What Is the Maximum Number of Vectors with Non-Positive Inner Products in R^n?

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1. Suppose v_1, v_2, . . . v_k are non-zero vectors is R^n such that (v_i,v_j)<= 0 for all i,j. Determine, with proof, the maximal possible k for n = 3, and also for arbitrary n.
2. A is 2x2, A(1,1)=A(2,2)=x-1, A(1,2)=1, A(2,1)=0. Find invertible P,Q such that
P*A*Q is diagonal. I tried singular value decomposition, but calculations are getting nasty.
Please, give me hints for these problems. Thanks a lot
 
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tohauz said:
1. Suppose v_1, v_2, . . . v_k are non-zero vectors is R^n such that (v_i,v_j)<= 0 for all i,j. Determine, with proof, the maximal possible k for n = 3, and also for arbitrary n.
2. A is 2x2, A(1,1)=A(2,2)=x-1, A(1,2)=1, A(2,1)=0. Find invertible P,Q such that
P*A*Q is diagonal. I tried singular value decomposition, but calculations are getting nasty.
Please, give me hints for these problems. Thanks a lot

OK, I got the first one, hopefully somebody can help me 2nd.
1) Answer is 2n.
Use <u,v>=|u|*|v|*cosa, it is nonpositive if a>=90 degrees.
So in R^3 take i,j,k,-i,-j,-k. If you want to squeeze in other vector the angle
between that and 3 of those is less than 90.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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