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If a $2\times3$ matrix $A$ has full row rank, then it can be expressed as $E_k\cdots E_1\begin{bmatrix}1&0&0\\0&1&0\\ \end{bmatrix}$. So its one right inverse can be expressed as $\begin{bmatrix}1&0\\0&1\\0&0\\ \end{bmatrix}E^{-1}_1\cdots E^{-1}_k$ where $E_i$s are elementary matrices. How do we get other right inverses?

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2 Answers 2

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Set $G := E_k \cdots E_1$.

Claim: For any $a, b \in \mathbb{R}$,

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ a & b\end{bmatrix}G^{-1} = \begin{bmatrix} G^{-1} \\ \begin{bmatrix} a & b \end{bmatrix} G^{-1} \end{bmatrix} $$

is a right inverse of $A$.

Proof: Just do the matrix multiplication.

Claim: For any matrix $B$ not of the form above, $B$ is not a right inverse.

Proof: Suppose, $B$ does not have the above form. Parition $B$ as

$$ B = \begin{bmatrix} B_1 \\ B_2 \end{bmatrix} $$

where $B_1$ are the first two rows of $B$ and $B_2$ the third. Since $B$ does not have the above form, we have $B_1 \ne G^{-1}$. Thus, carrying out the matrix multiplication

$$ AB = G \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix} B_1 \\ B_2 \end{bmatrix} = GB_1 \ne I \text{ since } B_1 \ne G^{-1} $$

Thus $B$ is not a right inverse for $A$.


Thus the right inverse of $A$ are precisely those of the form

$$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ a & b\end{bmatrix}G^{-1}. $$

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  • $\begingroup$ Thanks for your elaborate reply. $\endgroup$
    – Silent
    Commented Nov 4, 2017 at 8:22
  • $\begingroup$ One question: When you say $B$ has not above form, how do you assume that $B_1\ne G^{-1}$? Isn't it possible without affecting that part, make change in $B_2$? $\endgroup$
    – Silent
    Commented Nov 4, 2017 at 12:27
  • $\begingroup$ I think it is because for any $B_2=\begin{bmatrix} x & y \end{bmatrix}$ can be represented as $\begin{bmatrix} a & b \end{bmatrix}G^{-1}$ because $G$ is invertible. Am I right? $\endgroup$
    – Silent
    Commented Nov 4, 2017 at 12:34
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    $\begingroup$ @Silent Yes exactly. Any possible $B_2$ can be represented as $\begin{bmatrix} a & b \end{bmatrix} G^{-1}$. $\endgroup$
    – eepperly16
    Commented Nov 5, 2017 at 4:56
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We want to find the matrix $B$ such that $AB = I_2$.

$$A\begin{bmatrix} b_1 & b_2\end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

We can solve $Ab_1 = e_1$, notice that there will be a free parameter. $b_1 = b_1^*+sv$ where $v$ is a vector in the nullspace of $A$ and $b_1^*$ is a particular solution.

We can also solve $Ab_2 = e_2$, there will be another free parameter.$b_2 = b_2^*+tv$ where $v$ is a vector in the nullspace of $A$ $b_2^*$ is a particular solution.

$$B=B^*+\begin{bmatrix} v & v\end{bmatrix}\begin{bmatrix}s & 0 \\ 0 & t \end{bmatrix}$$

Remark: the pivot columns need not be at the first two columns.

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  • $\begingroup$ Can I conclude from this that if $A$ has unique right inverse then $A$ is invertible, because I can get right inverse of $A$ only if it has full row rank, but if $A$ does not have full column rank, then it has multiple right inverses, so only way $A$ has unique right inverse is when $A$ is square, then from [here](math.stackexchange.com/q/3852) $A$ has to be invertible. $\endgroup$
    – Silent
    Commented Nov 4, 2017 at 8:21
  • $\begingroup$ you mean if $A$ is a square? then yes. $\endgroup$ Commented Nov 4, 2017 at 8:23
  • $\begingroup$ No no, I did not assume it outright, but it seems that if $A$ has unique right inverse then $A$ has to be square. Is this true? $\endgroup$
    – Silent
    Commented Nov 4, 2017 at 8:24
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    $\begingroup$ yes, if it exists and unique, then it has to be a square. $\endgroup$ Commented Nov 4, 2017 at 8:28

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