## Matrix is invertible - proof. Tikhonov regularization and least squares problem.

Hello,
I have a big problem with proofing, that :

$A^TA+\lambda I$
is invertible.

$A^TA$ is invertible only if $rank(A^TA) = M$, where M is size of a matrix. Now i have to proof that, adding lambda times identity matrix will make the whole expression invertible.
I have a small advice from my professor: if A is a positive-definite matrix then there is a $A^(-1)$.
I've tried the standard approach
$x \epsilon N(A^TA+\lambda I) \Rightarrow x^T(A^TA+\lambda I)x = x^T 0$ but I've failed.
N means null space.
The whole task is to proof that least square with tikhonov regularization has only one exact solution, but I figure out that when above statement is invertible, we can compute exact solution.
I would really be great full, if someone could give me some advice.