8
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Have you done a small case to make sure it is plausible? $\endgroup$ – Trevor Feb 19 '14 at 16:19
  • $\begingroup$ The determinant does not behave well under sums, so your line of reasoning might not be fruitful. However, your diagonal grows pretty rapidly, so it might be possible to show that the product of the diagonal terms dominates all other diagonals. That is, you would show that $2\cdot4\cdot\cdots\cdot2n(i+1)^n$ dominates the other diagonals for all values of $n$. $\endgroup$ – Trevor Feb 19 '14 at 16:45
2
$\begingroup$

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 $c_1=c_2=0$.

$\endgroup$
3
  • $\begingroup$ You are talking about a different $A$. The $A$ in the OP is complex symmetric, yours is not. $\endgroup$ – user1551 Feb 12 '15 at 8:38
  • $\begingroup$ Right. The matrix is not symmetric. But how does this change anything? $\endgroup$ – Yiorgos S. Smyrlis Feb 12 '15 at 17:12
  • $\begingroup$ Hmm, the OP asked whether his/her $A$ is invertible, but you showed that another $A$ is invertible. It is not clear how your answer addresses the OP. $\endgroup$ – user1551 Feb 12 '15 at 17:48
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.