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def basis(c):
    return [prod([(x-c[j])/(c[i]-c[j]) for j in range(len(c)) if j != i]) for i in range(len(c))]

def rep(p, c):
    return matrix([p(x = c[i]) for i in range(len(c))]).T

def A(c, d): # matrtix converting from c to d
    B = basis(c)
    return matrix([vector(rep(B[i],d)) for i in range(len(d))]).T

%md
Change $c$ an $d$ below to determine two basis.

Change cc an dd below to determine two basis.

c = [-2,-1,0,1,2]
d = [1,2,3,4,5]
Bc = basis(c)
Bd = basis(d)
A = A(c, d)


str1 = "\\begin{align}"
str1 += ' '.join([f"q_{i}^c &= {latex(Bc[i])}={latex(Bc[i].simplify_full())}\\\\" for i in range(len(c))])
str1 += "\\end{align}"
html(str1)
q0c=124 (x+1)(x−1)(x−2)x=124 x4−112 x3−124 x2+112 xq1c=−16 (x+2)(x−1)(x−2)x=−16 x4+16 x3+23 x2−23 xq2c=14 (x+2)(x+1)(x−1)(x−2)=14 x4−54 x2+1q3c=−16 (x+2)(x+1)(x−2)x=−16 x4−16 x3+23 x2+23 xq4c=124 (x+2)(x+1)(x−1)x=124 x4+112 x3−124 x2−112 x\begin{align}q_0^c &= \frac{1}{24} \, {\left(x + 1\right)} {\left(x - 1\right)} {\left(x - 2\right)} x=\frac{1}{24} \, x^{4} - \frac{1}{12} \, x^{3} - \frac{1}{24} \, x^{2} + \frac{1}{12} \, x\\ q_1^c &= -\frac{1}{6} \, {\left(x + 2\right)} {\left(x - 1\right)} {\left(x - 2\right)} x=-\frac{1}{6} \, x^{4} + \frac{1}{6} \, x^{3} + \frac{2}{3} \, x^{2} - \frac{2}{3} \, x\\ q_2^c &= \frac{1}{4} \, {\left(x + 2\right)} {\left(x + 1\right)} {\left(x - 1\right)} {\left(x - 2\right)}=\frac{1}{4} \, x^{4} - \frac{5}{4} \, x^{2} + 1\\ q_3^c &= -\frac{1}{6} \, {\left(x + 2\right)} {\left(x + 1\right)} {\left(x - 2\right)} x=-\frac{1}{6} \, x^{4} - \frac{1}{6} \, x^{3} + \frac{2}{3} \, x^{2} + \frac{2}{3} \, x\\ q_4^c &= \frac{1}{24} \, {\left(x + 2\right)} {\left(x + 1\right)} {\left(x - 1\right)} x=\frac{1}{24} \, x^{4} + \frac{1}{12} \, x^{3} - \frac{1}{24} \, x^{2} - \frac{1}{12} \, x\\\end{align}
str2 = "\\begin{align}"
str2 += ' '.join([f"q_{i}^d &= {latex(Bd[i])}={latex(Bd[i].simplify_full())}\\\\" for i in range(len(d))])
str2 += "\\end{align}"
html(str2)
q0d=124 (x−2)(x−3)(x−4)(x−5)=124 x4−712 x3+7124 x2−7712 x+5q1d=−16 (x−1)(x−3)(x−4)(x−5)=−16 x4+136 x3−596 x2+1076 x−10q2d=14 (x−1)(x−2)(x−4)(x−5)=14 x4−3 x3+494 x2−392 x+10q3d=−16 (x−1)(x−2)(x−3)(x−5)=−16 x4+116 x3−416 x2+616 x−5q4d=124 (x−1)(x−2)(x−3)(x−4)=124 x4−512 x3+3524 x2−2512 x+1\begin{align}q_0^d &= \frac{1}{24} \, {\left(x - 2\right)} {\left(x - 3\right)} {\left(x - 4\right)} {\left(x - 5\right)}=\frac{1}{24} \, x^{4} - \frac{7}{12} \, x^{3} + \frac{71}{24} \, x^{2} - \frac{77}{12} \, x + 5\\ q_1^d &= -\frac{1}{6} \, {\left(x - 1\right)} {\left(x - 3\right)} {\left(x - 4\right)} {\left(x - 5\right)}=-\frac{1}{6} \, x^{4} + \frac{13}{6} \, x^{3} - \frac{59}{6} \, x^{2} + \frac{107}{6} \, x - 10\\ q_2^d &= \frac{1}{4} \, {\left(x - 1\right)} {\left(x - 2\right)} {\left(x - 4\right)} {\left(x - 5\right)}=\frac{1}{4} \, x^{4} - 3 \, x^{3} + \frac{49}{4} \, x^{2} - \frac{39}{2} \, x + 10\\ q_3^d &= -\frac{1}{6} \, {\left(x - 1\right)} {\left(x - 2\right)} {\left(x - 3\right)} {\left(x - 5\right)}=-\frac{1}{6} \, x^{4} + \frac{11}{6} \, x^{3} - \frac{41}{6} \, x^{2} + \frac{61}{6} \, x - 5\\ q_4^d &= \frac{1}{24} \, {\left(x - 1\right)} {\left(x - 2\right)} {\left(x - 3\right)} {\left(x - 4\right)}=\frac{1}{24} \, x^{4} - \frac{5}{12} \, x^{3} + \frac{35}{24} \, x^{2} - \frac{25}{12} \, x + 1\\\end{align}
p(x) = 1 - 2*x^2 + 4*x^4
html(f"$$[p]_{{\\cal q^c}}={latex(rep(p, c))};\\quad[p]_{{\\cal q^d}}={latex(rep(p, d))}$$")
[p]qc=(5731357);[p]qd=(3573079932451)[p]_{\cal q^c}=\left(\begin{array}{r} 57 \\ 3 \\ 1 \\ 3 \\ 57 \end{array}\right);\quad[p]_{\cal q^d}=\left(\begin{array}{r} 3 \\ 57 \\ 307 \\ 993 \\ 2451 \end{array}\right)


str3 = f"$$A_{{\\cal Q^c,Q^d}}[p]_{{\\cal q^c}} = {latex(A)}{latex(rep(p, c))}={latex(A*matrix(rep(p, c)))}=[p]_{{\\cal q^d}}$$"
html(str3)
AQc,Qd[p]qc=(00010000011−510−1055−2445−401515−70126−10535)(5731357)=(3573079932451)=[p]qdA_{\cal Q^c,Q^d}[p]_{\cal q^c} = \left(\begin{array}{rrrrr} 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & -5 & 10 & -10 & 5 \\ 5 & -24 & 45 & -40 & 15 \\ 15 & -70 & 126 & -105 & 35 \end{array}\right)\left(\begin{array}{r} 57 \\ 3 \\ 1 \\ 3 \\ 57 \end{array}\right)=\left(\begin{array}{r} 3 \\ 57 \\ 307 \\ 993 \\ 2451 \end{array}\right)=[p]_{\cal q^d}
n = 4
str4 = f"\\begin{{split}}{latex(Bc[n])}&={latex(Bc[n].simplify_full())}\\\\&={latex(vector(Bd)*A[:,n])}\\\\&={latex((vector(Bd)*A[:,n])[0].simplify_full())}\\end{{split}}"
html(str4)
124 (x+2)(x+1)(x−1)x=124 x4+112 x3−124 x2−112 x=(3524 (x−1)(x−2)(x−3)(x−4)−52 (x−1)(x−2)(x−3)(x−5)+54 (x−1)(x−2)(x−4)(x−5)−16 (x−1)(x−3)(x−4)(x−5))=124 x4+112 x3−124 x2−112 x\begin{split}\frac{1}{24} \, {\left(x + 2\right)} {\left(x + 1\right)} {\left(x - 1\right)} x&=\frac{1}{24} \, x^{4} + \frac{1}{12} \, x^{3} - \frac{1}{24} \, x^{2} - \frac{1}{12} \, x\\&=\left(\frac{35}{24} \, {\left(x - 1\right)} {\left(x - 2\right)} {\left(x - 3\right)} {\left(x - 4\right)} - \frac{5}{2} \, {\left(x - 1\right)} {\left(x - 2\right)} {\left(x - 3\right)} {\left(x - 5\right)} + \frac{5}{4} \, {\left(x - 1\right)} {\left(x - 2\right)} {\left(x - 4\right)} {\left(x - 5\right)} - \frac{1}{6} \, {\left(x - 1\right)} {\left(x - 3\right)} {\left(x - 4\right)} {\left(x - 5\right)}\right)\\&=\frac{1}{24} \, x^{4} + \frac{1}{12} \, x^{3} - \frac{1}{24} \, x^{2} - \frac{1}{12} \, x\end{split}
latex(A)
\left(\begin{array}{rrrrr} 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & -5 & 10 & -10 & 5 \\ 5 & -24 & 45 & -40 & 15 \\ 15 & -70 & 126 & -105 & 35 \end{array}\right)