>>7961975
Yeah. Math always finds a way.

A * At * A = (n/3) * A
A * At * A * At = (n/3) * A * At
(A * At)^2 = (n/3) * (A * At)
A*At = (n/3)

>>7962022
Illegal, AAt is not invertable if A is non-invertable, and A can only be invertable when n = 3; which means AAt cannot be a scalar. Instead it would be only true when At is its own inverse.

>>7962040
Sorry, At has to be the inverse of A.

>>7961975
a lot

>>7961975
Can you find one such solution? It's good to know if the set you're looking for is nonempty. Also, in finding one explicit solution, you can generate conjectures as to the general solution.

>>7962071
>>7962022

So it works when n=3 and At is the inverse of A?

>>7962086
Yeah. At least. There's probably more solutions, because A can also be an eigenvector of AAt with eigenvalue n/3

>>7962086
Just thought of one, when A is 1x 3 and AAt = 1/3

>>7962071

One solution is

[math]\begin{bmatrix}
\sqrt{\frac{2}{3}}& -\sqrt{\frac{2}{3}}& 0 & 0\\
0 & 0 & \sqrt{\frac{2}{3}} & -\sqrt{\frac{2}{3}}\\
-\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}\end{bmatrix} [/math]

This is a solution for n=4

>>7961975

I am the original poster A is a 3Xn matrix not a nX3 matrix.

>>7961975
if A^T A is n/3*I , then you have a solution. If A is nx3, just take any 3 orthogonal vectors and scale them to length sqrt(n/3).

>>7962133
If A is 3xn, then look for A A^T = (n/3)*I...
take 3 orthogonal n-vectors of euclidean length sqrt(n/3), put them in the rows of A.

>>7962142
>If A is nx3, just take any 3 orthogonal vectors and scale them to length sqrt(n/3).

This is true; However, I made a mistake in the original post. The matrix A is 3Xn.

>>7962150

I agree. I was hoping the suction space might be larger. How to find a sition to this? Could power iteration work? The lanczos algorithm?

>>7962150

n=10;
A=randn(3,n); % A random matrix

% Graham-Schmidt to make A orthogonal
A(1,:)=A(1,:)/norm(A(1,:));
A(2,:)=A(2,:)-(A(1,:)*A(2,:)')*A(1,:);
A(2,:)=A(2,:)/norm(A(2,:));
A(3,:)=A(3,:)-(A(1,:)*A(3,:)')*A(1,:);
A(3,:)=A(3,:)-(A(2,:)*A(3,:)')*A(2,:);
A(3,:)=A(3,:)/norm(A(3,:));

% scale to length sqrt(n/3)
A=A*sqrt(n/3);

% test that A satisfies equation
max(max(abs( A*A'*A - n/3*A )) )

>>7962166
If A*A'*A - n/3*A = 0, and B=A*A', then I think you get
B*(B - n/3*I)^2 = 0
Maybe you could find some such B, then work backward to get some A.

>>7961975
There's an infinite number of solutions for n=1 alone. Any solution A = (a, b, c) where a^2 + b^2 + c^2 = 1/3 works. In other words, the n=1 case is the set of points on the sphere of radius 1/sqrt(3).

>>7962204
This works, it seems. More general. I am pleased. Enjoy, OP!

clear all
E=randn(3,3); % E random 3x3 matrix

% Graham-Schmidt to make E orthogonal
E(1,:)=E(1,:)/norm(E(1,:));
E(2,:)=E(2,:)-(E(1,:)*E(2,:)')*E(1,:);
E(2,:)=E(2,:)/norm(E(2,:));
E(3,:)=E(3,:)-(E(1,:)*E(3,:)')*E(1,:);
E(3,:)=E(3,:)-(E(2,:)*E(3,:)')*E(2,:);
E(3,:)=E(3,:)/norm(E(3,:));

n=5;

% find symmetric B
% such that F(B)=0, where F(B)= B*(B - n/3*I)^2
% make matrix with characteristic polynomial F
B=E'*diag([0,n/3,n/3])*E;

% test
max(max( abs( B*(B - n/3*eye(3))^2 )))

% now find (nx3) A such that A*A' = C*C',
% where C*C'=B. We can do this by finding
% nx3 orthogonal E, then setting A = C*E

C=E'*diag([0,sqrt(n/3),sqrt(n/3)])*E;

A=randn(3,n); % A random matrix

% Graham-Schmidt to make A orthogonal
A(1,:)=A(1,:)/norm(A(1,:));
A(2,:)=A(2,:)-(A(1,:)*A(2,:)')*A(1,:);
A(2,:)=A(2,:)/norm(A(2,:));
A(3,:)=A(3,:)-(A(1,:)*A(3,:)')*A(1,:);
A(3,:)=A(3,:)-(A(2,:)*A(3,:)')*A(2,:);
A(3,:)=A(3,:)/norm(A(3,:));

A=C*A;

max(max(abs( A*A'*A - n/3*A )) )