In Linear regression, we have &#x03B8;<!-- θ --> = (

Carley Haley

Carley Haley

Answered question

2022-05-08

In Linear regression, we have θ = ( X T X ) 1 X T y .
In Ridge regression, we have θ = ( λ I + X T X ) 1 X T y .
I learnt somewhere that while X T X is not guaranteed to be invertible, λ I + X T X is guaranteed to be invertible.
Is this true? If so, why?

Answer & Explanation

Marco Meyer

Marco Meyer

Beginner2022-05-09Added 16 answers

Can you write the eigenvalues of λ I + X X in terms of the eigenvalues of X X?
The eigenvalues of X X are nonnegative. (Why?) It is possible for it to be not invertible (specifically, if some of the eigenvalues are zero). However, the eigenvalues of λ I + X X are of the form λ + μ where μ is an eigenvalue of X X. (Why?) Since λ > 0 and μ 0, the eigenvalues of λ I + X X are all strictly positive, so it is invertible.
Direkotogbkmn

Direkotogbkmn

Beginner2022-05-10Added 5 answers

First, replace λ λ ~ . Then the given expression
X T X λ ~ I
is similar to the eigenvalue equation of the matrix X T X. As X T X is a positive semi-definite matrix we know that its eigenvalues are all 0. This implies that for λ the corresponding values are 0. If you state that λ > 0 then you rule out all possible eigenvalues. Hence, the eigenvalue equation
det [ X T X λ ~ I ] = 0
does not have any solutions. Hence,
[ X T X λ ~ I ] = [ X T X + λ I ] is invertible for λ > 0

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