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.

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