I don't understand how to prove that a matrix is definite positive using Sylvester's criterion or Gershgorin's theorems

Question

I have the following matrix: $$A=\left[ \begin{array}{cc} 1 & i & \\ 3i & 2 & i & \\ & 3i & 3 & i & \\ & & \ddots & \ddots & \ddots & \\ & & & 3i & n-1 & i \\ & & & & 3i & n \end{array} \right]$$ And the following exercise:

Locate the eigenvalues of $\operatorname{Re}(A)$ as precisely as possible using only Gershgorin’s theorems, and state whether the matrix $A$ is positive definite, justifying your answer.

Now, I managed to solve the first point by calculating the matrix $\operatorname{Re}(A)=\frac{A+A^\ast}2$ and the corresponding Gershgorin's circles. After doing so, I found that the eigenvalues of this matrix are located in the $[0, n+1]$ interval, so $0\le\lambda_i\le n+1$ but I am having a lot of trouble proving the second point.

Now, in my course, we only used Sylvester's criterion to prove that a matrix is positive but I don't really understand how to apply it here. Our professor told us that if a matrix $A$ is not Hermitian, in order to prove that it is positive definite I need to apply Sylvester's criterion to $\operatorname{Re}(A)=\frac{A+A^\ast}2$ but again, it didn't help me here.

I also know that if a matrix is Hermitian and its eigenvalues are real and positive, the matrix is also definite positive. Now, $\operatorname{Re}(A)$ is Hermitian but since the eigenvalues are in the range $0$ to $n+1$, I can't say they are all positive since I can't prove that $0$ is not one of the eigenvalues.

I don't really know what else to try. Any kind of help is appreciated.

Answer 1

For the $n\in \big\{1,2\big\}$ case, one can easily apply Sylvester's Determinant Criterion so I assume $n\geq 3$ and apply a Max Modulus type of argument. Consider $\frac{1}{2}\cdot Re(A) =D +M$ where $D$ is a positive diagonal matrix, with $d_{k,k} =\frac{k}{2}$, and $M$ contains the off-diagonal elements and the modulus of entries in each row sums to one, except the 1st and nth rows where the modulus sums to $\frac{1}{2}$. Suppose $\mathbf x \in \ker Re(A)\implies D \mathbf x = -M\mathbf x$

For $k=1$: $\vert x_1\vert =\vert x_2\vert$
For $2\leq k \leq n$, by application of triangle inequality:
$\vert x_k\vert\leq d_{k,k} \cdot\vert x_k\vert =\vert \mathbf e_k^T D\mathbf x\vert = \vert \mathbf e_k^T M\mathbf x_k\vert \leq \frac{1}{2}\big(\vert x_{k-1}\vert +\vert x_{k+1}\vert \big)$
[with the convention $x_{n+1}:=0$]

if $x_{k^*}$ is a coordinate with maximum modulus, then it immediately follows that its 'neighbors' have the same modulus'-- e.g. $\vert x_{k^*}\vert\leq \frac{1}{2}\big(\vert x_{k^*-1}\vert +\vert x_{k^*+1}\vert \big)\leq \vert x_{k^*}\vert$ and the $k^*=1$ case is immediate. Thus $\vert x_{r}\vert =\vert x_{k^*}\vert $ for some $r\geq 3$
$\implies \vert x_{k^*}\vert\leq d_{r,r}\cdot \vert x_{k^*}\vert= d_{r,r}\cdot \vert x_r\vert \leq \frac{1}{2}\vert x_{r-1}\vert+\frac{1}{2}\vert x_{r+1}\vert \leq \vert x_{k^*}\vert$
but $d_{r,r}\gt 1$ so the 1st inequality being met with equality implies $x_{k^*}=0$. Conclude that $\mathbf x =\mathbf 0\implies \dim \ker Re(A)=0$ so $Re(A)$ is invertible.

a better argument
$\det\Big( Re(A)\Big)\neq 0$ follows immediately from Taussky's refinement of Gerschgorin Discs, since $Re(A)$ is irreducible, weakly diagonally dominant and the dominance is strict in at least one row. I gave a proof under "Optional Second" here: Prove that this block matrix is positive definite

Answer 2

As pointed out by user8675309, the best way to prove the positive definiteness of $H:=\frac{A+A^\ast}{2}$ is to simply observe that $H$ is irreducibly diagonally dominant (i.e., $H$ is irreducible, diagonally dominant, and strictly diagonally dominant on at least one row). By Taussky’s refinement of Gerschgorin disc theorem, every irreducibly diagonally dominant matrix is nonsingular.

Alternatively, when $n\ge 3$, note that \begin{align*} H=\frac{A+A^\ast}{2} &=\left[\begin{smallmatrix}1&-i\\ i&2&-i\\ &i&3&-i\\ &&\ddots&\ddots&\ddots\\ &&&i&n-1&-i\\ &&&&i&n\end{smallmatrix}\right] =\left[\begin{smallmatrix}1&-i\\ i&2&-i\\ &i&1\\ &&&0\\ &&&&\ddots\\ &&&&&0\end{smallmatrix}\right] +\left[\begin{smallmatrix}0\\ &0\\ &&2&-i\\ &&i&4&\ddots\\ &&&\ddots&\ddots&-i\\ &&&&i&n\end{smallmatrix}\right] =:X+Y \end{align*} where $H,X$ and $Y$ are positive semidefinite, by Gerschgorin disc theorem. Therefore, $$ v\in\ker(H) \iff v^\ast Hv=0 \iff v^\ast Xv=0 \text{ and } v^\ast Yv=0 \iff v^\ast Xv=0 \text{ and } Yv=0. $$ By Gerschgorin disc theorem, the trailing principal $(n-2)\times(n-2)$ submatrix of $Y$ is positive definite. Therefore $Yv=0$ if and only if the last $n-2$ entries of $v$ are zero, i.e., iff $v$ is in the form of $(v_1,v_2,0,\ldots,0)^\top$. However, for this $v$, as the leading principal $2\times2$ submatrix of $X$ is positive definite, $v^\ast Xv=0$ if and only if $v=0$. Hence $\ker(H)=0$.

Answer 3

The real part of the matrix $A$ is

$$ H := Re(A) = (h_{jk}) = \left( \begin{array}{cc} 1 & - i & \\ i & 2 & -i & \\ & i & 3 & -i & \\ & & \ddots & \ddots & \ddots & \\ & & & i & n-1 & -i \\ & & & & i & n \end{array} \right)$$

There are a few slightly different ways to see that $0$ is not an eigenvalue of $Re(A)$. (This is clear for $n=1, 2, 3$ e.g. by computing the determinant of $H$.)

(1) The most straightforward approach is merely to show that the only vector in the kernel of $Re(A)$ is the null vector. Solving the equations associated with the first $n-1$ rows inductively, it is not too difficult to show that if the vector $X := (x_1 \; x_2 \; \dots \; x_n)^T$ is in the kernel, then for $1 \le j \le n$, we have $x_j = c_j x_1$ where $c_j$ are complex numbers whose moduli increase with $j$. However, the $n$th row implies $i x_{n-1} + nx_n = 0$, hence $(i c_{n-1} + n c_n) x_1 = 0$. Since $| i c_{n-1} + nc_n| \ge |n c_n | - |c_{n-1}| > 0$, it follows that $x_1 = 0$, i.e. $X = 0$.

(2) Let $n \ge 4$. Another approach uses some informations from the proof of Gershgorin's theorem. Indeed, let $\lambda$ be an eigenvalue with eigenvector $X$. Let $k$ be an index for which $|x_k| > 0$ is maximal. Hence $$| \lambda - h_{kk} | = \left| \sum_{j \neq k} h_{jk} x_j/x_k \right| \le \sum_{j \neq k} |h_{jk}| |x_j|/|x_k| \le \sum_{j \neq k} |h_{jk}| =: R_k $$

In the case of the matrix $H$ above, $R_k$ is either $1$ or $2$. Hence if $\lambda = 0$ were possible, we would have $|h_{kk}| \le 2$, which is possible only with $k = 1$ or $k=2$, in which case $|h_{kk}| = k = R_k$. The case $k=2$ is in fact impossible, since it would require $|x_1| = |x_2| = |x_3|$ (which is indeed the case based on approach (1)) and $|h_{21} + h_{23}| = |h_{21}| + |h_{23}|$ (which is false).

Hence $k=1$, so for any vector $X$ in the kernel of $H$, its first component has maximal modulus. But arguing as in approach (1), we see this is false for $n > 3$ (since $x_4 = 2i x_1$.)

(3) More in the spirit of Sylvester's criterion, one can prove that $det(H) \neq 0$. Since $H$ is a tridiagonal matrix, this can be done using this recurrence relation, with $a_j = j$ for $1 \le j \le n$ and $b_j = - c_j = -i$ for $1 \le j \le n-1$. Hence the recurrence relation here is $$ f_0 = 1, f_1 = 1, f_j = a_j f_{j-1} - c_{j-1}b_{j-1} f_{j-2} = j f_{j-1} - f_{j-2}$$ from which it is easy to prove (by induction) that the $f_j$s form an increasing sequence. Hence $det(H) = f_n > 0$.