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Suppose that $A$ is an $n \times m$ ($n$ less than $m$) full rank matrix. Apply Gram-Schmidt orthogonalization to the rows of $A$, then we get an $n \times m$ matrix $B$ with orthonormal columns. According to the Gram-Schmidt algorithm, there exists a lower triangular matrix $L_m$, such that $L_m A=B$.

Let $C=AA^*$, if the Frobenius norm of $C-I_n$, $\|C-I_n\|=a$, can we calculate the Frobenius norm of $L_m C^{1/2}-I_n, \|L_m C^{1/2}-I_n\|$, where $C^{1/2}$ is the square root of $C$, and $I_n$ is the $n$ order identity matrix?

At least, can we prove that $\|L_m C^{1/2}-I_n\|$ is smaller that $\|C-I_n\|$?

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1 Answer

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HINT: Try running this in Matlab (or Octave).

iter=1;
n=3;
m=4;
while true A=randn(n,m); [Q R]=qr(A',0); L=inv(R)'; % B in question is Q' C=A*A'; stuff=[norm(C-eye(n),'fro'), norm(L*C^(1/2)-eye(n),'fro')]; if stuff(2) > stuff(1) fprintf('Found a counter-example at step %d!\n', iter); break; end; iter=iter+1;
end

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