I'm a second-year undergraduate from China. This is my first time posting a question on Mathematics Stack Exchange. If there's anything I've overlooked or if my question doesn't quite adhere to the site's etiquette, please let me know in the comments.
Well,recently, I've been reading a book on wavelet theory titled "Linear Algebra, Signal Processing, and Wavelets - A Unified Approach: Python Version (Springer Undergraduate Texts in Mathematics and Technology)". I've encountered a problem related to matrix induction in the book. Logically, I should probably ask this question in the Signal Processing SE section. However, as indicated in my title, I'm more interested in an answer from a mathematical perspective.
The following context will give a detailed description of my problem:
Define: \begin{align} &\phi_{0,0}(t) = \begin{cases} 1-|t| & \text{if } -1 \le t \le 1 \\ 0 & \text{otherwise} \end{cases} \\ \\ &\phi_{m,n}(t) = 2^{m/2}\phi_{0,0}(2^mt-n)\\ \\ &\psi_{0,0}(t)=\frac{1}{\sqrt{2}}\phi_{1,1}(t)\\ \\ &\psi_{m,n}(t) = 2^{m/2}\psi_{0,0}(2^mt-n)\\ \end{align}
Let $\phi_m=\{\phi_{m,0}(t),\phi_{m,1}(t)...\phi_{m,2^{\mkern+2mu m}N-1}(t)\}$,where $\phi_m$ is a set,each of its element is a function $\phi_{m,n}(t),n\in[0,2^mN-1],N\in Z^+$.When $m=0$,$\phi_m=\phi_0=\{\phi_{0,0}(t),\phi_{0,1}(t)...\phi_{0,N-1}(t)\}$
The set $\phi_m$ may span a function space,denoted by $V_m$,has a dimensionality of $2^mN-1$.
Let $C_m=\{\phi_{m-1,0}(t),\psi_{m-1,0}(t)...\phi_{m-1,2^{m-1}\hspace{+0.4em}\times{\hspace{+0.4em}N}-1}(t),\psi_{m-1,N2^{m-1}\hspace{0.3em}-1}(t)\}$, The set $C_m$ may also be a basis of $V_m$.
Now we know that : \begin{align} &\phi_{0,0}(t)=\frac{1}{\sqrt{2}}(\frac{1}{2}\phi_{1,-1}(t)+\phi_{1,0}(t)+\frac{1}{2}\phi_{1,1}(t))\\ &ψ_{0,0}(t)=\frac{1}{\sqrt{2}}\phi_{1,1}(t)\\ &\phi_{m,k}(t)=\frac{1}{\sqrt{2}}(\frac{1}{2}\phi_{m+1,2k-1}(t)+\phi_{m+1,2k}(t)+\frac{1}{2}\phi_{m+1,2k+1}(t))\\ &\psi_{m,n}(t)=\frac{1}{\sqrt{2}}\phi_{m+1,2k+1}(t)\\ \end{align} if we have a vector $\boldsymbol{v_1}(c_{m-1,0},w_{m-1,0},c_{m-1,1},w_{m-1,1}...c_{m-1,2^{m-1}\hspace{+0.3em}N-1},w_{m-1,2^{m-1}\hspace{+0.3em}N-1})$,where $c_{m,n}$means the coordinates of some function f(t) in the basis of $\phi_{m}$,and $w_{m,n}$ means the coordinates in the basis of $\psi_{m}$
i.e. $f(t) = \sum_{n=0}^{2^{m-1}\hspace{+0.3em}N-1} c_{m-1,n} \phi_{m-1,n}(t) + \sum_{n=0}^{2^{m-1}\hspace{+0.3em}N-1} w_{m-1,n}(t) \psi_{m,n}(t)$.
Now we have a new vector $\boldsymbol{v_2}(c_{m,0},c_{m,1}...c_{m,2^{m}\hspace{+0.3em}N-1})$,where $f(t) = \sum_{n=0}^{2^{m}\hspace{+0.3em}N-1} c_{m,n} \phi_{m,n}(t)$.We need to find a matrix A satisfying the equation $A\boldsymbol{v_1}={v_2}$.This step is where I got stuck. The answer on the book is: $\begin{pmatrix} 1 & 0 & 0 & 0 &\ldots & 0 & 0 & 0\\ 1 & 1/2 & 1/2 & 0 &\ldots & 0 & 0 & 0\\ 0 & 0 & 1 & 0 &\ldots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots &\vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 &\ldots & 0 & 1 & 0\\ 1/2 & 0 & 0 & 0 &\ldots & 0 & 1/2 & 1\\ \end{pmatrix}$ without any other hints,and I couldn't understand how to fetch it.
Well,I did have a few tries.These two equations
$\begin{array}{l}{{\phi_{m,k}(t)=\frac{1}{\sqrt{2}}(\frac{1}{2}\phi_{m+1,2k-1}(t)+\phi_{m+1,2k}(t)+\frac{1}{2}\phi_{m+1,2k+1}(t))}}\\ {{\psi_{m,n}(t)=\frac{1}{\sqrt{2}}\phi_{m+1,2k+1}(t)}}\end{array}$
remind me of the definition of the basis transformation matrix(from wiki):
"Let $B_{\mathrm{old}}=(v_1,v_2,...,v_n)$be a basis of a finite-dimensional vector space $V$ over field $F$.For $j=1,2,....,n$,one can define a vector $w_j$(Here and in what follows, the index$i$ always refer to the rows of C and the $v_{i}$while the index $j$ refers always to the columns of C and the $w_{j}$ such a convention is useful for avoiding errors in explicit computations.) Setting $B_{\mathrm{new}}=(w_1,..,w_n)$,one has that $B_{\mathrm {new}}$is a basis of $V$ if and only if the matrix $C$ is invertible, or equivalently if it has a nonzero determinant. In this case, $C$ is said to be the change-of-basis matrix from the basis $B_{\mathrm {old} }$ to the basis $B_{\mathrm {new} }$.Given a vector $z\in V$,let$(x_1.x_2....x_n)$be the coordinates of $z$ over $B_{\mathrm {old} }$, and $(y_{1},\ldots ,y_{n})$ its coordinates over $B_{\mathrm {new} }$ that is:
$z=\sum_{i=1}^{n}x_{i}v_{i}=\sum_{j=1}^{n}y_{j}w_{j}.$ (One could take the same summation index for the two sums, but choosing systematically the indexes i for the old basis and j for the new one makes clearer the formulas that follows, and helps avoiding errors in proofs and explicit computations.)
The change-of-basis formula expresses the coordinates over the old basis in term of the coordinates over the new basis. With above notation, it is ${\displaystyle x_{i}=\sum _{j=1}^{n}c_{i,j}y_{j}\qquad {\text{for }}i=1,\ldots ,n.}$ In terms of matrices, the change of basis formula is , ${\displaystyle \mathbf {x} =C\,\mathbf {y} ,}$ where $\mathbf{x}$ and $\mathbf{y}$ are the column vectors of the coordinates of z over ${\displaystyle B_{\mathrm {old} }}$ and ${\displaystyle B_{\mathrm {new} },}$ respectively."
"I am aware that there are two imprecisions in this association. The two equations we have describe the linear relationship between functions (infinite-dimensional vectors) rather than finite-dimensional vectors. Moreover, the matrix extracted from the equations is non-invertible. However, I believe that this indeed provides a certain direction for approaching the problem. Therefore, I roughly understand that the matrix $A$ we want to obtain is likely a repetition of the matrix in some direction. But I truly cannot comprehend the '1' in the top-left corner of the first row of the matrix.
