91Ó°ÊÓ

If the vector \(\mathbf{b}\) is in the range of matrix \(A\), then there exists a vector \(\mathbf{x} \in \mathbb{R}^n\) such that: \( A\mathbf{x} = \mathbf{b} \) Now, we need to show that \(\mathbf{b}=A A^{+} \mathbf{b}\). To do this, multiply both sides of the above equation with \(A^{+}\): \( A^{+}A\mathbf{x} = A^{+}\mathbf{b} \) Since \(A^{+}\) is the Moore-Penrose inverse of \(A\), we know that \(A^{+}A\) is an idempotent matrix, which means \((A^{+}A)^2 = A^{+}A\). Therefore, \( A^{+}A\mathbf{x} = A^{+}A (A^{+}\mathbf{b}) \) Now replace \(\mathbf{x}\) with \(A^{+}\mathbf{b}\): \( A^{+}A(A^{+}\mathbf{b}) = A^{+}\mathbf{b} \) This means, \( A (A^{+}\mathbf{b}) = \mathbf{b} \) Thus, if \(\mathbf{b} \in R(A)\), then \(\mathbf{b}=A A^{+} \mathbf{b}\).

Now we need to show that if \(\mathbf{b}=A A^{+} \mathbf{b}\), then \(\mathbf{b}\) is in the range of \(A\). We have the given equation: \( \mathbf{b} = A (A^{+}\mathbf{b}) \) Let \(\mathbf{y} = A^{+}\mathbf{b}\). Then we have: \( \mathbf{b} = A \mathbf{y} \) This means there exists a vector \(\mathbf{y} \in \mathbb{R}^n\) such that \(A\mathbf{y} = \mathbf{b}\). This implies that \(\mathbf{b}\) is in the range of \(A\) (i.e., \(\mathbf{b} \in R(A)\)).

Thus, we have shown that the vector \(\mathbf{b}\) is in the range of matrix \(A\) (i.e., \(\mathbf{b} \in R(A)\)) if and only if the equation \(\mathbf{b}=A A^{+} \mathbf{b}\) holds true.