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I is an identity matrix with appropriate dimension. How to prove for any given matrix A ∈ R^m×n and any positive number µ :

Both $A^TA + \mu I$ and $AA^T + \mu I$ are invertible.

One way to prove matrix is invertible is to prove the det of matrix is not equal to zero, but i have no idea about it, really need some help.

P Vanchinathan
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Haoming Li
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    Use correct formatting. Put a "$" at the beginning and at the end of every math notation. Please. – petru Feb 14 '17 at 02:39

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Negation of the statement is that $A^TA+\mu I$ is NOT invertible for $\mu>0$.

That means $A^TA (v)=-\mu I(v)$ for some non-zero vector $v$. So your problem boils down to showing that $A^TA$ cannot have real negative eigenvalue.

Now note that $A^TA $ (and $AA^T$) is a symmetric matrix. For a given vector $v$ calculate the norm of $A^TAv$. This norm is is $v^TA^TA A^TAv\geq0$. Can you complete the proof now, choosing $v$ as an eigenvector?

P Vanchinathan
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  • I did not say norm of a matrix. Read it again: I said norm of $A^TAv$ which is a vector. To get its norm, the inner product (with itself) is obtained as $v^tA^TAA^tv$. – P Vanchinathan Feb 15 '17 at 04:51