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\textbf{Exercises on orthogonal matrices and Gram-Schmidt}
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\begin{problem} (4.4 \#10.b  {\em Introduction to Linear Algebra:} Strang) 

Orthonormal vectors are automatically linearly independent. 

Matrix Proof: Show that $Q \vb x = \vb 0$ implies $\vb x=\vb 0.$ Since $Q$ may be rectangular, you can use $Q^T$ but not $Q^{-1}.$
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\solution{ By definition, $Q$ is a matrix whose columns are orthonormal, and so we know that $Q^T Q = I$ (where $Q$ may be rectangular). Then:
$$Q \vb x = \vb 0 \Longrightarrow Q^T Q \vb x = Q^T \vb 0
  \Longrightarrow I \vb x = \vb 0 \Longrightarrow \vb x = \vb 0. $$
  Thus the nullspace of $Q$ is the zero vector, and so the columns of
  $Q$ are linearly independent. There are no non-zero linear
  combinations of the columns that equal the zero vector. Thus, 
  orthonormal vectors are automatically linearly
  independent.
}

\begin{problem} (4.4 \#18) Given the vectors $\vb a$, $\vb b$ and $\vb c$ listed below, use the Gram-Schmidt process to find orthogonal vectors $\vb A$, $\vb B$, and $\vb C$ that span the same space.
$$\vb a = (1,-1,0,0 ), \vb b = (0,1,-1,0 ), \vb c = (0,0,1,-1 ).$$
Show that $\{\vb A, \vb B, \vb C\}$ and $\{\vb a, \vb b,\vb c\}$ are bases for the space of vectors perpendicular to $\vb d = (1,1,1,1)$.
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\solution{ We apply Gram-Schmidt to $\vb a, \vb b, \vb c.$ First, we set
$$\vb A = \vb a = \mathbf{(1,-1,0,0)}. $$
Next we find $\vb B:$
$$\vb B = \vb b - \frac{\vb{A^T} \vb b}{\vb {A^T} \vb A} \vb A =  (0,1,-1,0) + \frac{1}{2} (1,-1,0,0) = \mathbf{\left(\frac{1}{2}, \frac{1}{2}, -1,0\right)}.$$
And then we find $\vb C:$
$$\vb C = \vb c - \frac{\vb{A^T} \vb c}{\vb {A^T} \vb A} \vb A - \frac{\vb{B^T} \vb c}{\vb {B^T} \vb B} \vb B = (0,0,1,-1) + \frac{2}{3}\left(\frac{1}{2}, \frac{1}{2}, -1,0 \right) = \mathbf{\left (\frac{1}{3} , \frac{1}{3} , \frac{1}{3}, -1 \right)}.$$

We know from the first problem that the elements of the set $\{\vb A,
\vb B,\vb C\}$ are linearly independent, and each vector is orthogonal
to (1,1,1,1).  The space of vectors perpendicular to $\vb
d$ is three dimensional (since the row space of $(1,1,1,1)$ is
one-dimensional, and the number of dimensions of the row space added
to the number of dimensions of the nullspace add to 4). Therefore
$\{\vb A, \vb B,\vb C\}$ forms a basis for the space of vectors
perpendicular to $\vb d$.

Similarly, $\{\vb a , \vb b, \vb c \}$ is a basis for the space of vectors perpendicular to $\vb d$ because the vectors are linearly independent,  orthogonal to (1,1,1,1), and because there are three of them.
}

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