How to prove this $A$ is an invertible matrix

Question

let Symmetric matrix $A=(a_{ij})_{n\times n},n\ge 2$,and $$\begin{cases} a_{jk}=j+k\cdot i&j< k\\ a_{jj}=2j\cdot(i+1) \end{cases}$$ where $i^2=-1$

show that :$A$ is Invertible matrix

My idea: I want to find this value $\det(A)=?$, or maybe don't have closed form?

when $n=2$,then $$A=\begin{bmatrix} 2(i+1)&1+2i\\ 1+2i&4(i+1) \end{bmatrix}$$

so $$det(A)=8(i+1)^2-(1+2i)^2=16i-(4i^2+4i+1)=3+12i?$$

and this problem is from china linear algebra problem book ,and this book most of problem is very hard.and this problem is last at this book.

Thank you for you help me

Answer 1

Note that $$ A=\boldsymbol{n}\boldsymbol{1}^t+i\boldsymbol{1}\boldsymbol{n}^t+(1+i)\mathrm{diag}(\boldsymbol{n}), $$ where $\boldsymbol{n}=(1,2,\ldots,n)$, $\boldsymbol{1}=(1,1,\ldots,1)$.

We need to show that $A\boldsymbol{u}=0$, implies that $\boldsymbol{u}=0$.

We have $$ A\boldsymbol{u}=\boldsymbol{n}\boldsymbol{1}^t\boldsymbol{u}+i\boldsymbol{1}\boldsymbol{n}^t \boldsymbol{u}+(1+i)\mathrm{diag}(\boldsymbol{n})\boldsymbol{u}= (\boldsymbol{1},\boldsymbol{u})\boldsymbol{n}+i(\boldsymbol{n}, \boldsymbol{u})\boldsymbol{1}+(1+i)\mathrm{diag}(\boldsymbol{n})\boldsymbol{u}. \tag{1} $$ If $\boldsymbol{u}=(u_1,\ldots,u_n)$, then $A\boldsymbol{u}=0$, implies that $$ \mathrm{diag}(\boldsymbol{n})\boldsymbol{u}=(u_1,2u_2,\ldots,nu_n), $$ is a linear combination of $\boldsymbol{1}$ and $\boldsymbol{n}$, i.e., $$ (u_1,2u_2,\ldots,nu_n)=c_1(1,1,\ldots,1)+c_2(1,2,\ldots,n), \tag{2} $$ with $c_1$ and $c_2$ not both vanishing. Plugging $(2)$ to $(1)$ we get $$ 0=A\boldsymbol{u}=(\boldsymbol{1},c_1\boldsymbol{1}+c_2\boldsymbol{n})\boldsymbol{n} +i(\boldsymbol{n}, c_1\boldsymbol{1}+c_2\boldsymbol{n})\boldsymbol{1}+(1+i)\big(c_1\boldsymbol{1}+c_2\boldsymbol{n}\big). $$ Equivalently $$ i(\boldsymbol{n},c_1\boldsymbol{1}+c_2\boldsymbol{n})+c_1(1+i)=0 \tag{$A_1$} $$ and $$ (\boldsymbol{1},c_1\boldsymbol{1}+c_2\boldsymbol{n})+c_2(1+i)=0. \tag{$A_2$} $$ It is not hard to calculate the $2\times 2$ determinant of the system $(A_1)-(A_2)$ and obtain that the only solution is $c_1=c_2=0$.

Answer 2

Since $A$ is complex symmetric, we have $\operatorname{Re}(u^\ast Au)=u^\ast \operatorname{Re}(A)u$ for any complex vector $u$. So, to prove that $A$ is invertible, it suffices to show that its real part is positive definite. Let $$ X=\pmatrix{ 2&1&1&1&\cdots\\ 1&4&2&2&\cdots\\ 1&2&6&3&\cdots\\ 1&2&3&8&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots}, \ L=\pmatrix{ 1&0&0&0&\cdots\\ 1&1&0&0&\cdots\\ 1&1&1&0&\cdots\\ 1&1&1&1&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots} $$ and $D=\operatorname{diag}(1,2,3,\ldots)$. The real part of $A$ is given by $$ X = LD+(LD)^T. $$ By Sylvester's law of inertia, $X$ is positive definite if and only if $Y = L^{-1}X(L^{-1})^T = D(L^{-1})^T + L^{-1}D$ is positive definite. It is easy to see that $Y$ is a symmetric tridiagonal matrix whose main diagonal is $(2,4,6,8,\ldots)$ and whose superdiagonal is $(-1,-2,-3,\ldots)$: $$ Y=D(L^{-1})^T + L^{-1}D =\pmatrix{ 2&-1\\ -1&4&-2\\ &-2&6&-3\\ &&-3&8&\ddots\\ &&&\ddots&\ddots}. $$ Therefore, by Gershgorin disc theorem and the symmetry of $Y$, all eigenvalues of $Y$ are real numbers bounded below by $1$. Hence $Y$ and in turn $X$ are positive definite.