# Why the sum of cosines between "v" and any vector =1?

1. The problem statement, all variables and given/known data
Given that matrix, A can be decomposed using SVD (Singular Value Decomposition) into $A=USV^T$, why does always the sum of the square of cosines between v vectors and any other column vector q representation of arbitrarily column vector Q vector sum up to 1?

2. Relevant equations
$A=USV^T$. $Q=USq$

3. The attempt at a solution
I tried a simple 2x2 matrix but even with this simple matrix, the calculation goes missy. In addition, I seek a rigorous proof.

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#### Stephen Tashi

the sum of the square of cosines between v vectors and any other column vector q representation of arbitrarily column vector Q vector sum up to 1?
It isn't clear (to me) what those words mean.

$1 = \sum_{k =1}^N \frac{ \overrightarrow{v_k} \cdot \overrightarrow{q_j}}{ |\overrightarrow{v_k}||\overrightarrow{q_j}|}$ (?)

Exactly, $1 = \sum_{k =1}^N (\frac{ \overrightarrow{v_k} \cdot \overrightarrow{q_j}}{ |\overrightarrow{v_k}||\overrightarrow{q_j}|})^2$
I discovered this fact by coincidence but it turns out that it may have a nice link to the quantum mechanics.
For example, if the cosine of the angle represents the inner products (the eigen function which is also the inner product between the system state and the eigen state in, say position representation), then the sum of the cosine square is equal to 1. In other words, if the quantum system is complete, then all information of the system is encoded in the wave-function(s). Consequently, it is natural to think about the square of the wave function (or the square of the cosine) as a probability amplitude and because the system is complete, all probabilities should sum up to 1.

#### marcusl

Gold Member
You are projecting an arbitrary unit vector onto a complete orthonormal basis set, and then have defined (or "discovered") the standard Euclidean vector norm.

"Why the sum of cosines between "v`" and any vector =1?"

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