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\begin{center}
\textbf{Exercises on left and right inverses; pseudoinverse}
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\begin{problem}  Find a right inverse for $\mb A = \small \twobythree 101010$. 
\end{problem}

\solution{
We apply the formula $\mb A^{-1}_{\text{right}} = \mb A^T (\mb A \mb A^T)^{-1}$:
\begin{eqnarray*}
\mb A^T &=& \threebytwo 100110\\
\mb A \mb A^T &=& \twobytwo 2001\\
(\mb A \mb A^T)^{-1} &=& \twobytwoc{1/2}001\\
\mb A^T(\mb A \mb A^T)^{-1} &=& \threebytwoc {1/2}001{1/2}0.
\end{eqnarray*}
Thus, $\mb A^{-1}_{\text{right}} = \small \threebytwoc {1/2}001{1/2}0$ is one right inverse of $\mb A$.  We can quickly check that $\mb A \mb A^{-1}_{\text{right}} = \mb I$.
}

\begin{problem} Does the matrix $\mb A = \twobytwo 4386$ have a left inverse? A right inverse? A pseudoinverse? If the answer to any of these questions is "yes", find the appropriate inverse. 
\end{problem}

\solution{
The second row of $\mb A$ is a multiple of the first row, so $\mb A$
has rank $1$ and $\det \mb A = 0$.  Because $\mb A$ is a square matrix
its determinant is defined, and we can use the fact that $\det \mb C
\cdot \det \mb D = \det (\mb C \mb D)$ to prove that $\mb A$ can't
have a left or right inverse.  (If $\mb A \mb B = \mb I$, then $\det
\mb A \det \mb B = \det \mb I$ implies $0=1$.)

We {\em can} find a pseudoinverse $\mb A^+ = \mb V \mb \Sigma^+\mb
U^T$ for $\mb A$.  We start by finding the singular value
decomposition $\mb U \mb \Sigma \mb V^T$ of $\mb A$.

The SVD of $\mb A$ was calculated in the lecture on singular value decomposition, so we know that
$$
\begin{array}{ccccc}
\twobytwo 4386 &=& \frac{1}{\sqrt{5}} \twobytwo 122{-1} &
\twobytwo {\sqrt{125}}000 & \twobytwo {.8}{.6}{.6}{-.8}.\\
\mb A && \mb U & \mb \Sigma & \mb V^T
\end{array}
$$

Hence, $\mb \Sigma^+ = \twobytwo{1/\sqrt{125}}000$ and
\begin{eqnarray*}
\mb A^+ &=& \mb V \mb \Sigma^+\mb U^T\\
&=&  \twobytwo {.8}{.6}{.6}{-.8}
\twobytwo {1/\sqrt{125}}000 \left(\frac{1}{\sqrt{5}} \twobytwo 122{-1}\right) \\
&=&  \twobytwo {.8}{.6}{.6}{-.8}
\twobytwo {1/25}000 \twobytwo 122{-1} \\
&=&  \twobytwo {.8}{.6}{.6}{-.8}
\twobytwo {1/25}000 \twobytwo 122{-1} \\
&=& \frac{1}{125}\twobytwo 4836.
\end{eqnarray*}

To check our work, we confirm that $\mb A^+$ reverses the operation of
$\mb A$ on its row space using the bases we found while computing its SVD.
Recall that
$$\mb A \vb v_j = \left\{ \begin{array}{ll} 
\sigma_j \vb u_j & \text{for } j \leq r\\
\vb 0 & \text{for }j > r.
\end{array}\right.$$
Here $\vb u_1 = \small \twobyone {1/\sqrt{5}}{2/\sqrt{5}}$ and $\mb A^+ \vb u_1 = \frac {1}{\sqrt{125}} \small \twobyone 43 = \frac{1}{\sigma_1}\vb v_1$.
We can also check that $\mb A^+ \vb u_2 = \frac{1}{125} \small \twobytwo 4836 \twobyone 2{-1} = \vb 0$.
}

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