37.4: Inverse of a Matrix
- Page ID
- 70391
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- Recall the four fundamental subspaces of a \(m \times n\) matrix \(A\)
- The rowspace and nullspace of \(A\) in \(R^n\)
- The columnspace and the nullspace of \(A^{\top}\) in \(R^m\)
- The two-sided inverse gives us the following \( {A}{A}^{-1}=I={A}^{-1}{A} \)
- For this we need \(r = m = n\), here \(r\) is the rank of the matrix.
- For a left-inverse, we have the following
- Full column rank, with \(r = n \leq m\) (but possibly more rows)
- The nullspace contains just the zero vector (columns are independent)
- The rows might not all be independent
- We thus have either no or only a single solution to \(Ax=b\).
- \(A^{\top}\) will now also have full row rank
- From \((A^\top A)^{-1}A^\top A = I\) follows the fact that \((A^\top A)^{-1}A^\top\) is a left-sided inverse
- Note that \((A^\top A)^{-1}A^\top\) is a \(n \times m\) matrix and \(A\) is of size \(m \times n\), theire mulitiplication \((A^\top A)^{-1}A^\top A\) results in a \(n \times n\) identity matrix
- The \(A(A^\top A)^{-1}A^\top\) is a \(m \times m\) matrix. BUT \(A(A^\top A)^{-1}A^\top\neq I\) if \(m\neq n\). The matrix \(A(A^\top A)^{-1}A^\top\) is the projection matrix onto the column space of \(A\).
What is the projection matrix that projects any vector onto the subspace spanned by \([1,2]^{\top}\). (What matrix will give the same result as projecting any point onto the vector \([1,2]^{\top}\).)
If \(m=n\), is the left inverse the same as the inverse?
For a matrix \(A\) with \(r=n<m\), the columnspace of \(A\) has dimension \(r(=n)\). The linear transfrom \(A: R^n\mapsto R^m\) is one-to-one. In addition, the linear transformation \(A\) from \(R^n\) to the columnspace of \(A\) is one-to-one and onto (it means that for any element in the columnspace of \(A\), we can find \(x\) in \(R^n\) such that it equals \(Ax\).) Then the left inverse of \(A\) is a one-to-one mapping from the columnspace of \(A\) to \(R^n\), and it can be considered as an inverse transform of \(A\).