solutions to 2.9

Exercise: consider the mapping that takes a quadratic polynomial $$q$$ to its values at $$0,1,2$$ and $$3$$. Find the normal form of this operator.
Solution: let $$V$$ denote the space of quadratic polynomials with the standard basis $$\{1,x,x^2\}$$, in which $$p(x) = a_0+a_1 x + a_2 x^2$$. The mapping $$T:V\to \mathbb{R}^4$$ has the form $$Tp = \begin{pmatrix} p(0) \\ p(1) \\ p(2) \\p(3) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 3 & 9 \end{pmatrix} \begin{pmatrix} a_0 \\ a_1 \\ a_2 \end{pmatrix}$$

All we need to do is to find bases in which the matrix representation is block identity. We can simply use the columns of the above matrix, and complete with a fourth linear independent vector, for example:
$$w_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix},~~ w_2 =\begin{pmatrix} 0 \\ 1 \\ 2 \\ 3 \end{pmatrix},~~ w_3 = \begin{pmatrix} 0 \\ 1 \\ 4 \\ 9 \end{pmatrix},~~~ w_4 =\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \end{pmatrix}$$

Then, the representation of $$T$$ with respect to the basis $$\{w_1,w_2,w_3,w_4\}$$ has the normal form

$$\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \hline 0 & 0 & 0 \end{array} \right)$$

Exercise: Find the Jordan normal form for the operator of differentiation acting on the polynomials of degree at most $$d$$. Find all eigenvectors.
Solution: let $$V$$ denote the polynomial space of degree $$\leq d$$. Note that the differentiation operator $$D:V\to V$$ is nilpotent, that is $$D^d=0$$ (take a $$d$$ derivative of a polynomial of degree $$d$$), or $$D$$ is the zero of the polynomial $$p(z) = z^d$$. This means that its only eigenvalue is $$0$$, hence its Jordan form consists of a single Jordan block, which must be (in this case) $$J = \left(\begin{array}{c|cccc} 0 & 1 & \\ & & 1 \\ \vdots & & & \ddots \\ & & & & 1 \\ \hline 0 & & \ldots & & 0 \end{array}\right)$$ which is a shift operator.
From this point on, finding the eigenvectors is a question of picking the right polynomial basis, in which differentiation amounts to shifting. The standard polynomial basis $$u_k:= x^k$$ is not so good for that, as $$Du_k = (x^k)’ = k u_{k-1}$$ which is almost a shift – but with an unwanted scaling by $$k$$. We can mitigate it by re-normalizing the basis elements. Let $$\tilde u_k:= \alpha_k x^k$$ for $$k\geq 1$$, so that the differentiation becomes a shift, or $$(\alpha_k x^k)’ = \alpha_{k-1} x^{k-1}$$ so $$k \alpha_k = \alpha_{k-1}$$. The solution of this regression is of course the factorial $$\alpha_k = \frac{1}{k!}$$, leading to the basis $$\tilde u_k = \frac{1}{k!} x^k$$ in which $$D \tilde u_k = u_{k-1}$$, and those are the eigenvectors of $$D$$.
For the sake of completeness, note that the matrix representation of $$D$$ in the standard basis is $$D= \left(\begin{array}{c|cccc} 0 & 1 & \\ & & 2 \\ \vdots & & & \ddots \\ & & & & d \\ \hline 0 & & \ldots & & 0 \end{array}\right)$$ which is almost natively in a Jordan form, up to the elements of the sub-diagonal being not $$1$$.

Exercise: Find eigenvalues and eigenvectors for the matrix $$\begin{pmatrix} 0 & 1 \\ 0.01 & 0 \end{pmatrix}$$
Solution: the eigvenvalues are the roots of the characteristic polynomial $$p(z) = \left| \begin{matrix} z & -1 \\ -0.01 & z \end{matrix}\right| = z^2 -0.01 = (z-0.1)(z+0.1),$$ which are $$\lambda_1 = 0.1$$ and $$z_2=-0.1$$. The eigenvectors corresponding to the eigenvalues $$z=\lambda$$ are the nullvectors of the matrix $$\lambda E-A$$. For $$\lambda_1 = 0.1$$, solve $$(\lambda_1 E – A)w_1 = \begin{pmatrix} 0.1 & -1 \\ -0.01 & 0.1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0$$ and get $$x = 10y$$. We can pick $$w_1 = \begin{pmatrix} 10 \\ 1 \end{pmatrix}$$. For $$\lambda_2 = 0.1$$, the constraint becomes and get $$x = -10y$$, so we can pick $$w_2 = \begin{pmatrix} 10 \\ -1 \end{pmatrix}$$, and indeed
$$\begin{pmatrix} 0 & 1 \\ 0.01 & 0 \end{pmatrix} = \begin{pmatrix} 10 & 10 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 0.1 & 0 \\ 0 & -0.1 \end{pmatrix} \begin{pmatrix} 10 & 10 \\ 1 & -1 \end{pmatrix}$$