Invariance of Pseudoinverse under Addition of Degenerate Transformation
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Theorem
Let $U, V$ be vector spaces over a field $K$.
Let $S: U \to V$ be a linear transformation.
Let $T: V \to U$ be a linear transformation.
Let $S$ and $T$ are pseudoinverse to each other.
Then $S + G_1$ and $T + G_2$ are pseudoinverse to each other, where:
- $G_1: U \to V$ is an arbitrary degenerate linear transformation
- $G_2: V \to U$ is an arbitrary degenerate linear transformation
Proof
Let:
\(\text {(1)}: \quad\) | \(\ds G_3\) | \(:=\) | \(\ds T \circ S - I_U\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds G_4\) | \(:=\) | \(\ds S \circ T - I_V\) |
By Definition of Pseudoinverse of Linear Transformation, $G_3, G_4$ are degenerate.
Then:
\(\ds \paren {T + G_2} \circ \paren {S + G_1} - I_U\) | \(=\) | \(\ds T \circ S + T \circ G_1 + G_2 \circ \paren {S + G_1} - I_U\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {T \circ S - I_U} + T \circ G_1 + G_2 \circ \paren {S + G_1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds G_3 + T \circ G_1 + G_2 \circ \paren {S + G_1}\) | by $\paren 1$ |
which is degenerate in view of:
- Product with Degenerate Linear Transformation is Degenerate
- Right Product with Degenerate Linear Transformation is Degenerate
- Product with Degenerate Linear Transformation is Degenerate
Similarly:
\(\ds \paren {S + G_1} \circ \paren {T + G_2} - I_V\) | \(=\) | \(\ds S \circ T + S \circ G_2 + G_1 \circ \paren {T + G_2} - I_V\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {S \circ T - I_V} + S \circ G_2 + G_1 \circ \paren {T + G_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds G_4 + S \circ G_2 + G_1 \circ \paren {T + G_2}\) | by $\paren 2$ |
which is degenerate.
$\blacksquare$
Sources
- 2002: Peter D. Lax: Functional Analysis: $2.2$: Index of a Linear Map