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2.8: Bases as Coordinate Systems

  • Page ID
    77007
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    Objectives
    1. Learn to view a basis as a coordinate system on a subspace.
    2. Recipes: compute the \(\mathcal{B}\)-coordinates of a vector, compute the usual coordinates of a vector from its \(\mathcal{B}\)-coordinates.
    3. Picture: the \(\mathcal{B}\)-coordinates of a vector using its location on a nonstandard coordinate grid.
    4. 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.

    Fact \(\PageIndex{1}\)

    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 \(c_1=c_1',\: c_2=c_2',\cdots ,c_m=c_m'\).

    Example \(\PageIndex{1}\)

    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.

    Definition \(\PageIndex{1}\): \(\mathcal{B}\)-Coordinate Vector

    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 \((3,\:5,\:−2)\).”

    Note \(\PageIndex{1}\)

    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 \(\mathbb{R}^2\).

    Example \(\PageIndex{2}\): A Nonstandard Coordinate System of \(\mathbb{R}^2\))

    Define

    \[v_1=\left(\begin{array}{c}1\\1\end{array}\right),\quad v_2=\left(\begin{array}{c}1\\-1\end{array}\right),\quad\mathcal{B}=\{v_1,\:v_2\}.\nonumber\]

    1. Verify that \(\mathcal{B}\) is a basis for \(\mathbb{R}^2\).
    2. If \([w]_{\mathcal{B}}=\left(\begin{array}{c}1\\2\end{array}\right)\), then what is \(w\)?
    3. Find the \(\mathcal{B}\)-coordinates of \(v=\left(\begin{array}{c}5\\3\end{array}\right)\).
    Solution
    1. 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.
    2. 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:
      \[w=v_1+2v_2=\left(\begin{array}{c}1\\1\end{array}\right)+2\left(\begin{array}{c}1\\-1\end{array}\right)=\left(\begin{array}{c}3\\-1\end{array}\right).\nonumber\]
    3. 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 \(\left(\begin{array}{c}1\\2\end{array}\right)\).

    clipboard_e7d0e441efc56f2c8b4008721824087c9.png

    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 \(w\).

    clipboard_e7f6968a79281ff91feff7ba761a12132.png

    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 \(w\).

    Example \(\PageIndex{3}\)

    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”:

    clipboard_e79e06e9ea7c6dffdd50518031d9ed280.png

    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

    \[ [\color{blue}{u_2}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}-1 \\ \frac{1}{2}\end{array}\right)\quad [\color{Green}{u_3}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}\frac{3}{2} \\ -\frac{1}{2}\end{array}\right)\quad [\color{orange}{u_4}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}0\\ \frac{3}{2}\end{array}\right).\nonumber\]

    clipboard_e7ceecf7663ca781c87ad23a2684f5893.png

    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.

    Example \(\PageIndex{4}\): A Coordinate System on a Plane

    Define

    \[v_1=\left(\begin{array}{c}1\\0\\1\end{array}\right),\: v_2=\left(\begin{array}{c}1\\1\\1\end{array}\right),\quad\mathcal{B}=\{v_1,\: v_2\},\quad V=\text{Span}\{v_1,\: v_2\}.\nonumber\]

    1. Verify that \(\mathcal{B}\) is a basis for \(V\).
    2. If \([w]_{\mathcal{B}}=\left(\begin{array}{c}5\\2\end{array}\right)\), then what is \(w\)?
    3. Find the \(\mathcal{B}\)-coordinates of \(v=\left(\begin{array}{c}5\\3\\5\end{array}\right)\).
    Solution
    1. 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.
    2. 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:
      \[w=5v_1 +2v_2=5\left(\begin{array}{c}1\\0\\1\end{array}\right)+2\left(\begin{array}{c}1\\1\\1\end{array}\right)=\left(\begin{array}{c}7\\2\\7\end{array}\right).\nonumber\]
    3. 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 \([v]_{\mathcal{B}}=\left(\begin{array}{c}2\\3\end{array}\right)\).

    clipboard_e6b7190f6e5e9134b366052a8081d3794.png

    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 \(v\).

    Example \(\PageIndex{5}\): A Coordinate System on Another Plane

    Define

    \[v_1=\left(\begin{array}{c}2\\3\\2\end{array}\right),\quad v_2=\left(\begin{array}{c}-1\\1\\1\end{array}\right),\quad v_3=\left(\begin{array}{c}2\\8\\6\end{array}\right),\quad V=\text{Span}\{v_1,\: v_2,\: v_3\}.\nonumber\]

    1. Find a basis \(\mathcal{B}\) for \(V\).
    2. Find the \(\mathcal{B}\)-coordinates of \(x=\left(\begin{array}{c}4\\11\\8\end{array}\right).\)
    Solution
    1. 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 \(V\).
    2. 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 \([x]_{\mathcal{B}}=\left(\begin{array}{c}3\\2\end{array}\right)\).

    clipboard_ea14ac9d528da68c68c56a56cfa711fdb.png

    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 \(x\).

    Recipes: \(\mathcal{B}\)-Coordinates

    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

    \[\left(\begin{array}{cccc|c} |&|&\quad&|&| \\ v_1&v_2&\quad &v_m&x \\ |&|&\quad &|&|\end{array}\right).\nonumber\]

    Remark

    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 \(V\).


    2.8: Bases as Coordinate Systems is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by LibreTexts.

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