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

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