linear algebra - Is the pseudoinverse of a singular, lower triangular matrix itself lower triangular?

Suppose $L\in\mathbb{R}^{n\times n}$ is a singular, lower triangular matrix. Is its psuedoinverse, $L^\dagger\in\mathbb{R}^{n\times n}$, also lower triangular? I have already proved by induction that the product of two lower triangular matrices is lower triangular, and I also proved that the inverse of a (non-singular) lower triangular matrix is lower triangular. I have shown by working out an example with lower triangular $L\in\mathbb{R}^{2\times 2}$ that the pseudoinverse, $L^\dagger\in\mathbb{R}^{2\times 2}$, does not have to be lower triangular, but I was wondering if there might be a better way to prove this than just by showing through an example in $\mathbb{R}^{2\times 2}$ that $L^{\dagger}$ is not necessarily lower triangular.

Useful information:
The pseudoinverse of a matrix $A$ satisfies the following properties:
1) $(A^\dagger A)^*=A^\dagger A$
2) $(AA^\dagger)^*=AA^\dagger$
3) $AA^\dagger A=A$
4) $A^\dagger AA^\dagger = A^\dagger$
Note: here I will be considering only the reals, so the conjugate transpose becomes just a transpose.

Attempt at Solution:
For $L\in\mathbb{R}^{n\times n}$, where $n=1$, it is easy to show that $L^\dagger$ is lower triangular. Since $L$ is singular, we have $L=0$, which is also lower triangular, by the definition of a lower triangular matrix. Then, properties 1) through 3) above are trivially satisfied by any $L^\dagger\in\mathbb{R}$, but for property 4) to be satisfied, we need $L^\dagger 0L^\dagger=L^\dagger\implies L^\dagger=0$. But, $L^\dagger$ is lower triangular by the definition of a lower triangular matrix, so for $n=1$, the answer to the original question is "yes".

Now, consider the case of $n=2$. Let
$$L=\left(\begin{array}{cc}a & 0 \\ b & \tilde{L} \end{array}\right)$$ for $a$ and $b\in\mathbb{R}$, and with $\tilde{L}$ a singular, lower triangular matrix $\in\mathbb{R}^{1 \times 1}$, as before in the example with $n=1$. So $L$ is a singular, lower triangular matrix, and $$L=\left(\begin{array}{cc}a&0\\b&0\end{array}\right).$$ Let $L^\dagger\in\mathbb{R}^{2\times 2}$ be the pseudoinverse of $L$, and define $L^\dagger$ as $$L^\dagger=\left(\begin{array}{cc}c&d\\e&\tilde{L}^\dagger\end{array}\right),$$ with $c,d,e\in\mathbb{R}$ and with $\tilde{L}^\dagger$ being the psuedoinverse of $\tilde{L}$ as derived in the above example for $n=1,$ so $\tilde{L}^\dagger = 0$ and so $$L^\dagger=\left(\begin{array}{cc}c&d\\e&0\end{array}\right).$$

Now, property 1) above requires $ea=0$. Property 2) requires $ad=bc$. Property 3) requires that ($a=0$ or $ca+db=1$) and ($b=0$ or $ca+db=1$). Property 4) requires $e=0$ and ($c=0$ or $ca+db=1$) and ($d=0$ or $ca+db=1$). I chose $d\neq0$, which would then make $L^\dagger$ not a lower triangular matrix. Then, with $ca+db=1$ and $d\neq0$, I found $b=\frac{1-ca}{d}$. Further, from $ad=bc$, I found (substituting the previous expression for $b$), $a=\frac{c}{c^2+d^2}$, which then gives $b$ in terms of $c$ and $d$ as $b=\frac{d}{c^2+d^2}.$ My $L$ and $L^\dagger$ then become $$L=\frac{1}{c^2+d^2}\left(\begin{array}{cc}c&0\\d&0\end{array}\right),$$ and $$L^\dagger=\left(\begin{array}{cc}c&d\\0&0\end{array}\right).$$ $L^\dagger$ satisfies all four of the properties above, but it is not lower triangular, so my answer to the original question is, in general, "no".

However, surely their is a smarter way to do this, in which I don't have to resort to doing all this algebra, isn't there? If so, how might I approach the proof? Thanks a lot!