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
$$
\begin{pmatrix} 0 & 1 \\ 0.01 & 0 \end{pmatrix} = \begin{pmatrix} 10 & 10 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 0.1 & 0 \\ 0 & -0.1 \end{pmatrix} \begin{pmatrix} 10 & 10 \\ 1 & -1 \end{pmatrix}
$$

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