Given Information
We are given with some statements that we have to prove whether they are True or False.

Step-1: (a)
The term least squares means that $\| \mathbf { b } - A \mathbf { x } \|$ is the square root of a sum of squares. In other words, the smallest distance between $b$ and $Ax$ is given by: $\| \mathbf { b } - A \mathbf { x } \|$. Therefore,

The statement is True

Step-2: (b)
If $b$ is the orthogonal projection of $b$ onto Col A, then $Ax=b$. Also, by the orthogonal projection theorem, $b - \hat { b }$ is orthogonal to Col A.
\[ \begin{aligned} A ^ { T } ( \mathbf { b } - A \hat { \mathbf { x } } ) & = 0 \\ A ^ { T } \mathbf { b } - A ^ { T } A \hat { \mathbf { x } } & = 0 \\ A ^ { T } A \hat { \mathbf { x } } & = A ^ { T } \mathbf { b } \end{aligned} \] Thus,

The statement is True

Step-3: (c)
By the definition of least squares solution, a least-squares solution of $Ax=b$ is a vector $x$ such that
\[ \| \mathbf { b } - A \mathbf { x } \| \geq \| \mathbf { b } - A \hat { \mathbf { x } } \| \] Thus,

The statement is True

Step-4: (d)
The set of least-squares solutions of $Ax=b$ coincides with the nonempty set of solutions of the normal equations $A ^ { T } A \mathbf { x } = A ^ { T } \mathbf { b }$ Thus,

The statement is True

Step-5: (e)
By the least squares solution theorem, If the columns of are linearly independent. then the equation $Ax=b$ has only one least squares solution and it is given by \[ \hat { \mathbf { x } } = \left( A ^ { T } A \right) ^ { - 1 } A ^ { T } \mathbf { b } \] Thus,

The statement is True