Show that the diagonal entries of symmetric & idempotent matrix must be in [$0,1$].

Let $A$ be a symmetric and idempotent $n \times n$ matrix. By the definition of eigenvectors and since $A$ is an idempotent,
$Ax=\lambda x \implies A^2x=\lambda Ax \implies Ax=\lambda Ax=\lambda^2 x.$
So $\lambda^2=\lambda$ and hence $\lambda \in \{0,1\}$. To show the part about the "diagonal matrix" I use the fact that every symmetric matrix is diagonalizable.

Is this a complete proof?

  • $\begingroup$ Look at where the upper left entry in $A^2$ comes from. $\endgroup$ – Gerry Myerson Sep 15 '13 at 9:50
  • $\begingroup$ @GerryMyerson I don't get it, but if you mean the typo that I made in second equation I fix it now. $\endgroup$ – James C. Sep 15 '13 at 10:00
  • $\begingroup$ No, I mean what I said. When you compute $A^2$, how do you get the upper left entry? What the formula for it, in terms of the entries of $A$. $\endgroup$ – Gerry Myerson Sep 15 '13 at 10:02
  • $\begingroup$ Do you mean $[0,1]$ or $\{0,1\}$? $\endgroup$ – Vishal Gupta Sep 15 '13 at 10:25
  • $\begingroup$ @Vishal It means interval from $0$ to $1$ $\endgroup$ – James C. Sep 15 '13 at 10:28

Expanding my comment to an answer, as OP appears to have lost interest:

Recall the hypotheses: $A$ is $n\times n$, idempotent (so $A^2=A$), and symmetric (so $a_{ij}=a_{ji}$, if we let $a_{ij}$ be the entry in row $i$, column $j$ of $A$).

Looking at the entry in row $i$, column $i$ on both sides of $A=A^2$ we get $$a_{ii}=a_{i1}^2+a_{i2}^2+\cdots+a_{ii}^2+\cdots+a_{in}^2\ge a_{ii}^2$$ But the inequality $a_{ii}\ge a_{ii}^2$ is equivalent to $0\le a_{ii}\le1$.


Let $Q$ be a real symmetric and idempotent matrix of "dimension" $n \times n$. First, we establish the following: The eigenvalues of $Q$ are either $0$ or $1$.

proof. note that if $(\lambda,v)$is an eigenvalue- eigenvector pair of $Q$ we have

$\lambda v=Qv= Q^{2} v=Q(Qv)=Q(\lambda v) = \lambda^{2} v$. Since $v$ is nonzero then the result follows immediately.

With this result at hand the following observation gets us to the desired answer:

Let $e_{i}$ and $q_{ii}$ denote the standard unit vector and $i_{th}$ diagonal element of $Q$, respectively. Then we have

$$ 0= min \{\lambda_1 ,...,\lambda_{n} \} \leq q_{ii} = e_{i}' Q e_{i} \leq max \{\lambda_{1},...,\lambda_n\}=1 $$


All you have shown till now (using idempotency) is that the eigenvalues are either $0$ or $1$.

Now we use symmetry to say that your matrix (let us call it $A$) is unitarily similar to the diagonal matrix consisting of eigenvalues of $A$ on its diagonal. That is

$$A = UDU^{-1}$$

for some unitary matrix $U$ and the diagonal matrix $D$ where $D$ has eigenvalues of $A$ on its diagonal. We already know that these eigenvalues are either $0$ or $1$. So now expand the above representation of $A$ to get the diagonal entries of $A$.

  • $\begingroup$ I think this is the hard way. $\endgroup$ – Gerry Myerson Sep 15 '13 at 10:41
  • $\begingroup$ I read your comment. That is a indeed an infinitely simpler and neat proof. Why don't you make it an answer so that we can upvote and OP can accept it? $\endgroup$ – Vishal Gupta Sep 15 '13 at 10:54
  • 1
    $\begingroup$ I'm hoping that OP will understand what I'm getting at, and then I will encourage OP to post an answer. I'm willing to give OP a few days to think it over. $\endgroup$ – Gerry Myerson Sep 15 '13 at 10:57
  • $\begingroup$ @GerryMyerson Nice of you! Do you think it is in general a bad idea to post answers to such questions? So did I make a mistake by posting a solution? $\endgroup$ – Vishal Gupta Sep 15 '13 at 11:03
  • $\begingroup$ I take it on a case-by-case basis. If I think someone can be led to discover the answer, I usually try to do that. I don't expect everyone to take that approach. I occasionally get upset when someone robs a questioner of the pleasure of discovering an answer, but I am not upset with what you have done here. $\endgroup$ – Gerry Myerson Sep 15 '13 at 11:08

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