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Sry too simple question probably.
But I have the matrix X'X (=A) given, and have to find X. X is symmetric. Is there an easy trick? thanks
What is X′? Is it the transpose of X? If it is, then you should know that if X is symmetric, its transpose is equal to X itself.
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hi yes thats clear.
X' is the transpose of X.
but (X'X) is given. how to get the single X or X', respectively?
thanks
Whats the size of X? If X is only 2×2, you can let
and try and solve for X. If X is a bigger matrix, well its a bit more complicated.
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it's actually a 4x4 matrix
What's A? It's a hard problem to solve in general, but if A is nice we might have some shortcuts.
Especially if it's got a lot of 0's.
Why did the vector cross the road?
It wanted to be normal.
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I guess you should take primary manipulation both horizontally and vertically until you reduce it to simple I matrix, that is
A=P[sub]n[/sub]...P[sub]3[/sub]P[sub]2[/sub]P[sub]1[/sub]IP'[sub]1[/sub]P'[sub]2[/sub]P'[sub]3[/sub]...P'[sub]n[/sub]
X'(y-Xβ)=0
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