Calculation Results
Pseudo Inverse Calculation: Definition and Formula
The pseudo inverse of a matrix, also known as the Moore–Penrose inverse, is a generalized matrix inverse that works for both square and non-square matrices. For a matrix \( A \), its pseudo inverse is denoted as \( A^{+} \).
The pseudo inverse satisfies the four Moore–Penrose conditions:
\[ AA^{+}A = A \] \[ A^{+}AA^{+} = A^{+} \] \[ (AA^{+})^{T} = AA^{+} \] \[ (A^{+}A)^{T} = A^{+}A \]
Closed-form formulas:
- If \(A\) has full column rank: \[ A^{+} = (A^{T}A)^{-1}A^{T} \]
- If \(A\) has full row rank: \[ A^{+} = A^{T}(AA^{T})^{-1} \]
- General case: computed using SVD \[ A = U\Sigma V^{T} \quad \Rightarrow \quad A^{+} = V\Sigma^{+}U^{T} \]
Examples
Example 1
Compute the pseudo inverse of:
\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \]
Since this is a full column rank matrix:
\[ A^{+} = (A^{T}A)^{-1}A^{T} \]
Steps:
\[ A^{T}A = \begin{bmatrix} 10 & 14 \\ 14 & 20 \end{bmatrix} \]
\[ (A^{T}A)^{-1} = \begin{bmatrix} 5 & -3.5 \\ -3.5 & 2.5 \end{bmatrix} \cdot \frac{1}{(10)(20)-(14)(14)} \]
Final result:
\[ A^{+} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix} \]
Example 2
Compute the pseudo inverse of:
\[ B = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \]
This is a row vector, so:
\[ B^{+} = \frac{1}{BB^{T}} B^{T} \]
\[ BB^{T} = 1 \]
\[ B^{+} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \]
Common Mistakes
- Confusing pseudo inverse with regular inverse—they are not the same.
- Forgetting that non-square matrices can also have pseudo inverses.
- Using the wrong formula when the matrix does not have full row/column rank.
- Ignoring that SVD is required for the general case.
Practice Problems
Problem 1
Find the pseudo inverse of:
\[ A = \begin{bmatrix} 2 & 0 \\ 0 & 1 \\ \end{bmatrix} \]
Show Answer
\[ A^{+} = \begin{bmatrix} 1/2 & 0 \\ 0 & 1 \end{bmatrix} \]
Problem 2
Find \(A^{+}\) for:
\[ A = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \]
Show Answer
\[ A^{+} = \frac{1}{14} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \]
Problem 3
Compute the pseudo inverse:
\[ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} \]
Show Answer
\[ A^{+} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \]
Problem 4
Compute the pseudo inverse using SVD:
\[ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix} \]
Show Answer
\[ A^{+} = \frac{1}{4} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \]
The pseudo inverse is especially useful together with RREF, eigenvalues, and QR decomposition.
Learn matrix multiplication more easily in just 2 minutes with this free game.
