Recipe

Build a spectral decomposition, and test simultaneous diagonalizability

Use when: you have a normal operator T that needs to be written as \sum_i\lambda_iP_i, or two such operators that need to be checked for a shared eigenbasis

Goal: the projections P_i, the decomposition T=\sum_i\lambda_iP_i, and a definite answer, when asked, on whether two operators share a common eigenbasis

Taught in Spectral decomposition as a sum of projections, and simultaneous diagonalization

1. 1

   Find T's eigenvalues and an orthonormal eigenbasis, then sort the basis vectors into groups, one group per distinct eigenvalue.

   Example

   For A=\begin{pmatrix}2&1\\1&2\end{pmatrix}: \lambda_1=3 has eigenvector \mathbf u_1=(1,1)/\sqrt2, and \lambda_2=1 has eigenvector \mathbf u_2=(-1,1)/\sqrt2, one vector in each group.

   Why: the spectral theorem guarantees this orthonormal eigenbasis exists for any normal operator, so nothing here is assumed beyond what's already proven.

2. 2

   For each eigenvalue \lambda_i, form P_i=\sum\mathbf u\mathbf u^{\mathsf T}, summed over the orthonormal eigenvectors in that eigenvalue's own group.

   Example

   P_1=\mathbf u_1\mathbf u_1^{\mathsf T}=\begin{pmatrix}1/2&1/2\\1/2&1/2\end{pmatrix} and P_2=\mathbf u_2\mathbf u_2^{\mathsf T}=\begin{pmatrix}1/2&-1/2\\-1/2&1/2\end{pmatrix}.

   Why: P_i built this way is exactly the orthogonal projection onto that eigenspace, since \mathbf u\mathbf u^{\mathsf T} is how an earlier chapter builds a projection from an orthonormal subspace basis.

3. 3

   Add every projection together and confirm the result is exactly I.

   Example

   P_1+P_2=\begin{pmatrix}1/2&1/2\\1/2&1/2\end{pmatrix}+\begin{pmatrix}1/2&-1/2\\-1/2&1/2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.

   Why: the eigenvectors together form a basis for the whole space, so their combined projections must reconstruct every vector exactly, leaving nothing uncovered.

4. 4

   Weight each projection by its own eigenvalue and add them together: T=\sum_i\lambda_iP_i.

   Example

   3P_1+1\cdot P_2=\begin{pmatrix}3/2&3/2\\3/2&3/2\end{pmatrix}+\begin{pmatrix}1/2&-1/2\\-1/2&1/2\end{pmatrix}=\begin{pmatrix}2&1\\1&2\end{pmatrix}=A.

   Why: this is exactly what the theorem states, and the previous step's identity check already confirms these are the right projections to weight.

5. 5

   To find whether S and T share the same eigenbasis, compute both products ST and TS directly and compare them entry by entry.

   Example

   S=\begin{pmatrix}5&3\\3&5\end{pmatrix} against T=A: both ST and TS equal \begin{pmatrix}13&11\\11&13\end{pmatrix}, so they commute.

   • the two products match — S and T share the same eigenspaces and projections, and are simultaneously diagonalizable.
   • the two products differ — S and T do NOT share an eigenbasis, and no single basis diagonalizes both.

   Why: this section's theorem shows commuting is exactly the condition for two normal operators to share their whole family of eigenspaces.