As usual with such problems, it is most insightful to forget about *matrices* for a while and think about abstract vector spaces instead.

Let $V$ and $W$ vector spaces and $A: V\to W$ linear^{†}. Furthermore let $d_V$ and $d_W$ be metrics on each of the spaces.

Now, given $b\in W$, we have two convex functions $\zeta_V, \zeta_b: V\to\mathbb{R}^+$:
$$\begin{align}
\zeta_V(v) =& d_V(v,\vec0)
\\ \zeta_b(v) =& d_W(A\:v, b)
\end{align}$$

The point of the Tikhonov problem is to make a tradeoff between these two cost functions, i.e. you minimise
$$
\zeta_\lambda := \zeta_b(v) + \lambda\cdot\zeta_V(v).
$$
But why would you want that? Basically, $\zeta_b$ is what we really care for, because it tells us by how much we're *missing our target point*. The problem is when $A$ fails to be injective, because in that case $\zeta_b$ will not be *strictly* convex and you have a whole *set* of solutions $\Xi_b\subset V$ on which $\zeta_b$ is minimised – some of which are very bad solutions, in the sense of, *unbounded* as visible by huge $d_V$ values.

Those solutions can be eliminated by even an arbitrarily small $\lambda$, because $\zeta_b$ is constant on $\Xi_b$. So, in the limit $\lambda\to 0$, you're only really minimising $\zeta_b$, but still preventing solutions that have a needlessly big norm in $V$.

In actual applications though, you're already in trouble even if $A$ *is* injective but badly conditioned, i.e. when there are $v\in V$ for which $d_W(A\:v,\vec0)$ happens to be very small. Because then, just a small bit of *measurement noise* on $b$ could cause the minimum of $\zeta_b$ to be thrown off by a big amount, even though the actual cost is barely changed. That can still be prevented by the $\zeta_V$ contribution, but in this case you can't make $\lambda$ *arbitrarily* small anymore but have to select an application-appropriate finite value.

^{†}_{Really, there's no reason for $W$ to actually be a vector space, it could as well be any metric space – but only for affine spaces and affine mappings can the problem be solved so easily.}