Find $\min(\operatorname{trace}(AA^T))$ for invertible $A_{n\times n}$ For invertible $A_{n\times n}$ find $\min(\operatorname{trace}(AA^T))$
(a) $0$
(b) $1$
(c) $n$
(d) $n^2$
Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it.

I will assume all the entries are integers.  By multiplying out the matrices, you see that
$$ \text{trace}(A^TA) = \sum_{i,j} a_{i,j}^2 .$$
Suppose $\text{trace}(A^TA) < n$.  Then at most $n-1$ entries must be non-zero, meaning that at least one column of the matrix is all zeros.  Hence $A$ is not invertible.
Hence $\text{trace}(A^TA) \ge n$.  We know that the value $n$ can be obtained (e.g. by the identity, or any permutation matrix).
