Mathematics Prelims

June 13, 2009

Change of Basis

Filed under: Algebra,Linear Algebra — cjohnson @ 4:19 pm

In any non-trivial vector space there will be several possible bases we could pick.  In $\mathbb{R}^3$, for instance, we could use $\{ [1, 0, 0]^T, [0, 1, 0]^T, [0, 0, 1]^T \}$ or $\{ [1, 2, 0]^T, [3, 0, 1]^T, [4, 2, -1]^T \}$.  This first basis is known as the standard basis, and in general for an $n$-dimensional vector space over $\mathcal{F}$, we’ll refer to $\{ [1, 0, 0, ..., 0]^T, [0, 1, 0, ..., 0]^T, ..., [0, ..., 0, 1]^T \}$ as the standard basis, and will let $e_i$ denote the vector with a 1 in the $i$-th position, and zeros elsewhere.

When we write down a vector like $[1, 2, 3]^T$ we implicitly mean that these are the coordinates to use with the vectors in the standard basis; these are the coefficients we multiply the $e_i$ basis vectors by to get describe our vector.

$\displaystyle \left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right] = 1 \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right]+ 2 \left[ \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right] + 3 \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]$

But how would we find the appropriate coordinates if we were to use $\{ [1, 2, 0]^T, [3, 0, 1]^T, [4, 2, -1]^T \}$ as our basis?  Let’s suppose our coefficients are $\alpha_1, \alpha_2, \alpha_3$.  Then what we want to do is find the values of the $\alpha_i$ such that

$\displaystyle \alpha_1 \left[ \begin{array}{c} 1 \\ 2 \\ 0 \end{array} \right] + \alpha_2 \left[\begin{array}{c}3 \\ 0 \\ 1\end{array}\right] + \alpha_3\left[\begin{array}{c}4 \\ 2 \\ -1\end{array}\right] = \left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right]$

So what we have is a system of equations:

$\displaystyle \left[ \begin{array}{ccc} 1 & 3 & 4 \\ 2 & 0 & 2 \\ 0 & 1 & -1 \end{array} \right] \left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array} \right]= \left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right]$

Notice that since the columns of this matrix form a basis for $\mathbb{R}^3$, the matrix is invertible, and so we can easily solve for the $\alpha_i$.

$\displaystyle \left[ \begin{array}{c} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{array} \right] = \left[ \begin{array}{ccc} 1 & 3 & 4 \\ 2 & 0 & 2 \\ 0 & 1 & -1 \end{array} \right]^{-1} \left[ \begin{array}{c} 1 \\ 2 \\ 3 \end{array} \right] = \left[ \begin{array}{c} 5/2 \\ 3/2 \\ -3/2\end{array} \right]$

In general, if we have a basis $A = \{ a_1, ..., a_n \}$, we will write

$\displaystyle \left[ \begin{array}{c} \alpha_1 \\ \vdots \\ \alpha_n \end{array} \right]_A$

to mean that the $\alpha_i$ are the coefficients of the $a_i$ vectors in our $A$ basis, and will let $_AI$ be the matrix converts the coordinates for the standard basis to coordinates in our $A$ basis.  Likewise, $I_A$ will be the matrix which converts coordinates from the $A$ basis to coordinates with the standard basis.

Generalizing on the argument above we see that we can calculate $I_A$ by simply using our basis vectors as our columns.  For $_AI$ we take the inverse of this matrix.

$\displaystyle I_A = \left[ \, \left. a_1 \, \right| \, \left. a_2 \, \right| \, \left. \cdots \, \right| \, a_n \, \right]$

$\displaystyle _AI = \left[ \, \left. a_1 \, \right| \, \left. a_2 \, \right| \, \left. \cdots \, \right| \, a_n \, \right]^{-1}$

Now suppose we have two bases, $A = \{ a_1, ..., a_n \}$ and $B = \{ b_1, ..., b_n \}$.  The matrix which will take coordinates with respect to the $A$ basis and convert them into coordinates for the $B$ basis is denoted $_BI_A$.  One way to calculate $_BI_A$ is to convert our $A$-coordinates into standard coordinates, and then convert those into $B$-coordinates:

$\displaystyle _BI_A = _BI \, I_A$

Alternatively, we could note that

$\displaystyle [ a_i ]_B = _BI_A [ a_i ]$

$\displaystyle \, = _BI_A \left[ \begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{array} \right]_A$ (with the 1 in the $i$-th spot)

$\displaystyle \, = \left(_BI_A\right)_{*i}$ (recall $M_{*i}$ is the notation I use for the $i$-th column of $M$)

So the $i$-th column of $_BI_A$ is simply $i$-th vector from our $A$ basis, but in $B$-coordinates.