Originally Posted by

**HallsofIvy** If r< n- that is if the rank is less than the number of rows, you cannot use this because A is not invertible- $\displaystyle A^{-1}$ does not exist.

Does the book say what $\displaystyle A^\~$, $\displaystyle X^\~$, $\displaystyle A_r$, and $\displaystyle B_r$ **mean**?

You could do this: If A:$\displaystyle R^n\to R^m$ does not have an inverse then it maps all of $\displaystyle R^n$ into a **sub** space of $\displaystyle R^m$. There is no solution to Ax= B unless B happens to lie in that subspace (in which case there may be an infinite number of solutions). If B is not in the subspace then the best you can do is find x so that Ax is "closest" to B. To do that, drop a perpendicular from B to the subspace. Call vector, the vector in the range of A closest to B, B' and look at B- B'= B- Ax where x is such that Ax= B'. That vector is perpendicular to the subpace: <B- Ax, u>= 0 for all u in the subspace. But every vector in that subspace is of the form u= Av for some v because that subspace **is** the range of A, that is the same as <B- Ax, Av>= 0. Now define A~ to be the "adjoint" of A, the linear operator such that <v, Au>= <A*v, u> for all u in $\displaystyle R^n$ and all v in $\displaystyle R^m$ (for real matrices this is the transpose).

Then we have <B- Ax, Av>= <A*(B- Ax), v> where this inner product is on $\displaystyle R^n$ and v can be any vector in $\displaystyle R^n$/ In particular, v couls be A*(B- Ax) itself so we must have A*(B- Ax)= A*B- A*Ax= 0 or A*Ax= A*B. Now it **may** happen that A*A is invertible even if A itself is not. In that case we have [itex]x= (A*A)^{-1}A*B[/itex], and [itex](A*A)^{-1}A*[/itex] is called the "generalized inverse". If A itself has an inverse, then [itex](A*A)^{-1}= A^{-1}A*^{-1}[/itex] so that [itex](A*A)^{-1}A*= A^{-1}A*^{-1}A*= A^{-1}[/itex] the true inverse.

Note that * throughout means the "adjoint", not multiplication.