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\textbf{Exercises on projections onto subspaces}
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\begin{problem} (4.2 \#13.  {\em Introduction to Linear Algebra:} Strang) Suppose $A$ is the four by four identity matrix with its last column removed; $A$ is four by three. Project $\vb b = (1,2,3,4)$ onto the column space of $A$. What shape is the projection matrix $P$ and what is $P$? 
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\solution{ $P$ will be four by four since we are projecting a 4-dimensional vector to another 4-dimensional vector. We will have:
$$P = \left [ \begin{array}{cccc} 
1 & 0&0 & 0\\
0& 1& 0& 0\\
0& 0& 1& 0\\
0&0 &0 & 0 \end{array} \right ]$$
This can be seen by observing that the column space of $A$
is the $wxy-$space, so we just need to subtract the $z$ coordinate from
the 4-dimensional vector $(w,x,y,z)$ we're projecting. The projection
of $\vb b$ is therefore:
$$\vb p = P \vb b = \left [ \begin{array}{cccc} 
1 & 0&0 & 0\\
0& 1& 0& 0\\
0& 0& 1& 0\\
0&0 &0 & 0 \end{array} \right ] \fourbyone 1 2 3 4 = \fourbyone 1 2 3 0.$$
}

\begin{problem} (4.2 \#17.)  If $P^2 = P,$ show that $(I-P)^2 = I-P.$ For the matrices $A$ and $P$ from the previous question, $P$ projects onto the column space of $A$ and $I-P$ projects onto the \rule{20mm}{0.4pt}.
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\solution{ $$(I-P)^2 = I^2 - IP - PI + P^2 = I - 2P + P^2 = I - 2P + P = I - P. $$
Using the matrices $A$ and $P$ from the previous question, $$I-P = \left [ \begin{array}{cccc} 
0 & 0&0 & 0\\
0& 0& 0& 0\\
0& 0& 0& 0\\
0&0 &0 & 1 \end{array} \right ]$$ projects onto the \textbf{left nullspace} of $A.$ 
}

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