Where $X$ is an $N{\times}1$ vector, $A^{-1}$ is a $N{\times}N$ symmetric positive semi-definite matrix, and the elements of A=x[n] @x[-n],where A is the auto-correlation matrix, where @ means the convolution operator. $\sigma^2$ is a constant and $I$ is the $N{\times}N$ identity matrix. What I tried to do is to argue that only the diagonal element of $A$ will change with an addition of a constant so that should not affect the maximum of the function itself. As a Note: In fact for my problem $A$ also depends on $X$ hence the function $f$ is a function of the vector $X$ and the matrix $A$ is also depends on the vector values in $X$.
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$f$ depends on $X$? – Maximilian Janisch Oct 05 '19 at 17:47
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Yes actually in fact A also depends on X hence the function f is a function of the vector X and the matrix A is also depends on the vector values in X. – Jordan Oct 05 '19 at 17:53
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1You left out one important information: How does $A$ depend on $X$? Also, is $X\in\mathbb R^n$ arbitrary? I mean, can the values in $X$ be arbitrarily large in absolute value? – amsmath Oct 05 '19 at 18:00
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I added all the information needed thank you for pointing that out – Jordan Oct 05 '19 at 20:24