2.8 Bases as Coordinate Systems¶

The standard basis vectors are the columns of the \(n\times n\) identity matrices which are generally referred to as \(I_n\).

Identity matrix and standard basis for \(\mathbb R^2\):¶

I2 = eye(2)

We refer to the standard basis vectors as \(\vec e_1\) and \(\vec e_2\).

e1 = I2(:,1)

e2 = I2(:,2)

Identity matrix and standard basis for \(\mathbb R^3\)¶

I3 = eye(3)

Notice that we still use the vectors \(\vec e_1\) and \(\vec e_2\) even though we are referring to 3-component vectors now.

e1 = I3(:,1)

e2 = I3(:,2)

e3 = I3(:,3)

Alternate basis \(\mathcal B\)¶

Any set of linearly independent vectors \(\{\vec v_1,\vec v_2, \dots , \vec v_m \}\) may serve as a basis for the subset it spans. For example, suppose that the vectors \(\{\vec v_1,\vec v_2, \vec v_3 \}\) all live in \(\mathbb R^3\) and that they are linearly independent. Then, we can use

\[\mathcal B = \{\vec v_1,\vec v_2, \vec v_3 \}\]

as the basis for \(\mathbb R^3\). To see why, let’s create a linearly independent set.

Example: Alternate basis for \(\mathbb R^3\)¶

Suppose that we have the following vectors:

\[\begin{split}\vec v_1 =\left[\begin{array}{r}-1\\1\\0\\\end{array}\right],\vec v_2=\left[\begin{array}{r}1\\9\\5\\\end{array}\right],\vec v_3=\left[\begin{array}{r}-1\\-1\\-2\\\end{array}\right] \hspace{1cm}\text{and} \hspace{1cm}\vec w=\left[\begin{array}{r}-3\\5\\2\\\end{array}\right]\end{split}\]

Verifying \(\{\vec v_1,\vec v_2, \vec v_3 \}\) is a basis for \(\mathbb R^3\)¶

Let’s create a matrix whose columns are the candidates for the basis vectors, then row-reduce.

B = [-1 1 -1 ; 1 9 -1 ; 0 5 -2 ]

B =

    -1     1    -1
     1     9    -1
     0     5    -2

rref(B)

Because we have 3 pivots, we know the \(\text{Span}\{\vec v_1,\vec v_2, \vec v_3 \}\) is a 3-dimensional subspace of \(\mathbb R^3\), and the only 3D subspace of \(\mathbb R^3\) is \(\mathbb R^3\) itself.

Finding the \(\mathcal B\)-coordinates of a point¶

We want to determine how to write the vector \(\vec w\) as a linear combination of the basis vectors. The coordinates \(c_1, c_2, c_3\) will be the coefficients of the linear dependence relation when we solve the vector equations

\[c_1\vec v_1+c_2\vec v_2+c_3 \vec v_3=\vec w\]

which is equivalent to the matrix equations

\[B\vec c = \vec w\]

where the matrix \(B\) has as its columns the vectors from the basis \(\mathcal B\). To do so, we row reduce the augmented matrix.

B = [-1 1 -1 ; 1 9 -1 ; 0 5 -2 ]
w = [-3 ; 5 ; 2]

B =

    -1     1    -1
     1     9    -1
     0     5    -2


w =

    -3
     5
     2

We augment the matrix and row reduce \([B,\vec w]\).

A = [ B, w ]

A =

    -1     1    -1    -3
     1     9    -1     5
     0     5    -2     2

rref(A)

ans =

   0     0     4
   1     0     0
   0     1    -1

The linear dependence relation tells use the linear combination of the basis vectors that will produce \(\vec w\).

\[\begin{split}-4\vec v_1 +(0)\vec v_2 +\vec v_3 = w \hspace{1cm}\implies \hspace{1cm} \vec c = \left[\begin{array}{r}-4\\0\\1\\\end{array}\right]\end{split}\]

We can verify using the columns of matrix \(B\).

b1 = B(:,1)
b3 = B(:,3)
-4 * b1 + b3

The \(\mathcal B\)-coordinates of the vector \(\vec w\) are the coefficients of the linear dependence relation. This means we have

\[\begin{split}\vec w_\mathcal B = \left[\begin{array}{r}-4\\0\\1\\\end{array}\right]_\mathcal B\end{split}\]

where the subscript indicates the corrdinates used to represent the vector. To convert the vector back to standard coordinates, we simply multiply the matrix \(B\) by the coordinates from the linear dependence relation.

B * [ 4 ; 0 ; -1]

Linear Algebra with MATLAB