Spectral Decomposition for matrices with eigenvalue(s) 0 When expressing the spectral decomposition of a matrix that has eigenvalues of 0, do you express the corresponding matrices with factor '0', or leave them out altogether?
EDIT: let's consider some matrix M, where
$$
M =
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix}
$$
Then, would the spectral decomposition of this matrix M be
(i)
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix}
=
3
\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\end{bmatrix}
+
0
\begin{bmatrix}
\frac{1}{2} & -\frac{1}{2} & 0 \\
-\frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
+
0
\begin{bmatrix}
\frac{1}{6} & \frac{1}{6} & -\frac{1}{3} \\
\frac{1}{6} & \frac{1}{6} & -\frac{1}{3} \\
-\frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \\
\end{bmatrix}
$$
or (ii)
$$
\begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{bmatrix}
=
3
\begin{bmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\end{bmatrix}
$$
I hope this clarifies the problem.
EDIT 2: Maybe the terminology in my textbook is incorrect, but the form I am talking about looks like this: if the matrix M was orthogonally diagonalized by
$$
P =
\begin{bmatrix}
\bar{u_1} & \bar{u_2} & \bar{u_3} \\
\end{bmatrix}
$$
and
$$
D =
\begin{bmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3 \\
\end{bmatrix}
$$
where $\bar{u_1}$, $\bar{u_2}$, and $\bar{u_3}$ are unit eigenvectors of M, then M could be rewritten as
$$M = \lambda_1\bar{u_1}\bar{u_1}^T + \lambda_2\bar{u_2}\bar{u_2}^T + \lambda_3\bar{u_3}\bar{u_3}^T$$
If the answerer happens to know the name of this form, I'd be much obliged.
 A: What your book calls spectral decomposition is just the equation you obtain after carrying out the matrix product in what is usually called a spectral decomposition. This seems very odd to me, because usually the term decomposition is used to refer as a way to write a matrix as a product of other matrices, not as a sum.
Anyway, a spectral decomposition of a matrix $M$ is an equation of the form
$$
M = Q \Delta Q^{-1} \label{eq:1} \tag{1}
$$
where $\Delta$ is the diagonal matrix with diagonal elements $\lambda_1,\dotsc,\lambda_n$, the eigenvalues of $M$ counted with their respective multiplicity, and $Q$ is a matrix whose columns $q_1,\dotsc,q_n$ form a basis of eigenvectors of $M$ such that $q_i$ corresponds to $\lambda_i$.
Now, if you choose an orthonormal basis of eigenvectors, then $Q$ is an orthogonal matrix and $Q^{-1} = Q^T$. Thus $\eqref{eq:1}$ becomes
$$
\begin{align}
M &=
\left(
\begin{array}{c | c | c}
q_1 & \dotsc & q_n
\end{array}
\right)
\begin{pmatrix}
\lambda_1 & & \\
& \ddots & \\
& & \lambda_n
\end{pmatrix}
\left(
\begin{array}{c}
q_1^T \\
\hline
\vdots \\
\hline
q_n^T
\end{array}
\right) \\
&= \lambda_1 q_1 q_1^T + \dotsc + \lambda_n q_n q_n^T
\end{align}
$$

TL;DR: No, you don't need to include the summands with eigenvalue $0$.
