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$(A^TA+aI)^{-1}A^T = A^TX$

where A and X are real-matrices and a is a positive non-zero real number. A is non-zero and you may assume that A is such that there is a unique X that satisfies the above equation.

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up vote 2 down vote accepted

Using the inverse of the inverse so to speak:

$$A^T=(A^TA+aI)A^TX$$ $$A^T=A^T(AA^T+aI)X$$

and now if $X=(AA^T+aI)^{-1}$ it is solved.

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$(A^TA+aI)A^T=(A^TAA^T+aIA^T)=A^T(AA^T+aI)$ – adam W Oct 4 '12 at 0:52
Sorry, i forgot to put -1 in the power – Karan Oct 4 '12 at 0:54
I have changed the question now – Karan Oct 4 '12 at 0:54
Thanx, this is perfect – Karan Oct 4 '12 at 1:16
Glad I could help! – adam W Oct 4 '12 at 1:19

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