Transition matrix
Kapitoly: Vector spaces, Examples of vector spaces, Vector subspace, Linear combinations of vectors, Linear wrapper, Bases of vector space, Dimensions of vector space, Transition matrix
If we have a vector space V and two of its bases, we can express one vector from the space as a combination of vectors from the two bases. The transition matrix then helps us convert one expression to the other.
Motivation
Let us have a n-dimensional vector space V and two different bases E = {e1, …, en} and F = {f1, …, fn}. Next, let us choose some vector x from V. Since x is from the space V and since E and F are both bases of this space, it must hold that there are coefficients of a1, …, an such that:
$$ \mathbf{x} = a_1 \cdot \mathbf{e}_1 + \ldots + a_n \cdot \mathbf{e}_n $$
and at the same time there must exist coefficients of b1, …, bn such that:
$$ \mathbf{x} = b_1 \cdot \mathbf{f}_1 + \ldots + b_n \cdot \mathbf{f}_n $$
In the first case we express the vector x using the basis E, in the second case using the basis F. Using two different bases to express x, we get two different sets of coefficients ai and bi at the same time. The question arises: if we know the coefficients of ai, is there any way to get the coefficients of bi, i.e. the expression with respect to the other basis, from these coefficients?
Yes, there is a so-called transition matrix for this purpose.
How to obtain the transition matrix
Let's stay with the fact that we have the space V and two bases E = {e1, …, en} and F = {f1, …, fn}. Since E is a basis of the space V, as surely all vectors f1, …, fn can be expressed as linear combinations of vectors from the basis E. The vectors f1, …, fn are indeed part of the basis F, but they are also ordinary elements of the space V, so they must hold:
$$\begin{eqnarray} \mathbf{f}_1 &=& a_{11} \cdot \mathbf{e}_1 + a_{21} \cdot \mathbf{e}_2 + \ldots + a_{n1} \cdot \mathbf{e}_n\\ \mathbf{f}_2 &=& a_{12} \cdot \mathbf{e}_1 + a_{22} \cdot \mathbf{e}_2 + \ldots + a_{n2} \cdot \mathbf{e}_n\\ &\ldots&\\ \mathbf{f}_n &=& a_{1n} \cdot \mathbf{e}_1 + a_{2n} \cdot \mathbf{e}_2 + \ldots + a_{nn} \cdot \mathbf{e}_n \end{eqnarray}$$
We now rewrite these equations into a matrix so that the first column of the matrix contains all the coefficients of a1i, the second column contains all the coefficients of a2i, etc. We get the matrix:
$$P=\begin{pmatrix} a_{11}&a_{21}&\ldots&a_{n1}\\ a_{12}&a_{22}&\ldots&a_{n2}\\ \ldots\\ a_{1n}&a_{2n}&\ldots&a_{nn}\\ \end{pmatrix}$$
Such a matrix is then called the transition matrix from the basis E to the basis F. If we have a vector x = {x1,…, xn} of V, which has coefficients b1, …, bn with respect to the basis F, then the coefficients a1, …, an with respect to the basis E are obtained
$$ \begin{pmatrix} a_1\\vdots\\a_n \end{pmatrix} P\cdot \begin{pmatrix} b_1\vdots\b_n \end{pmatrix} $$Example
We will work over the vector space ℝ3. We will choose one standard basis E = {[1,0,0], [0,1,0], [0,0,1]} and another basis F = {[2,0,0], [3,2,0], [1,5,4]}. The transition matrix from basis E to basis F will be of the form
$$P=\begin{pmatrix} 2&3&1\\ 0&2&5\\ 0&0&4 \end{pmatrix}$$
Now we choose an arbitrary vector x∈ ℝ3, for example x = [12,41,28]. This has coefficients with respect to the basis F a1 = −2, a2 = 3, a3 = 7 . Thus we perform the product of matrices:
$$ \begin{pmatrix} 2&3&1\\\ 0&2&5\\\ 0&0&4 \end{pmatrix} \cdot \begin{pmatrix} -2\\3\\7 \end{pmatrix} \begin{pmatrix} 12\41\28 \end{pmatrix} $$Given the basis E, the vector x has coefficients a1 = 12, a2 = 41, a3 = 28. Thus:
$$ \left[12{,}41,28\right] = 12\cdot\left[1{,}0,0\right]+41\cdot\left[0{,}1,0\right]+28\cdot\left[0{,}0,1\right] $$