I presume you have recently been introduced to finding "least squares" approximations by using the "adjoint" matrix.
If A is a linear operator from vector space U to vector space V, and y is NOT in the image of A, then we cannot solve Ax= y. But we can find x such that Ax is "closest" to y. We argue that Ax will be closest to y when y- Ax is perpendicular to Ax. That means that where is the inner product in V. Since the "adjoint" of A is the operator, , from U back to V such that, for all u in U, v, in V, where the second is the inner product in U.
Specifically, we have . But now, x can be any vector in U so we must have . The "Ax" satisfying is the projection of y onto the image of A.