# A matrix being symmetric/orthogonal/projection matrix/stochastic matrix

I am trying to do some practice questions and wanted to check the following properties and confirm my definition of projection matrix:

Let $$A = \left[\begin{matrix} 1/2 & 0 & 1/2 \\ 0 & 1 & 0 \\ 1/2 & 0 & 1/2 \end{matrix}\right]$$

Is the following true or false ?

i) $A$ is orthogonal

ii) $A$ is a projection matrix

I think that $A$ is not orthogonal since $AA^T \neq I$ but is a projection matrix given that a projection matrix is defined as $P^2 = P$.

Also accordingly, a projection matrix is orthogonal iff $P^T = P$ and in that case, is an orthogonal projection matrix different from the idea of an orthogonal matrix itself ?

It appears that $A^T = A$ but yet $AA^T \neq A$.

Would someone mind helping me fill up the gap for this misconception, any advice is appreciated.

Thank you

• I would like to be corrected, but projection matrices has always been defined to be $P^2 = P$ in my courses. – user305860 Jun 18 '16 at 7:43
• Ah yes, you are right, that is correct. It turns out that the $P^* = P$ condition imposes an orthogonal property for $P$. Does that mean here that the matrix,$A$, is also an orthogonal matrix or is the orthogonal projection matrix different from an orthogonal matrix since I thought that for a matrix to be orthogonal, it must be that $QQ^T = I$ right ? – PutsandCalls Jun 18 '16 at 7:55

## 1 Answer

First of all, pick one: either $A^*$ or $A^T$. In this context, they mean the same thing.

i) $A$ is not orthogonal because $AA^* \neq I$.

ii) $A$ is a projection because $A^2 = A$. It is, in fact, an orthogonal projection because $A = A^*$, in addition to the fact that $A$ is already a projection. That is, a projection that is symmetric is an orthogonal projections.

Note that orthogonal projections are not generally orthogonal in the sense of an "orthogonal matrix". That is, a matrix satisfying $A^2 = A$ and $A = A^*$ will not usually satisfy $AA^* = I$. "Orthogonal projections" are given their name because they project orthogonally onto their image.