2
$\begingroup$

We know that if a square matrix $A$ is idempotent, then

$$A^2 = A$$

If $A$ is non-singular, then the only possible matrix that is idempotent is $A=I$.

But if $A$ is singular, then are there infinitely many idempotent matrices. I got two matrices by brute force such as

$$A=\begin{bmatrix} 1 &1 \\ 0& 0 \end{bmatrix}$$

and

$$A=\begin{bmatrix} 0 &0 \\ 1& 1 \end{bmatrix}$$

How can we find any other matrices which are singular and idempotent?

$\endgroup$

3 Answers 3

5
$\begingroup$

Any projection on a proper subspace is idempotent. In particular, for column vector $v$ and row vector $u^T$ such that $u^T v = 1$, $v u^T$ is idempotent.

In the $2 \times 2$ case, this gives you the examples

$$ \pmatrix{u_1 v_1 & u_2 v_1\cr u_1 v_2 & u_2 v_2\cr} $$ whenever $u_1 v_1 + u_2 v_2 = 1$.

we can write this as

$$ \pmatrix{a & b\cr \frac{a-a^2}{b} & 1-a\cr}$$ for any $b \ne 0$.

$\endgroup$
1
$\begingroup$

Equating entries, you can check that the following is idempotent for any $\;\theta\in\Bbb R\;$:

$$\frac12\begin{pmatrix}1-\cos\theta&\sin\theta\\\sin\theta&1+\cos\theta\end{pmatrix}$$

$\endgroup$
2
  • $\begingroup$ So is it true that Trace of any Singular idempotent matrix is unity? $\endgroup$ Jul 25, 2016 at 17:35
  • $\begingroup$ @Umeshshankar Yes, it is...unless, of course, it is zero (for example, the zero matrix. But this one is diagonal...) $\endgroup$
    – DonAntonio
    Jul 25, 2016 at 17:45
0
$\begingroup$

$$\boxed{\textbf{HINT}}$$ Take advantage of the fact that $$a_{ij}=\sum_k a_{ik}a_{kj}$$

Then start $$a_{11}=a_{11}^2+a_{12}a_{21}$$ $$a_{22}=a_{22}^2+a_{21}a_{12}$$ $$a_{12}=a_{11}a_{12}+a_{12}a_{22}$$ $$a_{21}=a_{21}a_{11}+a_{22}a_{21}$$

then $$a_{11}+a_{22}=1$$ etc..

$\endgroup$

You must log in to answer this question.

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