Condition for existence of an orthonormal matrix whose column space is orthogonal to the column spac

Raul Walker 2022-07-01 Answered
Condition for existence of an orthonormal matrix whose column space is orthogonal to the column space of another matrix
While I was reading a statistics paper, I came across one statement that I don't understand (I just have basic linear algebra knowledge).
Assume (in the context of regressions), we have a n × p data matrix X, assuming that X is invertible and n>p. The paper states
" U R n × p is an orthonormal matrix whose column space is orthogonal to that of X s.t. U T X = 0": such matrix exists if n 2 p. I don't understand where the last statement comes from.
I know that the nullspace of X has dimension n−rank(X)=n−p in full rank case and U is the orthonormal basis of the null space of X. But I don't get the link why U only exists, if n p + r a n k ( X ), i.e. n 2 p.
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Answers (1)

billyfcash5n
Answered 2022-07-02 Author has 17 answers
Note: it is not standard to refer to the matrix X as "invertible" unless X is a square matrix. Presumably, "X is invertible" refers in this case to the fact that X has full rank. In this case, because X is n × p with n>p, this means that X has rank p (i.e. has "full column rank").
As you say, " U R n × p is an orthonormal matrix whose column space is orthogonal to that of X". The fact that the column space is orthogonal to that of X is equivalent to the statement that U T X = 0. Because U is n × p with orthonormal columns, the dimension of its column space is p. Because the column space of U is orthogonal to that of X, the column space of U must be a subspace of the orthogonal complement to the column space of X. The column space of X is a p-dimensional subspace of R n , which means that its orthogonal complement has dimension n−p.
Putting all this together leads us to the following conclusion: col ( U ) col ( X ) dim ( col ( U ) ) dim ( col ( X ) ) p n p 2 p n .
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