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Let $v^tv = w^tw = 1$; $v,w \in\mathbb{R}^n$, and $I$ be the identity matrix. Calculate the pseudo inverse $A_k^{+}$ for:

  1. $A_1 = v$
  2. $A_2 = v^t$
  3. $A_3 = v^tw$
  4. $A_4 = vw^t$
  5. $A_5 = I - 2vv^t$

This is from a sample question for an exam I am going to have to take very soon, and even though I know how to calculate the pseudo inverse using the singular value decomposition (and for specific matrices that have full rank in one of their dimensions using special formulas), I am drawing completely blank here, I am not even sure how to approach this. I would appreciate any help!

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  • $\begingroup$ Try to come with matrices satisfying the properties. For instance, can you think of a matrix that satisfies $vv^+v=v$? Does it satisfy the rest of the properties? $\endgroup$
    – cangrejo
    Jan 21, 2020 at 21:09

1 Answer 1

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You say you have formulas in the case when $A$ has full rank in one of their dimensions. The first three questions fall into this category.


$A_4 A_4^+ A_4 = A_4$ can be written as $vw^t A_4^+ vw^t = vw^t$. If you can find a way to make $w^t A_4^+ v = 1$, that would work.

$A_4^+ = wv^t$


Hint: Note that $A_5 A_5=A_5$.

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  • $\begingroup$ You mean like this: For $A_1 = v$ we have full column rank, which means that $A_1^+ = (A^tA)^{-1}\cdot A^t = \frac{1}{1} \cdot v^t = v^t$? :) For $A_2 = v^t$ we have full row rank, so $A_2^+ = A^t \cdot (AA^t)^{-1}$, but I don't know how to calculate the inverse of $vv^t$, since its not a scalar value as $v^tv$ was in the first example. I know that $(A^+)^t = (A^t)^+$, so I think the Pseudoinverse is just $A_2^+ = v$, correct? $\endgroup$ Jan 21, 2020 at 21:24
  • $\begingroup$ For $A_3 = v^tw$ I know that it is a $1\times 1$ matrix, and such it is invertible, which means that its pseudoinverse is just $(v^tw)^{-1}$ and that value can't be simplified further, correct? $\endgroup$ Jan 21, 2020 at 21:49

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