1.1 To prove that function \(L\) is linear we need to show that \(L(aP+bQ)=a L(P)+bL(Q)\) for any 2 elements \(P, Q\) of \(V\) and 2 real scalars \(a, b\): \[ L(aP+bQ)= ( (x-1) (aP+bQ))' = (x-1)(aP'+bQ')+ (aP+bQ)= a((x-1)P)'+b((x-1)Q)' =a L(P)+bL(Q) \] Here \(V\) is the vector space of polynomials of degree \(\leq 3\). Also we note that the degree of polynomial \([(x-1)P]'\) is \(\leq 3\). Thus \(L\) is a linear mapping from \(V\) to \(V\).
1.2 Let \(P\in V\) such that \(L(P)=0\), that is, \([(x-1)P]' =0\): this implies that \(P(x)= a/(x-1)\) (\(a\) is a constant). For \(P\) to be in \(V\), we need to impose \(a=0\). We conclude that \(\text{ker}(L)=\{0\}\).
Since \(V\) is finite dimensional, \(\text{ker}(L)=\{0\}\) implies that \(L\) is invertible.
To find \(L^{-1}\), we need to solve \[ [(x-1)P(x)]' = Q(x) \] given \(Q\in V\). We can find \(P(x)\): \[ P(x)= \frac{1}{x-1}\int_a^x Q(t) dt \] as long as \(x\neq 1\). For \(P\) to be in \(V\), we need to choose \(a\) properly. If we write \(Q(x)=q_0 + q_1 (x-a)+q_2 (x-a)^2 + q_3 (x-a)^3\), we find \[ P(x)= \frac{1}{x-1} \Big[ q_0 (x-a) + \frac{1}{2}q_1 (x-a)^2 + \frac{1}{3}q_2 (x-a)^3 +\frac{1}{4}q_3 (x-a)^4 \Big] \] It is clear that we need to choose \(a=1\): \[ \boxed{L^{-1}(Q)= \frac{1}{x-1}\int_1^x Q(t) dt } \] as long as \(x\neq 1\). If \(x=1\), we choose \(L^{-1}(Q)(1)=Q(1)\) since \[ L(L^{-1}(Q)) (1)= (x-1) Q'(1)+Q(1)= Q(1) \]
1.3 We find \(L(x^k)= x^k + (x-1) kx^{k-1} = (k+1)x^k - kx^{k-1}\). We can build the matrix of \(L\) in basis \({\cal B}=(1,x,x^2,x^3)\): \[ [L]_{\cal B}= A= \begin{pmatrix} 1 & -1 & 0 & 0 \\ 0 & 2 & -2 & 0 \\ 0 & 0 & 3 & -3 \\ 0 & 0 & 0 & 4 \end{pmatrix} \] We note that this matrix is upper triangular.
1.4 We find \(\det (A-\lambda I)= (1-\lambda)(2-\lambda)(3-\lambda)(4-\lambda)\). This matrix \(A\) has 4 simple eigenvalues \(\lambda=1,2,3,4\): matrix \(A\) is diagonalizable.
We find eigenspace \(V_1\) by solving \(A\boldsymbol{v}=\boldsymbol{v}\). Let \(\boldsymbol{v}= (x,y,z,t)^T\), then: \[ -y=0, \quad y-2z=0, \quad 2z-3t=0, \quad 3t=0 \Longrightarrow y=z=t=0 \] giving the eigenspace \[ V_1 =\text{span}\{(1,0,0,0)^T\}\Longrightarrow P(x)=1 \] Likewise for \(V_2\), solving \(A\boldsymbol{v}=2\boldsymbol{v}\): \[ -x-y=0, \quad -2z=0, \quad z-3t=0, \quad 4t=0 \Longrightarrow x=-y, z=t=0 \] giving the eigenspace \[ V_2 =\text{span}\{(-1,1,0,0)^T\}\Longrightarrow P(x)=x-1 \] For \(V_3\), we solve \(A\boldsymbol{v}=3\boldsymbol{v}\): \[ -2x-y=0, \quad -y-2z=0, \quad -3t=0 \Longrightarrow y=-2z, x=z \] giving the eigenspace \[ V_3 = \text{span}\{(1,-2,1,0)^T\}\Longrightarrow P(x)=(x-1)^2 \] Finally, to find \(V_4\), we solve \(A\boldsymbol{v}=4\boldsymbol{v}\): \[ -3x-y=0, \quad -2y-2z=0, \quad -z-3t=0 \Longrightarrow z=-3t, y=3t, x=-t \] thus giving the eignespace \[ V_4 =\text{span}\{(-1,3,-3,1)^T\}\Longrightarrow P(x)=(x-1)^3 \]
2.1 To show that \(L\) is linear, we consider \(L(aM+bN)\) given 2 elements \(M\) and \(N\) of vector space \(V\) (the space of 2x2 real matrices) and 2 real coefficients\(a\) and \(b\): \[ L(aM+bN)= A(aM+bN)-(aM+bN)A= a (AM-MA)-b(AN-NA)= aL(M)+bL(N) \] using standard matrix algebra.
2.2 We want to find \(L(A_k)\) for \(k=1,2,3,4\) (the elements of the standard basis of \(V\)): we find \[ L(A_1)= \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}-\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}= \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}= A_3 -A_2 \] Likewise, we find \[ L(A_2)= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}= -A_1 +A_4 \] \[ L(A_3)= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}= A_1 -A_4 \] \[ L(A_4)= \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}= A_2 -A_3 \] Now we can build the matrix \(T\) of \(L\) on basis \((A_1,A_2,A_3,A_4)\): \[ T= \begin{pmatrix} 0 & -1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \end{pmatrix} \] We note that \(T\) is symmetric, and thus diagonalizable.
2.3 We are told that \(T^3= 4T\). Let \(\boldsymbol{v}\neq \boldsymbol{0}\) be an eigenvector of \(T\) for eigenvalue \(\lambda\): \(T\boldsymbol{v}= \lambda\boldsymbol{v}\). Then \(T^3 \boldsymbol{v}= T^2 (T\boldsymbol{v})= \lambda T^2 \boldsymbol{v}= ...= \lambda^3 \boldsymbol{v}\). Therefore \(T^3 \boldsymbol{v}= 4 T\boldsymbol{v}\) implies that \(\lambda^3 \boldsymbol{v}= 4\lambda\boldsymbol{v}\) and since \(\boldsymbol{v}\neq 0\), \(\lambda(\lambda^2 -4)=0\). This shows that \(\lambda=2, \pm 2\) are possible eigenvalues.
2.4 We first set \(\lambda=0\) and see if we can find nonzero solutions of \(T\boldsymbol{v}=\boldsymbol{0}\). Setting \(\boldsymbol{v}= (x,y,z,t)^T\), we find \[ -y+z=0, \quad -x+t=0, \quad x-t=0, \quad y-z=0 \] which gives \(x=t\) and \(y=z\). Hence \(\boldsymbol{v}= (x,y,y,x)^T= x(1,0,0,1)^T + y (0,1,1,0)^T\). This shows that \[ V_0 = \text{span}( (1,0,0,1)^T,(0,1,1,0)^T )= \text{span}(\boldsymbol{v}_1,\boldsymbol{v}_2) \]
2.5 We easily verify that \(T\boldsymbol{v}_3 = 2\boldsymbol{v}_3\) and that \(T\boldsymbol{v}_4= -2\boldsymbol{v}_4\). Hence \(\lambda=2\) is an eigenvalue and \(V_2 = \text{span}(\boldsymbol{v}_3)\). Likewise \(V_{-2}= \text{span}(\boldsymbol{v}_4)\).
We conclude that we can write \(T\) as follows \[ T= S D S^{-1}, \qquad D=\text{diag}(0,0,2,-2), \qquad S=(\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3,\boldsymbol{v}_4) \] If we set \(\boldsymbol{w}_1 = \boldsymbol{v}_1/\sqrt{2}\), \(\boldsymbol{w}_2= \boldsymbol{v}_2 /\sqrt{2}\), \(\boldsymbol{w}_3= \boldsymbol{v}_3/2\) and \(\boldsymbol{w}_4= \boldsymbol{v}_4/2\) (after normalization), then matrix \[ Q= (\boldsymbol{w}_1,\boldsymbol{w}_2,\boldsymbol{w}_3,\boldsymbol{w}_4) \] is orthogonal, and we can then write \[ T= Q D Q^T \] since \(Q^{-1}=Q^T\).