Right unidimensional extension of Heisenberg Algebra

I'm reading a book on Mathemathical Physics and speaking of the Quantum harmonic oscillator says (this is a translation in english, hope is right):

The commutation relations between operators are: $[N, a] = 􀀀a$, $[N,a^*] = a^*$, $[a, a^*] = 1$ So this is an algebra generated by the operators $a$, $a^*$, $N$ e $1$. It's a solvable Lie Algebra which is a unidimensional right extension of the Heisenberg Algebra. Consequently N is a positive operator in H.

I'd like to understand better the phrase "unidimensional right extension of the Heisenberg Algebra" and why does it implies that N is a positive operator... Can you help me out?

The oscillator Lie algebra $\mathfrak{o}=\langle a^\dagger a, a^\dagger, a, 1\rangle$ is a $4$-dimensonal solvable non-nilpotent Lie algebra, such that $a$, $a^\dagger$, $1$ span an ideal isomorphic to the $3$-dimensional Heisenberg algebra. Hence $\mathfrak{o}$ is a right extension of the Heisenberg Lie algebra. This can be seen by taking a matrix representation with $$a=\left(\begin{array}{ccc}0&1&0\\0&0&0\\0&0&0\end{array}\right),\quad a^\dagger=\left(\begin{array}{ccc}0&0&0\\0&0&1\\0&0&0\end{array}\right),\quad 1=\left(\begin{array}{ccc}0&0&1\\0&0&0\\0&0&0\end{array}\right).$$ Now the adjoint action of $N=a^\dagger a$ on this ideal has this matrices as eigenvectors with respective eigenvalues $1$, $-1$, $0$, so we obtain $$N=\frac12\left(\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}\right).$$ There is a faithful $4$-dimensional matrix representation, given here, where $$N=\begin{pmatrix} 0 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 2\end{pmatrix}$$ has positive eigenvalues.