Does there exists a matrix who's product is itself?

[mscudder3]

To simplify this inquiry let us deal with R2. I know that if you have the identity matrix (let that be A), then AA=A. However, I recall the existence of a matrix with all nonzero entries that had the same property. Thinking of rotations, I cannot think of one, (since rotating by 0, or 2pi is equivalently the identity matrix). I'm sure that there must exist some type of orthogonal projection or reflection, but I'm unsure.

Thanks in advance!

 

[Hurkyl]

I'm sure that there must exist some type of orthogonal projection or reflection, but I'm unsure.

So instead of wondering, why not actually try an orthogonal projection or a reflection?


(aside -- you could always reduce the problem to an ordinary algebra problem with real numbers)

 

[mscudder3]

So instead of wondering, why not actually try an orthogonal projection or a reflection?


(aside -- you could always reduce the problem to an ordinary algebra problem with real numbers)

... I was wondering as in I cannot recall or figure it out.
Reducing it to algebra in terms of some a1 a2 a3 a4 (forming a 2x2), all that resolves is trivial solutions.

Would it be that difficult to show me one?

 

[micromass]

Would it be that difficult to show me one?

No, but you would learn more if you construct one yourself.

Now, take the line y=x in $\mathbb{R}^2$. Can you construct the matrix of the orthogonal projection on that line?

 

[mscudder3]

Let A=[a1 a2 ; a3 a4]. since y=x the vector x=(x1, x2)^t=(1, 1)^t
From the definition of a projection= [u1^2, u1*u2 ; u1*u2 u2^2]
I can see that if a1=a2=a3=a4=1/2 then,AA=A. However, I'm not quite sure on how/why?

Edit:
From Ax=b let x=(1 1)^t and A=projection and solve for u1 and u2 (which yields 1/2 for each entry). Then by the def of projection, b must be equal to x?

 

[micromass]

From the definition of a projection= [x1^2, x1*x2 ; x1*x2 x2^2]

Umm, I'm not sure where you got that formula... But the matrix

$$\left(\begin{array}{cc} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\\ \end{array}\right)$$

Certainly satisfies A²=A. So you've found your matrix!

 

[mscudder3]

Ya thanks! Now that you have me thinking I'd like to understand exactly how I arrived at this solution. I got that equation (or matrix) from my text: Linear Algebra 4e by Bretscher. Book is VERY heavy on application, thus my struggle with the general theory.

 

[micromass]

Ya thanks! Now that you have me thinking I'd like to understand exactly how I arrived at this solution. I got that equation (or matrix) from my text: Linear Algebra 4e by Bretscher. Book is VERY heavy on application, thus my struggle with the general theory.

It's actually very easy once you know it!

Just take the standard basis of $\mathbb{R}^2$, that is: (1,0) and (0,1). The key point is checking where they are projected. A little bit of drawing (or calculating) gives us that (1,0) is being sent to (1/2,1/2) and (0,1) is being sent to (1/2,1/2). Thus (1/2,1/2) and (1/2,1/2) make up the columns of our matrix.

Note that there was nothing special about projecting on the line y=x. You can pick any line through the origin, they will all give you a matrix such that A²=A.

For example, let me project on y=-x. Then (1,0) is being sent to (1/2,-1/2) and (0,1) is being sent to (-1/2,1/2). Thus the matrix becomes

$$\left(\begin{array}{cc}\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2}\\ \end{array}\right)$$

As you can see this matrix also satisfies A²=A.

 

[mscudder3]

Great! Thanks so much.
To briefly expand on your explanation, when you say it sends the basis (by the linear transform), how exactly do these numbers come about mathematically. I know I may be overlooking something, but I want to be complete.

From what I see, the length of the basis is clearly 1 (hence unit vector). Thus if that is projected onto y=x the new basis would be 1/sqrt(2), not 1/2.

 

[NaturePaper]

Hello mscudder3,
Take any unit column vector a (i.e., nx1 matrix with sum of square of moduli of its elements being 1, where n is the number of your choice). Then the matrix m=$$aa^\dagger$$ will satisfy $$m^2=m.$$