37.4: Inverse of a Matrix
- Page ID
- 70391
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- 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 \(A\).