Question 7.2.7: Determine the spectral projectors for A = (1 −4 −4 8 −11 −8 ......
Determine the spectral projectors for A = \begin{pmatrix}1 &−4& −4\\ 8 &−11& −8\\ −8 &8& 5\end{pmatrix}.
This is the diagonalizable matrix from Example 7.2.1 (p. 507). Since there are two distinct eigenvalues, λ_1 = 1 and λ_2 = −3, there are two spectral projectors,
G_1 = the projector onto N (A − 1I) along R(A − 1I),
G_2 = the projector onto N (A + 3I) along R(A + 3I).
There are several different ways to find these projectors.
1. Compute bases for the necessary nullspaces and ranges, and use (5.9.12).
P = [X | 0] [X | Y]^{-1} = [X | Y] \begin{pmatrix}I&0\\0&0\end{pmatrix}[X | Y]^{-1}, (5.9.12)
2. Compute G_i = X_iY^T_i as described in (7.2.11).
A = PDP^{−1} = \left(X_1 | X_2 | · · · | X_k\right) \begin{pmatrix}λ_1I& 0& · · ·& 0\\ 0 &λ_2I& · · ·& 0\\ \vdots&\vdots&\ddots&\vdots\\ 0 &0& · · ·& λ_kI\end{pmatrix} \begin{pmatrix}Y^T_1\\\hline Y^T_2\\\hline \vdots\\\hline Y^T_k\end{pmatrix}(7.2.11)
The required computations are essentially the same as those needed above. Since much of the work has already been done in Example 7.2.1, let’s complete the arithmetic. We have
P = \begin{pmatrix}\begin{array}{c | c c}1 &1& 1\\ 2 &1& 0\\ −2 &0& 1 \end{array}\end{pmatrix} = \left(X_1 | X_2\right), P^{-1} = \begin{pmatrix}1 &−1& −1\\\hline −2 &3& 2\\ 2 &−2& −1\end{pmatrix} = \begin{pmatrix}\frac{Y^T_1}{Y^T_2}\end{pmatrix},so
G_1 = X_1Y^T_1 = \begin{pmatrix}1 &−1& −1\\ 2 &−2& −2\\ −2 &2& 2 \end{pmatrix}, G_2 = X_2Y^T_2 = \begin{pmatrix}0 &1& 1\\ −2 &3& 2\\ 2 &−2& −1\end{pmatrix}.Check that these are correct by confirming the validity of (7.2.7)–(7.2.10).
A = λ_1G_1 + λ_2G_2 + · · · + λ_kG_k, (7.2.7)
G_i is the projector onto N (A − λ_iI) along R (A − λ_iI). (7.2.8)
G_i G_j = 0 whenever i ≠ j. (7.2.9)
G_1 +G_2 + · · · +G_k = I. (7.2.10)
3. Since λ_1 = 1 is a simple eigenvalue, (7.2.12) may be used to compute G_1 from any pair of associated right-hand and left-hand eigenvectors x and y^T.
G = xy^∗ / y^∗ x (7.2.12)
Of course, P and P^{−1} are not needed to determine such a pair, but since P and P^{−1} have been computed above, we can use X_1 and Y^T_1 to make the point that any right-hand and left-hand eigenvectors associated with λ_1 = 1 will do the job because they are all of the form x = αX_1 and y^T = βY^T_1 for α ≠ 0 ≠ β. Consequently,
G_1 = \frac{xy^T}{y^T x} = \frac{α \begin{pmatrix}1\\2\\-2\end{pmatrix} β (1 −1 −1 )}{αβ} = \begin{pmatrix}1 &−1& −1\\ 2 &−2 &−2\\ −2 &2 &2\end{pmatrix}.Invoking (7.2.10) yields the other spectral projector as G_2 = I − G_1.
4. An even easier solution is obtained from the spectral theorem by writing
A − I = (1G_1 − 3G_2) − (G_1 + G_2) = −4G_2,
A + 3I = (1G_1 − 3G_2) + 3(G_1 + G_2) = 4G_1,
so that
G_1 = \frac{(A + 3I)}{4} and G_2 = \frac{−(A − I)}{4}.
Can you see how to make this rather ad hoc technique work in more general situations?
5. In fact, the technique above is really a special case of a completely general formula giving each G_i as a function A and λ_i as
G_i = \frac{\prod\limits^{k}_{\substack{j=1\\j≠i}} (A − λ_jI)}{\prod\limits^{k}_{\substack{j=1\\j≠i}} (λ_i − λ_j)}.
This “interpolation formula” is developed on p. 529.
Below is a summary of the facts concerning diagonalizability.