end{array}\right)\quad [\color{orange}{u_4}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}0\\ \frac{3}{2}\end{array}\right).\nonumber\]
Figure \(\PageIndex{4}\): Left: the \(\mathcal{B}\)-coordinates of a vector \(x\). Right: the vector \(x\). The violet grid on the right is a picture of the coordinate system defined by the basis \(\mathcal{B}\); one set of lines measures the \(v_1\)-coordinate, and the other set measures the \(v_2\)-coordinate. Drag the heads of the vectors \(x\) and \([x]_{\mathcal{B}}\) to understand the correspondence between \(x\) and its \(\mathcal{B}\)-coordinate vector.
Define
\[v_1=\left(\begin{array}{c}1\\0\\1\end{array}\right),\: v_2=\left(\begin{array}{c}1\\1\\1\end{array}\right),\quad\mathcal{B}=\{v_1,\: v_2\},\quad V=\text{Span}\{v_1,\: v_2\}.\nonumber\]
- Verify that \(\mathcal{B}\) is a basis for \(V\).
- If \([w]_{\mathcal{B}}=\left(\begin{array}{c}5\\2\end{array}\right)\), then what is \(w\)?
- Find the \(\mathcal{B}\)-coordinates of \(v=\left(\begin{array}{c}5\\3\\5\end{array}\right)\).
Solution
- We need to verify that \(\mathcal{B}\) spans \(V\), and that it is linearly independent. By definition, \(V\) is the span of \(\mathcal{B}\); since \(v_1\) and \(v_2\) are not multiples of each other, they are linearly independent. This shows in particular that \(V\) is a plane.
- To say \([w]_{\mathcal{B}}=\left(\begin{array}{c}5\\2\end{array}\right)\) means that \(5\) is the \(v_1\)-coordinate of \(w\), and that \(2\) is the \(v_2\)-coordinate:
\[w=5v_1 +2v_2=5\left(\begin{array}{c}1\\0\\1\end{array}\right)+2\left(\begin{array}{c}1\\1\\1\end{array}\right)=\left(\begin{array}{c}7\\2\\7\end{array}\right).\nonumber\] - We have to solve the vector equation \(v=c_1v_1+c_2v_2\) in the unknowns \(c_1,\: c_2\). We form an augmented matrix and row reduce:
\[\left(\begin{array}{cc|c} 1&1&5\\0&1&3\\1&1&5\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{cc|c} 1&0&2\\0&1&3\\0&0&0\end{array}\right).\nonumber\]
We have \(c_1=2\) and \(c_2=3\), so \(v=2v_1+3v_2\) and \([v]_{\mathcal{B}}=\left(\begin{array}{c}2\\3\end{array}\right)\).
Figure \(\PageIndex{5}\): A picture of the plane \(V\) and the basis \(\mathcal{B}=\{v_1,\: v_2\}\). The violet grid is a picture of the coordinate system defined by the basis \(\mathcal{B}\); one set of lines measures the \(v_1\)-coordinate, and the other set measures the \(v_2\)-coordinate. Use the sliders to find the \(\mathcal{B}\)-coordinates of \(v\).
Define
\[v_1=\left(\begin{array}{c}2\\3\\2\end{array}\right),\quad v_2=\left(\begin{array}{c}-1\\1\\1\end{array}\right),\quad v_3=\left(\begin{array}{c}2\\8\\6\end{array}\right),\quad V=\text{Span}\{v_1,\: v_2,\: v_3\}.\nonumber\]
- Find a basis \(\mathcal{B}\) for \(V\).
- Find the \(\mathcal{B}\)-coordinates of \(x=\left(\begin{array}{c}4\\11\\8\end{array}\right).\)
Solution
- We write \(V\) as the column space of a matrix \(A\), then row reduce to find the pivot columns, as in Example 2.7.6, in Section 2.7.
\[A=\left(\begin{array}{ccc}2&-1&2\\3&1&8\\2&1&6\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{ccc}1&0&2\\0&1&2\\0&0&0\end{array}\right).\nonumber\]
The first two columns are pivot columns, so we can take \(\mathcal{B}=\{v_1,\:v_2\}\) as our basis for \(V\). - We have to solve the vector equation \(x=c_1v_1+c_2v_2\). We form an augmented matrix and row reduce:
\[\left(\begin{array}{cc|c} 2&-1&4 \\ 3&1&11\\2&1&8\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{cc|c} 1&0&3\\0&1&2\\0&0&0\end{array}\right).\nonumber\]
We have \(c_1=3\) and \(c_2=2\), so \(x=3v_1+2v_2\), and thus \([x]_{\mathcal{B}}=\left(\begin{array}{c}3\\2\end{array}\right)\).
Figure \(\PageIndex{6}\): A picture of the plane \(V\) and the basis \(\mathcal{B}=\{v_1,\:v_2\}\). The violet grid is a picture of the coordinate system defined by the basis \(\mathcal{B}\); one set of lines measures the \(v_1\)-coordinate, and the other set measures the \(v_2\)-coordinate. Use the sliders to find the \(\mathcal{B}\)-coordinates of \(x\).
If \(\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}\) is a basis for a subspace \(V\) and \(x\) is in \(V\), then
\[ [x]_{\mathcal{B}}=\left(\begin{array}{c}c_1 \\ c_2 \\ \vdots \\ c_m\end{array}\right)\quad\text{means}\quad x=c_1v_1+c_2v_2+\cdots c_mv_m.\nonumber\]
Finding the \(\mathcal{B}\)-coordinates of \(x\) means solving the vector equation
\[x=c_1v_1+c_2v_2+\cdots +c_mv_m\nonumber\]
in the unknowns \(c_1,\: c_2,\cdots ,c_m\). This generally means row reducing the augmented matrix
\[\left(\begin{array}{cccc|c} |&|&\quad&|&| \\ v_1&v_2&\quad &v_m&x \\ |&|&\quad &|&|\end{array}\right).\nonumber\]
Let \(\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}\) be a basis of a subspace \(V\). Finding the \(\mathcal{B}\)-coordinates of a vector \(x\) means solving the vector equation
\[x=c_1v_1+c_2v_2+\cdots +c_mv_m.\nonumber\]
If \(x\) is not in \(V\), then this equation has no solution, as \(x\) is not in \(V=\text{Span}\{v_1,\: v_2,\cdots ,v_m\}\). In other words, the above equation is inconsistent when \(x\) is not in \(V\).