# real matrix with $Tr((A-I)^{T}(A-I) )<1$

$A$ is a $n\times n$ real matrix, $$\operatorname{Tr}((A-I)^{T}(A-I) )<1$$ then $\det(A)\ne0$.

well, $$\sum_{i\ne j}a_{ij}^2+\sum (1-a_{ii})^2\lt1$$

How to derived $\det(A)\ne0$?

Thank you!

Note that $B=(A-I)^{T}(A-I)$ is a positive semi-definite Matrix and therefore all the eigenvalues $\lambda_i$'s are non-negative: $$\lambda_i\geq 0.$$ Now if $\det(A)=0$, then there is a non-zero $x$ such that: $$Ax=0\implies (A-I)x=-x$$ So $\mu=-1$ is an eigenvalue of $A-I$ and we have: $$x^TBx=\|x\|^2.$$ On the other hand we have: $$\operatorname{Tr}((A-I)^{T}(A-I) )<1\implies \sum_{i=1}^n \lambda_i<1 \implies 0\leq\lambda_i<1.$$ Now according to Spectral Theorem, there is an orthonormal basis $v_1,\dots,v_n$ of eigenvectors of $B$. This means that for $x$ chosen above we have: $$x=\alpha_1v_1+\dots+\alpha_n v_n\implies \|x\|^2=\sum_{i=1}^n \alpha_i^2.$$ But we can see that: $$\sum_{i=1}^n \alpha_i^2=\|x\|^2=x^TBx=\sum_{i=1}^n\lambda_i\alpha_i^2<\sum_{i=1}^n\alpha_i^2,$$ which is a contradiction.
We can actually prove that $\det(A)$ is positive (not just nonzero). The trace term in the given condition is the square of the Frobenius norm of $A-I$. So, from the given condition, we get $\rho(A-I)\le\|A-I\|_F<1$. Therefore, if $A$ has a real eigenvalue $\lambda(A)$, then $-1<\lambda(A-I)=\lambda(A)-1$, i.e $\lambda(A)$ is positive. Since nonreal eigenvalues of $A$ appear in conjugate pairs, and the product of every conjugate pair of nonreal complex numbers is positive, we conclude that $\det(A)>0$.