Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Coordinates

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Coordinates
Coordinates Relative to the Standard Basis
Wize Concept
Recall that every real vector space Rn\mathbb{R}^nRn has a standard basis SSS.
For example, in R2\mathbb{R}^2R2 the standard basis is S={[10],[01]}S=\left\{
\begin{bmatrix}
  1\\ 0\\
\end{bmatrix}
,
\begin{bmatrix}
  0\\ 1\\
\end{bmatrix}
\right\}S={[10​],[01​]}.

The vector v⃗=[35]\vec v = 
\begin{bmatrix}
3\\ 5\\
\end{bmatrix}
v=[35​] is a linear combination of the standard basis vectors.
The coefficients are called the coordinates of v⃗\vec vv relative to the basis SSS:
v⃗=[35]=3[10]+5[01]\vec v = 
\begin{bmatrix}
  \colorOne 3\\ \colorTwo 5\\
\end{bmatrix}
=
\colorOne 3
\begin{bmatrix}
  1\\ 0\\
\end{bmatrix}
+
\colorTwo 5
\begin{bmatrix}
  0\\ 1\\
\end{bmatrix}v=[35​]=3[10​]+5[01​]

We can use [  ]S[ \ \   ]_S[  ]S​ to write that this vector's coordinates are relative to the basis SSS : [v⃗]S=[35][\vec v]_S = 
\begin{bmatrix}
  \colorOne 3\\ \colorTwo 5\\
\end{bmatrix}
[v]S​=[35​]
In other words, the coordinates of v⃗\vec vv are the coefficients from the linear combination of the basis vectors that create v⃗\vec vv.

Wize Tip
We use the standard basis in Rn\reals^nRn by default, so we do not need to use the bracket notation [  ]S[ \ \   ]_S[  ]S​ in this case.

PAGE BREAK
Coordinates Relative to a General Basis
Suppose that B={v⃗1,v⃗2,…,v⃗n}B=\{\vec v_1,\vec v_2,\dots,\vec v_n\}B={v1​,v2​,…,vn​} is a basis for an nnn-dimensional vector space VVV. Let v⃗\vec vv be a vector in  VVV such that:
v⃗=c1v⃗1+c2v⃗2+⋯+cnv⃗n\vec v = \colorTwo{\mathbf{c_1}}\vec v_1
+
\colorTwo{\mathbf{c_2}}\vec v_2
+
\cdots
+
\colorTwo{\mathbf{c_n}}\vec v_nv=c1​v1​+c2​v2​+⋯+cn​vn​
Then the coordinates of v⃗\vec vv relative to the basis BBB are given by the coefficients:
[v⃗]B=[c1c2⋮cn][\vec v]_B = 
\begin{bmatrix}
\colorTwo{\mathbf{c_1}}\\ 
\colorTwo{\mathbf{c_2}}\\ 
\vdots\\
\colorTwo{\mathbf{c_n}}
\end{bmatrix}[v]B​=​c1​c2​⋮cn​​​
Example
The set B={[31],[1−1]}B=\left\{
\begin{bmatrix}
  3\\ 1\\
\end{bmatrix}
,
\begin{bmatrix}
  1\\ -1\\
\end{bmatrix}
\right\}B={[31​],[1−1​]} is a basis for R2\mathbb{R}^2R2.

Let u⃗=[35]\vec u =
\begin{bmatrix}
  3\\ 5\\
\end{bmatrix}u=[35​], then we can find coefficients a,b\textbf{a}, \textbf{b}a,b such that
[35]=a[31]+b[1−1]\begin{bmatrix}
  3\\ 5\\
\end{bmatrix}
=
\textbf{a}
\begin{bmatrix}
  3\\ 1\\
\end{bmatrix}
+
\textbf{b}
\begin{bmatrix}
  1\\ -1\\
\end{bmatrix}
[35​]=a[31​]+b[1−1​]

This is just a 2×22 \times 22×2 system of linear equations! Try solving and you will find that:

u⃗=[35]=2[31]+(−3)[1−1]\vec u=
\begin{bmatrix}
  3\\ 5\\
\end{bmatrix}

=\colorTwo{\mathbf{2}}
\begin{bmatrix}
  3\\ 1\\
\end{bmatrix}
+
(\colorTwo{\mathbf{-3}})
\begin{bmatrix}
  1\\ -1\\
\end{bmatrix}u=[35​]=2[31​]+(−3)[1−1​]

Therefore, the coordinate vector of u⃗\vec uu relative to the basis BBB is:

[u⃗]B=[2−3][\vec u]_B=
\begin{bmatrix}
  \colorTwo{\mathbf{2}}\\ 
  \colorTwo{\mathbf{-3}}\\
\end{bmatrix}[u]B​=[2−3​]

PAGE BREAK
Geometric Interpretation
A basis for a vector space defines a coordinate system.
In the standard basis for R2\mathbb{R}^2R2, the vectors [10]\begin{bmatrix}
  1\\ 0\\
\end{bmatrix}[10​] and [01]\begin{bmatrix}
  0\\ 1\\
\end{bmatrix}[01​] define the standard coordinate system with the xxx and yyy axes.
The basis vectors [31]\begin{bmatrix}
  3\\ 1\\
\end{bmatrix}[31​] and [1−1]\begin{bmatrix}
  1\\ -1\\
\end{bmatrix}[1−1​] define a different coordinate system with axes along these vectors.
These two different coordinate systems give us different ways of representing the same vector u⃗\vec uu:

          

Example: Coordinates
It can be shown that the set B={1+x2,  x+x2,  1+2x+x2}B=\{1+x^2,\ \ x+x^2,\ \ 1+2x+x^2\}B={1+x2,  x+x2,  1+2x+x2} is a basis for the polynomial vector space P2P_2P2​.
Find the coordinate vector of p(x)=1+4x+7x2p(x)=1+4x+7x^2p(x)=1+4x+7x2 relative to the basis BBB.

Wize Tip
The vector space P2P_2P2​ is said to be isomorphic to R3\reals^3R3, meaning they are essentially "equivalent".
Every polynomial p(x)=c+bx+ax2p(x)=c+bx+ax^2p(x)=c+bx+ax2 in P2P_2P2​ can be written as the vector [cba]\begin{bmatrix}
  c\\ b\\ a\\
\end{bmatrix}
​cba​​ in R3\mathbb{R}^3R3.
This lets us transform the problem into an equivalent one in R3\mathbb{R}^3R3.

Let's include the 0 coefficients in the basis BBB: 
B={1+0x+1x2,  0+1x+1x2,  1+2x+1x2}B=\{1+0x+1x^2,\ \ 0+1x+1x^2,\ \ 1+2x+1x^2\} B={1+0x+1x2,  0+1x+1x2,  1+2x+1x2}
We can then transform this into the following basis for R3\mathbb{R}^3R3:
B={[101],[011],[121]}B=\left\{
\begin{bmatrix}
  1\\ 0\\ 1\\
\end{bmatrix},
\begin{bmatrix}
  0\\ 1\\ 1\\
\end{bmatrix},
\begin{bmatrix}
  1\\ 2\\ 1\\
\end{bmatrix}
\right\}B=⎩⎨⎧​​101​​,​011​​,​121​​⎭⎬⎫​
Our given polynomial can be written as the following vector in R3\mathbb{R}^3R3:
p⃗=[147]\colorOne{
\vec p = 
\begin{bmatrix}
  1\\ 4\\ 7\\
\end{bmatrix}}
p​=​147​​
Let's find the coordinates of p⃗\vec pp​. These are the coefficients from the linear combination of the basis vectors of BBB that create p⃗\vec pp​:
[p⃗]B=[c1c2c3][\vec p]_B = 
\begin{bmatrix}
\colorTwo{\mathbf{c_1}}\\
\colorTwo{\mathbf{c_2}}\\
\colorTwo{\mathbf{c_3}}
\end{bmatrix}[p​]B​=​c1​c2​c3​​​ such that c1[101]+c2[011]+c3[121]=[147]\colorTwo{\bm c_1}
\begin{bmatrix}
  1\\ 0\\ 1\\
\end{bmatrix}
+
\colorTwo{\bm c_2}
\begin{bmatrix}
  0\\ 1\\ 1\\
\end{bmatrix}
+
\colorTwo{\bm c_3}
\begin{bmatrix}
  1\\ 2\\ 1\\
\end{bmatrix}
=
\colorOne{
\begin{bmatrix}
  1\\ 4\\ 7\\
\end{bmatrix}
}c1​​101​​+c2​​011​​+c3​​121​​=​147​​
This is equivalent to solving the system:
[101101241117]→RREF[10020106001−1]\left[\begin{array}{rrr|r}
1&0&1&\colorOne 1\\
0&1&2&\colorOne 4\\
1&1&1&\colorOne 7
\end{array}\right]

\quad
\xrightarrow{RREF}
\quad

\left[\begin{array}{rrr|r}
1&0&0&2\\
0&1&0&6\\
0&0&1&-1
\end{array}\right]​101​011​121​147​​RREF​​100​010​001​26−1​​
Therefore, [p⃗]B=[26−1][\vec p]_B = 
\begin{bmatrix}
\colorTwo{\mathbf{2}}\\
\colorTwo{\mathbf{6}}\\
\colorTwo{\mathbf{-1}}
\end{bmatrix}[p​]B​=​26−1​​.
We can check this:
2(1+x2)+6(x+x2)+(−1)(1+2x+x2)=2+2x2+6x+6x2−1−2x−x2=1+4x+7x2=p(x)✓\begin{array}{rcl}
&&\colorTwo{\mathbf{2}}(1+x^2) + \colorTwo{\mathbf{6}}(x+x^2) + (\colorTwo{\mathbf{-1}})(1+2x+x^2)\\[0.5em]

&=& 2+2x^2+6x+6x^2-1-2x-x^2\\[0.5em]

&=& 1+4x+7x^2\\[0.5em]
&=& p(x) \quad \colorThree{\checkmark}
\end{array}​===​2(1+x2)+6(x+x2)+(−1)(1+2x+x2)2+2x2+6x+6x2−1−2x−x21+4x+7x2p(x)✓​

Practice: Coordinates
Let B1={[−120],[3−52],[4−73]}B_1=
\left\{
\begin{bmatrix}
-1\\ 2 \\0\\
\end{bmatrix}
,
\begin{bmatrix}
3\\ -5\\ 2\\
\end{bmatrix}
,
\begin{bmatrix}
4\\ -7\\ 3\\
\end{bmatrix}
\right\}B1​=⎩⎨⎧​​−120​​,​3−52​​,​4−73​​⎭⎬⎫​ and B2={[2−31],[11−1],[−1−30]}B_2=
\left\{
\begin{bmatrix}
2\\ -3 \\1\\
\end{bmatrix}
,
\begin{bmatrix}
1\\ 1\\ -1\\
\end{bmatrix}
,
\begin{bmatrix}
-1\\ -3\\ 0\\
\end{bmatrix}
\right\}B2​=⎩⎨⎧​​2−31​​,​11−1​​,​−1−30​​⎭⎬⎫​ be two bases of R3\mathbb{R}^3R3.
Suppose [x⃗]B1=[−48−7][\vec x]_{\small B_1} = 
\begin{bmatrix}
  -4\\ 8\\ -7\\
\end{bmatrix}[x]B1​​=​−48−7​​.

What is x⃗\vec xx  (the coordinates relative to the standard basis)?

A) [01−5]\begin{bmatrix}
0\\ 1\\ -5\\
\end{bmatrix}​01−5​​

B) [105]\begin{bmatrix}
1\\ 0\\ 5\\
\end{bmatrix}​105​​

C) [11−1]\begin{bmatrix}
1\\ 1\\ -1\\
\end{bmatrix}​11−1​​

D) [015]\begin{bmatrix}
0\\ 1\\ 5\\
\end{bmatrix}​015​​

I don't know

Practice: Coordinates
Find the coordinates of the polynomial p(x)=5+x−3x2−3x3p(x)=5+x-3x^2-3x^3p(x)=5+x−3x2−3x3 relative to the following basis of P3P_3P3​:
B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}B = \{(1+x),\ \ (x+x^2),\ \ (x^2+x^3),\ \ (1-x^3)\}B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}

1st coordinate:  
2nd coordinate: 
3rd coordinate:  
4th coordinate:  

I don't know

Extra Practice

Coordinates

Practice: Coordinates
Find the coordinates of the polynomial p(x)=5+x−3x2−3x3p(x)=5+x-3x^2-3x^3p(x)=5+x−3x2−3x3 relative to the following basis of P3P_3P3​:
B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}B = \{(1+x),\ \ (x+x^2),\ \ (x^2+x^3),\ \ (1-x^3)\}B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}

Practice: Coordinates

Find the coordinates of the polynomial p(x)=5+x−3x2−3x3p(x)=5+x-3x^2-3x^3p(x)=5+x−3x2−3x3 relative to the following basis of P3P_3P3​:
{(1+x),(x+x2),(x2+x3),(1−x3)}\{(1+x),(x+x^2),(x^2+x^3),(1-x^3)\}{(1+x),(x+x2),(x2+x3),(1−x3)}

Verify if the following vectors form a basis for R3\mathbb{R}^3R3:
{[112],[1−21],[2−11]}\left\{
\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix},
\begin{bmatrix}
1 \\
-2 \\
1
\end{bmatrix},\begin{bmatrix}
2 \\
-1 \\
1
\end{bmatrix}
\right\}⎩⎨⎧​​112​​,​1−21​​,​2−11​​⎭⎬⎫​

19.4F_Mid_Builder_$\tkcth{8.5.}\tkco{9}$__$\tkco{Mock 3}$

Taken together,
 v⃗1=[11]\vv_1 = \colvec{1}{1}v1​=[11​] and v⃗2=[1−1]\vv_2 = \colvec{1}{-1}v2​=[1−1​] 
constitute a basis forR2\mathbb{R}^2R2. Determine the coordinates [x⃗]v=[xv1xv2][ \vx ]_{v} = \colvec{\bcth{x_{v_1}}}{\bct{x_{v_2}}}[x]v​=[xv1​​xv2​​​]of the vector x⃗=[21]\vx = \colvec{2}{1}x=[21​] relative to this basis.  

Concept Clarifier: Finding Coordinate Vectors

The following is a basis for R2\mathbb{R}^2R2 . 
{(11),(−20)}\left \{ \left ( \begin{array}{c}
1 \\ 1 
\end{array} \right ), \left ( \begin{array}{c}
-2 \\0 
\end{array} \right ) \right \}{(11​),(−20​)}
If v⃗=(2,3)\vec{v} = (2, 3)v=(2,3) is a coordinate vector in this basis, what is v⃗\vec{v}v in coordinates with respect to the basis

19.4F_Mid_Builder_$\tkcth{8.4.16}$_$\tkcth{mock 1 ✓}$

Find the coordinates (c1, c2)(c_1,\, c_2)(c1​,c2​) for the point P=(8,11)P = (8,11)P=(8,11) in terms of the nonstandard basis vectors e⃗1=<1, 1>\ve{1} = \rowvec{1}{1}e1​=⟨1,1⟩ and e⃗2=<6, 9>\ve{2} = \rowvec{6}{9}e2​=⟨6,9⟩. 

Coordinates

Find the coordinates of the polynomial p(x)=2−x+x2p(x)=2-x+x^2p(x)=2−x+x2 relative to the basis {1+x,1+x2,x+x2}\{1+x,1+x^2,x+x^2\}{1+x,1+x2,x+x2} of P2P_2P2​

19.4F_133_8.2_Mock_F1_$\tkco{eg6}$_$\key{Final}$_Builder_$\tkcth{8.2.}\tkcf{6}$_

The set {[21]}\left\{ \colvec{2}{1} \right\}{[21​]} is a basis for the subspace TTT, where T⊆R2T \subseteq \mathbb{R}^2T⊆R2.
Give the position vector r⃗P\vr_{_{P}}rP​​ (in the standard basis) corresponding to the point in TTT whose coordinate is given by PT=(3)P_{T} = (3)PT​=(3).
If r⃗=[xy]\vr = \colvec{\bco{x}}{\bct{y}}r=[xy​], then 

Coordinates

Practice: Coordinates
Find the coordinates of the polynomial p(x)=5+x−3x2−3x3p(x)=5+x-3x^2-3x^3p(x)=5+x−3x2−3x3 relative to the following basis of P3P_3P3​:
B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}B = \{(1+x),\ \ (x+x^2),\ \ (x^2+x^3),\ \ (1-x^3)\}B={(1+x),  (x+x2),  (x2+x3),  (1−x3)}

Coordinates

Practice: Coordinates
Let B1={[−120],[3−52],[4−73]}B_1=
\left\{
\begin{bmatrix}
-1\\ 2 \\0\\
\end{bmatrix}
,
\begin{bmatrix}
3\\ -5\\ 2\\
\end{bmatrix}
,
\begin{bmatrix}
4\\ -7\\ 3\\
\end{bmatrix}
\right\}B1​=⎩⎨⎧​​−120​​,​3−52​​,​4−73​​⎭⎬⎫​ and B2={[2−31],[11−1],[−1−30]}B_2=
\left\{
\begin{bmatrix}
2\\ -3 \\1\\
\end{bmatrix}
,
\begin{bmatrix}
1\\ 1\\ -1\\
\end{bmatrix}
,
\begin{bmatrix}
-1\\ -3\\ 0\\
\end{bmatrix}
\right\}B2​=⎩⎨⎧​​2−31​​,​11−1​​,​−1−30​​⎭⎬⎫​ be two bases of R3\mathbb{R}^3R3.
Suppose [x⃗]B1=[−48−7][\vec x]_{\small B_1} = 
\begin{bmatrix}
  -4\\ 8\\ -7\\
\end{bmatrix}[x]B1​​=​−48−7​​.

Example: Change of Basis

Example:  Change of Basis
Suppose we have 2 ordered bases: A=((1,−1),(−2,1)),B=((1,1),(1,2))A=((1,-1),(-2,1)),B=((1,1),(1,2))A=((1,−1),(−2,1)),B=((1,1),(1,2)) .
Compute the change of basis matrix CA→BC_{A\rightarrow B}CA→B​, and if[v⃗]A=(−3,−4)[\vec v]_A=(-3,-4)[v]A​=(−3,−4), then find [v⃗]B[\vec v]_B[v]B​ .

Practice Question: Change of Basis

Suppose we have an ordered basis B=((1,1,1),(2,0,−1),(−1,1,0))B=((1,1,1),(2,0,-1),(-1,1,0))B=((1,1,1),(2,0,−1),(−1,1,0)) .  Let S=((1,0,0),(0,1,0),(0,0,1))S=((1,0,0),(0,1,0),(0,0,1))S=((1,0,0),(0,1,0),(0,0,1)) be the standard ordered basis.  Compute the change of basis matrix CS→BC_{S\rightarrow B}CS→B​ , and use it to find (2,−1,2)B(2,-1,2)_B(2,−1,2)B​ 
(i.e. find [v⃗]B[\vec v]_B[v]B​ , if [v⃗]S=(2,−1,2)[\vec v]_S=(2,-1,2)[v]S​=(2,−1,2) ).

Let {1−x,x,1+x2,x−x3,x4}\{ 1 - x, x, 1 + x^2, x - x^3, x^4 \}{1−x,x,1+x2,x−x3,x4} be the basis for the vector space of polynomials with degree at most 4. What is the coordinate vector for vvv in this basis?
v=4−x2+x3−2x4v = 4 - x^2 + x^3 - 2x^4v=4−x2+x3−2x4

Consider the vector v=(1,2,3)v = (1, 2, 3)v=(1,2,3) with respect to the standard unit vector basis. What is the coordinate vector for vvv in the following basis:
{(110),(010),(−10−2)}\left \{ \left ( \begin{array}{c}
1 \\ 1\\ 0
\end{array} \right ), \left ( \begin{array}{c}
0 \\1 \\ 0
\end{array} \right ) , \left ( \begin{array}{c}
-1 \\ 0 \\ -2
\end{array} \right ) \right \}⎩⎨⎧​​110​​,​010​​,​−10−2​​⎭⎬⎫​

Wize University Linear Algebra Textbook > Vector Spaces and Subspaces

Coordinates

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Coordinates

Coordinates Relative to the Standard Basis

Coordinates Relative to a General Basis

Geometric Interpretation

Example: Coordinates

Practice: Coordinates

Practice: Coordinates

Extra Practice

Coordinates

Practice: Coordinates

19.4F_Mid_Builder_$\tkcth{8.5.}\tkco{9}$__$\tkco{Mock 3}$

Concept Clarifier: Finding Coordinate Vectors

19.4F_Mid_Builder_$\tkcth{8.4.16}$_$\tkcth{mock 1 ✓}$

Coordinates

19.4F_133_8.2_Mock_F1_$\tkco{eg6}$_$\key{Final}$_Builder_$\tkcth{8.2.}\tkcf{6}$_

Coordinates

Coordinates

Example: Change of Basis

Practice Question: Change of Basis