In Exercises 5 and 6, notice that (for each \(i\)) the space of characteristic vectors associated with the characteristic value \(c_i\) is spanned by the column vectors of the various matrices \(E_j\) with \(j \neq i\). Is that a coincidence?
The printed observation has the indices reversed. In the spectral resolution, the columns of \(E_i\) span the characteristic space for \(c_i\). The columns of the matrices \(E_j\) with \(j\ne i\) span the complementary sum of the other characteristic spaces.
Solution
It is not a coincidence; the corrected statement is a general property of spectral projections.
Let \(A\) be diagonalizable with distinct characteristic values \(c_1,\ldots,c_r\), and define the Lagrange polynomials \[ p_i(x)=\prod_{j\ne i}\frac{x-c_j}{c_i-c_j}, \qquad E_i=p_i(A). \] Then \(p_i(c_i)=1\) and \(p_i(c_j)=0\) for \(j\ne i\) [@HK71].
1. The range of \(E_i\) is exactly the \(c_i\)-characteristic space.
If \(v\) is a characteristic vector for \(c_j\), then \[ E_i v=p_i(A)v=p_i(c_j)v =\begin{cases} v,&j=i,\\ 0,&j\ne i. \end{cases} \] Because \(A\) is diagonalizable, the characteristic spaces span the whole space. Thus \(E_i\) acts as the identity on the \(c_i\)-space and as zero on all the others. Therefore \[ \operatorname{range}E_i=W_i, \] where \(W_i\) denotes the characteristic space for \(c_i\).
2. The columns of \(E_i\) span that characteristic space.
For any matrix representing a linear map, its column space is the range of the map. Hence the columns of \(E_i\) span \[ \operatorname{range}E_i=W_i. \] Likewise, for \(j\ne i\), the columns of \(E_j\) span \(W_j\), so the columns of all the other \(E_j\) together span \[ \bigoplus_{j\ne i}W_j, \] the complementary spectral subspace. Thus the printed statement is false as written, and the corrected column-space phenomenon follows directly from the spectral-projection construction.