1
$\begingroup$

I have two bases for $\Bbb{R^2}$, $C:=\{(2,-1)^T,(6,-2)^T\}$ and $B:=\{(-6,-1)^T,(2,0)^T\}$. To find the change of basis matrix $P_{B\to C}$ we row reduce the system $$\begin{bmatrix}2&6&-6&2 \\-1&-2&-1&0\end{bmatrix}$$

until we have $$\begin{bmatrix}1&0&9&-2 \\0&1&-4&1\end{bmatrix}$$

which gives us the coordinates of the basis vectors of $B$ with respect to basis $C$ on the columns of the rightwise $2\times 2$ matrix, i.e $[b_1]_C$ and $[b_2]_C$ - these are the columns of the change of basis matrix $P_{B\to C}$. I understand some of the connections here, the basis vectors of $C$ are just linear combinations of the natural basis of $\Bbb{R^2}$ - call it $E$. Thus the matrices of $C$ and $E$ are row equivalent. Why does the same sequence of row operations change the coordinates of the basis vectors of $B$ into $[b_1]_C$ and $[b_2]_C$?

$\endgroup$
2
  • $\begingroup$ What is it that you don’t understand? How row-reduction produces $C^{-1}B$ or what that is the correct change-of-basis matrix? $\endgroup$ – amd Mar 16 '20 at 0:52
  • $\begingroup$ I can't really grasp the theory behind this algorithm/process of finding the change of basis matrix, it seems analogous to the process of finding the inverse of a matrix - but not quite. I'm aware that it is a pretty vague question. $\endgroup$ – variations Mar 16 '20 at 7:38
1
$\begingroup$

It seems like there are two questions here, one about forming a change-of-basis matrix from two other matrices, and one about the mechanics of the specific method being used.

Taking the first one first, recall the definition of the coordinates of a vector $\mathbf v$ relative to some ordered basis $\mathcal B=\{\mathbf b_i\}$: they are the coefficients $a_i$ of the basis vectors in the unique linear combination $\mathbf v = a_1\mathbf b_1+\cdots+a_n\mathbf b_n$. We generally collect these coefficients into an $n$-tuple of scalars that your text denotes by $[\mathbf v]_{\mathcal B} = a_1[\mathbf b_1]_{\mathcal B}+\cdots+a_n[\mathbf b_n]_{\mathcal B}\in\mathbb F^n$, where $\mathbb F$ is the field over which the vector space is defined. I’ll call this a $\mathcal B$-tuple for brevity.

Now let $$M = \begin{bmatrix}[\mathbf b_1]_{\mathcal C}&\cdots&[\mathbf b_n]_{\mathcal C}\end{bmatrix},$$ that is, the matrix with columns equal to the coordinate tuples of the elements of $\mathcal B$ relative to some other basis $\mathcal C$. Since $[\mathbf b_j]_{\mathcal B}$ is just the $j$th column of the identity matrix, we have $$M[\mathbf v]_{\mathcal B} = a_1[\mathbf b_1]_{\mathcal C}+\cdots+a_n[\mathbf b_n]_{\mathcal C}.$$ This is a linear combination of $\mathcal C$-tuples, so is itself a $\mathcal C$-tuple, namely, $[\mathbf v]_{\mathcal C}$. Thus, $M=P_{\mathcal B\to\mathcal C}$. Since $M^{-1}M=I$, it should also be clear that $M^{-1}$ maps $[\mathbf b_j]_{\mathcal C}$ to $[\mathbf b_j]_{\mathcal B}$, so $P_{\mathcal C\to\mathcal B} = M^{-1}$.

We can also perform this change of basis in two steps, by first mapping to the standard basis, i.e., $$P_{\mathcal B\to\mathcal C} = P_{\mathcal E\to\mathcal C}P_{\mathcal B\to\mathcal E} = \begin{bmatrix}[\mathbf c_1]_{\mathcal E} & \cdots & [\mathbf c_n]_{\mathcal E}\end{bmatrix}^{-1} \begin{bmatrix}[\mathbf b_1]_{\mathcal E} & \cdots [\mathbf b_n]_{\mathcal E}\end{bmatrix}.$$ In your case, this is $C^{-1}B$, with $$B=\begin{bmatrix}-6&2\\-1&0\end{bmatrix}, C=\begin{bmatrix}2&6\\-1&-2\end{bmatrix}.$$

As to the second question regarding computing $C^{-1}B$ via row-reduction, remember that every elementary row operation corresponds to left-multiplication by a particular invertible matrix, and so the entire process of row-reduction is equivalent to left-multiplication by some invertible matrix $E$. If the matrix $C$ is invertible, its RREF is the identity matrix, i.e., $EC=I$, from which we have $E=C^{-1}$. Because of the way matrix multiplication works, if we augment $C$ and reduce it to its RREF, then whatever is on the right side also gets multiplied by $C^{-1}$: $$\left[C\mid B\right] \to C^{-1}\left[C\mid B\right] = \left[I\mid C^{-1}B\right],$$ which is exactly what was needed for $P_{\mathcal B\to\mathcal C}$. Comparing this to your specific case, the reduced augmented matrix is $$\left[\begin{array}{cc|cc}1&0 & 9&-2 \\ 0&1 & -4&1\end{array}\right]$$ so $P_{\mathcal B\to\mathcal C}$ is the submatrix on the right side.

Note that matrix inversion is a special case of this method in which we augment with the identity matrix: $$\left[C\mid I\right] \to C^{-1}\left[C\mid I\right] = \left[I\mid C^{-1}\right].$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.