Suppose y1 and y2 are the solution of the equation.
(1)
\(d^{3}\)(y1+y2)/d\(x^{3}\)-4\(d^{2}\)(y1+y2)/d\(x^{2}\)+d(y1+y2)/dx+6(y1+y2)
=\(d^{3}\)y\(\frac{1}{d}\)\(x^{3}\)+\(d^{3}\)y\(\frac{2}{d}\)\(x^{3}\)-4\(d^{2}\)y\(\frac{1}{d}\)\(x^{2}\)-4\(d^{2}\)y\(\frac{2}{d}\)\(x^{2}\)+dy\(\frac{1}{d}\)x+dy\(\frac{2}{d}\)x+6y1+6y2
=0+0=0
(2)
\(d^{3}\)(ky1)/d\(x^{3}\)-4\(d^{2}\)(ky1)/d\(x^{2}\)+d(ky1)/dx+6ky1
=k(\(d^{3}\)y\(\frac{1}{d}\)x-4\(d^{2}\)yd\(\frac{1}{d}\)\(x^{2}\)+dy\(\frac{1}{d}\)x+6y1)
Then V is a vector space

1)
f, g ∈ V, a, b は scalar, y = af + bg とすると
\(d^{3}\)\(\frac{y}{d}\)\(x^{3}\) - 4\(d^{2}\)\(\frac{y}{d}\)\(x^{2}\) + d\(\frac{y}{d}\)x + 6y
= af''' + bg''' - 4af'' -4bg''' + af' + bg' + 6af + 6bg
= a(f''' - 4f'' + f' + 6f) + b(g''' - 4g'' + g' + 6g) = 0.
だから y = af + bg ∈ V.

f が f''' - 4f'' + f' + 6f = 0 を満たせば, この両辺を微分すると
f'''' - f''' + f'' + 6f' = 0.
即ち Tf = f' ∈ V.

2)
\(A^{3}\) - A = A(\(A^{2}\) - I) = A(A - I)(A + I) = 0
但し I は単位行列。
従って固有値としては 0, \(\pm\)1 の何れか (或いはその全て)。
(A = 0 という場合や A = I, A = -I という場合もあるので,
どれか一つという可能性も落とせない)