Psedoinverse and Left Inverse The 'formula' to derive at pseudoinverse is same as Left inverse. And it is basically a projection matrix (projecting on the column space). Then is pseudoinverse synonymous to left inverse? 
 A: The pseudoinverse of a matrix $A$ exists for any matrix, and is uniquely determined.  I will work over the real numbers, though pseudoinverses also exist over complex numbers.
To explain what is going on here, let me take you back a bit to just regular functions (the matrices define linear transformations, which are special types of functions between vector spaces, and pseudoinverses and left inverses are related to their nature as functions).
Say $f\colon X\to Y$ is a function. A left inverse of $f$ is a function $g\colon Y\to X$ such that $g\circ f = \mathrm{id}_X$. A right inverse of $f$ is a function $h\colon Y\to X$ such that $f\circ h=\mathrm{id}_Y$. An inverse (or a two-sided inverse) is a function $\mathfrak{F}\colon Y\to X$ which is both a left and a right inverse of $f$: that is, $\mathfrak{F}\circ f = \mathrm{id}_X$, and $f\circ\mathfrak{F}=\mathrm{id}_Y$.
One can show that if $f$ has both a left inverse and a right inverse, then they are the same function and that function is a two-sided inverse, which is then unique, and so we denote it by $f^{-1}$: indeed, if $g$ is a left inverse to $f$, and $h$ is a right inverse to $f$, then
$$g = g\circ \mathrm{id}_Y = g\circ(f\circ h) = (g\circ f)\circ h = \mathrm{id}_X\circ h = h,$$
so $g=h$.
Now for functions between sets (when we don't ask any kind of structure preservation, continuity or anything like that) we have:

Theorem 1. Let $f\colon X\to Y$ be a function, with $X$ not empty. The following are equivalent:
  
  
*
  
*$f$ is one-to-one.
  
*$f$ has a left inverse.
  

and (assuming the Axiom of Choice; if you don't know what that means, don't worry about it, it's a technical logical issue):

Theorem 2. Let $f\colon X\to Y$ be a function. The following are equivalent:
  
  
*
  
*$f$ is onto.
  
*$f$ has a right inverse.
  

As a consequence, we get:

Theorem 3. Let $f\colon X\to Y$ be a function. The following are equivalent:
  
  
*
  
*$f$ is bijective (one-to-one and onto).
  
*$f$ has a two-sided inverse.
  

Now, it is instructive to see the proof that (1) implies (2) in Theorem 1, because it is relevant to your query.
Suppose $f\colon X\to Y$ is one to one. We construct a left inverse $g\colon Y\to X$ for $f$ as follows: 
Since $X$ is not empty, let $x_0\in X$ be an arbitrary element. If $y\in\mathrm{image}(f)$, then we know there exists $x_y\in X$ such that $f(x_y)=y$. And since $f$ is one-to-one, $x_y$ is uniquely determine by $y$. So define $g$ as follows:
$$g(y) = \left\{\begin{array}{ll}
x_y & \text{if }y\in\mathrm{image}(f)\\
x_0 & \text{if }y\notin \mathrm{image}(f).
\end{array}\right.$$
Now, if $x\in X$, then $f(x)\in Y$, and $x_{f(x)} = x$. So for all $x\in X$, $g(f(x)) = x_{f(x)} = x$, hence $g\circ f = \mathrm{id}_X$, showing that $g$ is a left inverse of $f$.
Now notice that there are a number of choices we can make here: 


*

*We can select $x_0$ arbitrarily; if $X$ has more than one element and $f$ is not onto, then picking different $x_0$s will give us different left inverses to $f$.

*If $X$ is has more than one element and $f$ is not onto, then we can even pick different values for different things not in $\mathrm{image}(f)$, arbitrarily, and get left inverses.

Now, let's go back to matrices and linear transformations. We can't just make arbitrary choices and hope to get a linear transformation (i.e., a matrix). For example, if we decide that $f(x) = a$ and that $f(y)=b$, then we better have $f(x+y) = a+b$. So it's not as simple to construct even one-sided inverses.
Now, what we can do is define the left inverse on a basis, arbitrarily, and extend it to a linear transformation. 
So, say $T\colon V\to W$ is a linear transformation that is one-to-one (has trivial nullspace); you can think of it as an $m\times n$ matrix $A$, mapping $\mathbb{R}^n$ to $\mathbb{R}^m$. Since we are assuming it is one-to-one, we must have $n\leq m$. 
So, we can look at the basis $\mathbf{e}_1,\ldots\mathbf{e}_n$ of $\mathbb{R}^n$, and since $T$ is one-to-one, the images form a linearly independent subset of $\mathbb{R}^n$; these are the columns $A$. To construct a left inverse, we can:


*

*Extend $A\mathbf{e}_1,\ldots, A\mathbf{e}_n$ to a basis for $\mathbb{R}^m$, by adding vectors $\mathbf{w}_{n+1},\ldots,\mathbf{w}_m$ to this list to get a basis $\gamma= [A\mathbf{e}_1,\ldots, A\mathbf{e}_n,\mathbf{w}_{n+1},\ldots,\mathbf{w}_m]$.

*Define a new linear transformation/matrix $U$ by defining it on $\gamma$ as $$\begin{align*}
U(A\mathbf{e}_1) &= \mathbf{e}_1\\
&\vdots\\
U(A\mathbf{e}_n) &= \mathbf{e}_n\\
\end{align*}$$
and then just picking any vectors $\mathbf{v}_{n+1},\ldots,\mathbf{v}_m$ in $\mathbb{R}^n$ and letting
$$\begin{align*}
U(\mathbf{w}_{n+1}) &= \mathbf{v}_{n+1}\\
&\vdots\\
U(\mathbf{w}_{m}) &= \mathbf{v}_m.
\end{align*}$$
Then extend linearly; then figure out what this linear transformation is on the standard basis for $\mathbb{R}^m$, and thus get a matrix $B$. Since $U\circ T = \mathrm{id}_{\mathbb{R}^n}$, it follows that $BA=I_n$. That is, $B$ is a left inverse of $A$.


Now again notice how much freedom you have in selecting $B$: you can arbitrarily extend your original list to the basis $\gamma$; and you can arbitrarily decide what happens to the new vectors.
So that if I go home and get a left inverse and you go home and get a left inverse, we have a hope of coming back with the same answer, we can reduce the choices by agreeing that some choices make more sense than others, or at least make things easier. One such thing we could do is to just agree that we will let those extra basis vectors go to $\mathbf{0}$. That is, we will always pick $\mathbf{v}_{n+1}=\cdots=\mathbf{v}_m = \mathbf{0}$. That's not an unreasonable choice.
However, it is very difficult to agree on how we will deal with the choices of $\mathbf{w}_{n+1},\ldots,\mathbf{w}_{m}$ in the abstract. In fact, there is no good way of having an a priori agreement in general without bringing in more structure.
So we bring in more structure: the inner product.
Given any subspace $W$ of $\mathbb{R}^k$, we let
$$W^{\perp} = \{\mathbf{x}\in\mathbb{R}^k\mid \langle \mathbf{x},\mathbf{w}\rangle = 0\text{ for all }\mathbf{w}\in W\},$$
where $\langle \mathbf{x},\mathbf{w}\rangle$ is the inner product (you may know it as the dot product; that's one type of inner product). Then $W^{\perp}$, the orthogonal complement of $W$, is a subspace of $W$.
Turns out, it is always the case that every vector in $\mathbb{R}^k$ can be written uniquely as a vector in $W$ plus a vector in $W^{\perp}$. This gives us a way of agreeing, ahead of time, on how we will define a left inverse:

Say $A$ is $n\times m$, and has trivial nullspace.
  
  
*
  
*Calculate the range of $A$, $R(A)$.
  
*Calculate the orthogonal complement $R(A)^{\perp}$ of the range of $A$.
  
*Define $U\colon\mathbb{R}^m\to \mathbb{R}^n$ as follows: 
  
  
  
i. if $\mathbf{y}\in R(A)$, let $\mathbf{x}_{\mathbf{y}}\in\mathbb{R}^n$ be the unique vector such that $A\mathbf{x}_{\mathbf{y}} = \mathbf{y}$.
    ii. If $\mathbf{w}\in \mathbb{R}^m$, write it uniquely as $\mathbf{w}=\mathbf{y} + \mathbf{z}$, where $\mathbf{y}\in R(A)$ and $\mathbf{z}\in R(A)^{\perp}$. 
    iii. Define $U(\mathbf{w}) = U(\mathbf{y})+U(\mathbf{z}) = \mathbf{x}_{\mathbf{y}} + \mathbf{0} = \mathbf{x}_{\mathbf{y}}$.


One can then check that $U$ is a left inverse to the linear transformation defined by $A$, and so if $B$ is the standard matrix representation of $U$, then $B$ is a left inverse of $A$.

What if $A$ is not one-to-one? Then we can't find a left inverse. Basically the problem is that if $\mathbf{x}$ is in the nullspace of $A$, then as soon as you apply $A$ you lose all information about $\mathbf{x}$. 
So instead we try to find something that will undo $A$ in some "essential part" of $A$: take the nullspace of $A$, $N(A)$, and find a basis for it, $\mathbf{z}_1,\ldots,\mathbf{z}_k$. Then extend to a basis for all of $\mathbb{R}^n$, $\mathbf{z}_1,\ldots,\mathbf{z}_k,\mathbf{x}_{k+1},\ldots,\mathbf{x}_{n}$. Then $A\mathbf{x}_{k+1},\ldots, A\mathbf{x}_{n}$ form a basis for $R(A)$, and now we can proceed as we did above to define $B$; this $B$ will have the property that $BA\mathbf{x}_{k+1}=\mathbf{x}_{k+1},\ldots, BA\mathbf{x}_{n}=\mathbf{x}_n$. Of course, $BA\mathbf{z}_i = \mathbf{0}$ for $i=1,\ldots,k$. 
Again, the problem is picking $\mathbf{x}_{k+1},\ldots, \mathbf{x}_n$; without more structure, we don't have good ways of making these arbitrary choices. But if we use the inner product, once again we have a good way of making those choices: start with $N(A)$, then construct its orthogonal complement $N(A)^{\perp}$, and let $\mathbf{x}_{k+1},\ldots,\mathbf{x}_n$ be a basis for $N(A)^{\perp}$. Now proceed.
This process actually works even if $A$ is one-to-one, since then $N(A)$ is trivial, so we can just take $k=0$ and $\mathbf{x}_1,\ldots,\mathbf{x}_n$ the standard basis.

This process produces a matrix $A^{\dagger}$. This matrix has the following properties, by design, as you can verify by following through the construction:


*

*If $A$ is invertible, then $A^{\dagger}$ is in fact the inverse of $A$.

*If $A$ is one-to-one but not onto, then $A^{\dagger}$ is one of the left inverses of $A$.

*If $A$ is onto but not one-to-one, then $A^{\dagger}$ is one of the right inverses of $A$.


The following properties also come out of the special choices we make for the construction:


*

*$AA^{\dagger}A = A$ and $A^{\dagger}AA^{\dagger}=A^{\dagger}$. (Not quite an inverse, but kind of close...)

*$AA^{\dagger}$ and $A^{\dagger}A$ are symmetric. 


These two properties turn out to uniquely determine $A^{\dagger}$: any other matrix that has those conditions is in fact equal to $A^{\dagger}$. 
It also turns out that, because of how $A^{\dagger}$ is constructed, we can say more about what $A^{\dagger}A$ and $AA^{\dagger} $ do:


*

*$A^{\dagger}A$ is the orthogonal projection onto $N(A)^{\perp}$.

*$AA^{\dagger}$ is the orthogonal projection onto $R(A)$. 


In particular: if $N(A)$ is trivial, then $A^{\dagger}A$ is the "orthogonal projection" onto all of $\mathbb{R}^n$, that is, the identity; and therefore $A^{\dagger}$ is indeed a (indefinite article) left inverse of $A$.  And if $R(A)$ is all of $\mathbb{R}^m$, then $AA^{\dagger}$ is the identity and so $A^{\dagger}$ is a (indefinite article) right inverse of $A$.
(Which we already knew, but now we have more information when they aren't just that).

Longwindedly, then:
The pseudoinverse $A^{\dagger}$ will work as a one-sided or two-sided inverse of $A$ when $A$ has that type of inverse (that is, if $A$ has left inverses, then $A^{\dagger}$ will be a very specific left inverse of $A$, carefully selected from among however many left inverses there may be; if $A$ has a right inverse, then $A^{\dagger}$ will be a very specific right inverse of $A$, carefully selected from among however many right inverses there may be; nd if $A$ has a two-sided inverse, then $A^{\dagger}$ will be that unique two-sided inverse).  If $A$ does not have a one-sided inverse, then $A^{\dagger}$ gets close, pretty much as close as we can hope for (via the identities $AA^{\dagger}A = A$ and $A^{\dagger}AA^{\dagger}=A^{\dagger}$). 
It turns out to have a lot of very good properties: you can use the pseudoinverse to solve systems of linear equations, in such a way that (i) if the system has a unique solution the pseudoinverse will find it; (ii) if the system has infinitely many solutions, the pseudoinverse will find the one of smallest size; (iii) if the system does not have solutions, the pseudoinverse will find the vector of smallest size that gives you the least squares solution; and (iv) you don't have to analyze the system to figure out which category you are in in order to get the solution (whereas the "usual" least squares and minimal solution procedures are different and so you need to know which case you are in first, before being able to solve it).
In that sense, it is way more useful than some random right inverse you found on the street (assuming one exists)...
And to finally answer your question: no, "pseudoinverse" is not synonymous with "left inverse". If left inverses exist for $A$, then the (definite article, there is just one) pseudoinverse is a (indefinite article, there may be many) left inverse; but not every left inverse need be the pseudoinverse. And certainly, if no left inverse exists, then the pseudoinverse still does exist, and so it cannot be synonymous with something that doesn't exist at all. 
