Suppose I have an n xx n matrix A with positive real entries. Is it possible to find an n xx 1 vector v such that vv^T=A where v^T is the transpose of v?

figoveck38

figoveck38

Answered question

2022-11-07

Suppose I have an n × n matrix A with positive real entries. Is it possible to find an n × 1 vector v such that
v v T = A
where v T is the transpose of v?
Is it possible to use singular value decomposition or any technique to achieve this?

Answer & Explanation

Calvin Maddox

Calvin Maddox

Beginner2022-11-08Added 15 answers

The rank of v v T is 1 or 0, so only square matrices of rank 1 or 0 can have such a representation.
If A=0 then v = 0
In general if v = ( a 1 , . . . , a n ) T then v v T has rows a 1 v T , . . . , a n v T , so you can find v by looking at the diagonal entries of A.
vidamuhae

vidamuhae

Beginner2022-11-09Added 3 answers

v v T has rank 1 , so such a decomposition exists only if A has rank 1.
If the decomposition exists, v is formed from taking square roots of the entries on the diagonal of A since a i , i = v i 2 . You will get 2 k possible v's where k is the number of non-zero entries.

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