OK, let me try again, maybe I'll get it right this time. I'll show (I hope) that $P$ is positive definite. This will imply the claim because if $(P+iQ)(x+iy)=0$ with $x,y\in\mathbb R^n$, then $Px=Qy$, $Py=-Qx$, and by taking scalar products with $x$ and $y$, respectively, we see that $\langle x, Px \rangle = -\langle y, Py\rangle$, which implies that $x=y=0$. Here I use that $Q$ is symmetric.

Let me first show that $\det P>0$. I'll use row operations (but I never exchange rows, so the determinant stays the same throughout). By subtracting multiples of the first row from all other rows, we get to the matrix $$ \begin{pmatrix} 2 & 1 & 1 & 1 & 1 & \ldots & 1& 1 \\ -3 & 2 & 0 & 0 & 0 & \ldots & 0 & 0\\ -5 & -1 & 3 & 0 & 0 & \ldots & 0 & 0 \\ -7 & -2 & -1 & 4 & 0 & \ldots & 0 & 0 \\ &&&\ldots &&&\\ -(2n-1) & -(n-2) & -(n-3) & -(n-4) & -(n-5) & \ldots & -1 & n \end{pmatrix} $$ In fact, the precise values don't matter: The key feature is that all entries in the lower triangular part are negative, and those on the diagonal and in the first row are positive.

This means that if we now subtract $1/n$ times the last row from the first row, then the remaining entries in the first row stay positive. So we now have to subtract (a positive multiple of) the $(n-1)$st row to get rid of the $(1,n-1)$ element, and again, this only makes the remaining entries in the first row bigger.

We continue in this way, always subtracting rows from the first row. When we're done with this, we are still looking at a positive $(1,1)$ element. Thus indeed $\det P>0$.

Finally, I can run the same computation to see that the determinants of all upper left $k\times k$ submatrices are also positive, and hence $P>0$ by Sylvester's criterion.

OK, let me try again, maybe I'll get it right this time. I'll show that $P$ is positive definite. This will imply the claim because if $(P+iQ)(x+iy)=0$ with $x,y\in\mathbb R^n$, then $Px=Qy$, $Py=-Qx$, and by taking scalar products with $x$ and $y$, respectively, we see that $\langle x, Px \rangle = -\langle y, Py\rangle$, which implies that $x=y=0$. Here I use that $Q$ is symmetric.

Let me now show that $P>0$. Following math110's suggestion, we can simplify my original calculation as follows: Let $ B=B_n = P -\textrm{diag}(1,2,\ldots , n)$. For example, for $n=5$, this is the matrix $$ B_ 5= \begin{pmatrix} 1 & 1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2 & 2\\ 1 & 2 & 3 & 3 & 3\\ 1 & 2 & 3 & 4 & 4\\ 1 & 2 & 3 & 4 & 5 \end{pmatrix} . $$ I can now (in general) subtract the $(n-1)$st row from the last row, then the $(n-2)$nd row from the $(n-1)$st row etc. This confirms that $\det B_n=1$. Moreover, the upper left $k\times k$ submatrices of $B_n$ are of the same type; they equal $B_k$. This shows that $B>0$, by Sylvester's criterion, and thus $P>0$ as well.

It suffices to show that either $P$ or $Q$ is non-singular, and I'll do it for $P$. By subtracting multiples of the first row from all other rows, we get to the matrix $$ \begin{pmatrix} 2 & 1 & 1 & 1 & 1 & \ldots & 1& 1 \\ -3 & 2 & 0 & 0 & 0 & \ldots & 0 & 0\\ -5 & -1 & 3 & 0 & 0 & \ldots & 0 & 0 \\ -7 & -2 & -1 & 4 & 0 & \ldots & 0 & 0 \\ &&&\ldots &&&\\ -(2n-1) & -(n-2) & -(n-3) & -(n-4) & -(n-5) & \ldots & -1 & n \end{pmatrix} $$ In fact, the precise values don't matter: The key feature is that all entries in the lower triangular part are negative, and those on the diagonal and in the first row are positive.

This means that if we now subtract $1/n$ times the last row from the first row, then the remaining entries in the first row stay positive. So we now have to subtract the $(n-1)$st row to get rid of the $(1,n-1)$ element, and again, this only makes the remaining entries in the first row bigger.

We continue in this way, always subtracting rows from the first row. When we're done with this, we are still looking at a positive $(1,1)$ element. In particular, it is not zero, and the transformed matrix has full rank.

OK, let me try again, maybe I'll get it right this time. I'll show (I hope) that $P$ is positive definite. This will imply the claim because if $(P+iQ)(x+iy)=0$ with $x,y\in\mathbb R^n$, then $Px=Qy$, $Py=-Qx$, and by taking scalar products with $x$ and $y$, respectively, we see that $\langle x, Px \rangle = -\langle y, Py\rangle$, which implies that $x=y=0$. Here I use that $Q$ is symmetric.

Let me first show that $\det P>0$. I'll use row operations (but I never exchange rows, so the determinant stays the same throughout). By subtracting multiples of the first row from all other rows, we get to the matrix $$ \begin{pmatrix} 2 & 1 & 1 & 1 & 1 & \ldots & 1& 1 \\ -3 & 2 & 0 & 0 & 0 & \ldots & 0 & 0\\ -5 & -1 & 3 & 0 & 0 & \ldots & 0 & 0 \\ -7 & -2 & -1 & 4 & 0 & \ldots & 0 & 0 \\ &&&\ldots &&&\\ -(2n-1) & -(n-2) & -(n-3) & -(n-4) & -(n-5) & \ldots & -1 & n \end{pmatrix} $$ In fact, the precise values don't matter: The key feature is that all entries in the lower triangular part are negative, and those on the diagonal and in the first row are positive.

This means that if we now subtract $1/n$ times the last row from the first row, then the remaining entries in the first row stay positive. So we now have to subtract (a positive multiple of) the $(n-1)$st row to get rid of the $(1,n-1)$ element, and again, this only makes the remaining entries in the first row bigger.

We continue in this way, always subtracting rows from the first row. When we're done with this, we are still looking at a positive $(1,1)$ element. Thus indeed $\det P>0$.

Finally, I can run the same computation to see that the determinants of all upper left $k\times k$ submatrices are also positive, and hence $P>0$ by Sylvester's criterion.