I study the properties of real symmetric idempotent matrices presented here and here. Relevant for my question are the assumptions that

(a) $X_i$, $i= 1,2,...p$, are symmetric, idempotent matrices;

(b) $X = \sum_{i}^PX_i$, which is also a symmetric, idempotent matrix.

In their outline of the proofs of some of those properties, Searle & Gruber (2017, p. 84) state that $X − X_i − X_j = \sum_{r\ne i\ne j}X_r$ can be assumed to be a non-negative definite matix. Since they state that $X − X_i − X_j$ is non-negative definite, I assume they imply that this matrix is idempotent, too. The eigenvalues of a, idempotent matrix are either $1$ or $0$. However, how to prove that $X − X_i − X_j$ is an idempotent (i.e., non-negative definite) matrix?

My approch would have been that

$\begin{align} \ (X − X_i − X_j) & = (X − X_i − X_j)(X − X_i − X_j) \\ & = (X^2 − 2XX_i − 2XX_j + X_iX_i + 2X_iX_j + X_jX_j) \\ & = (X − 2XX_i − 2XX_j + X_i + 2X_iX_j + X_j) \\ \end{align}$

To complete this proof, one would have to show that all the cross terms are equal to zero. However, this is not covered by the assumptions of the proof. In fact, I found the above mentioned statement in a proof of the property that if (a) and (b) above are given then $X_iX_J = 0$ for $i \ne j$.

My question therefore is whether or not the sum of symmetric, idempotent matrices always is an idempotent matrix and how to show this?

  • $\begingroup$ Have you tried some examples? Say, $I$ and some nonzero symmetric idempotent matrix? $\endgroup$ Mar 3, 2020 at 17:11
  • $\begingroup$ @RobertIsrael, you are absolutely right and I am an idiot. I tried different diagonal matrices whose diagonal elements were 1 and 0. In those, admittedly, selected cases it worked. Thanks for this quite obvious counter example. $\endgroup$
    – DomB
    Mar 3, 2020 at 18:07

1 Answer 1


No it is not. Take \begin{equation*} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation*}

Then $I$ is clearly symmetric and idempotent but

\begin{equation*} I + I = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \end{equation*}

Is not idempotent, as $(2I)^2 = 4I$

  • $\begingroup$ In nimber arithmetic, $I+I=0$. That's where the claim works. $\endgroup$ Mar 3, 2020 at 17:42
  • $\begingroup$ You mean Grundy numbers ? $\endgroup$ Mar 3, 2020 at 17:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .