Find the explicit change of basis matrix from the standard basis \(E\) of \(\mathbb R^3\) to the orthonormal basis \(B\)
\[\left( \begin{pmatrix} \frac{-1}{\sqrt 2} \\ \frac{-1}{\sqrt 3} \\ \frac{-1}{\sqrt 6} \end{pmatrix}, \begin{pmatrix} \frac{1}{\sqrt 2} \\ \frac{-1}{\sqrt 3} \\ \frac{-1}{\sqrt 6} \end{pmatrix}, \begin{pmatrix} 0 \\ \frac{-1}{\sqrt 3} \\ \frac{2}{\sqrt 6} \end{pmatrix} \right)\]The change of basis matrix which changes between two orthonormal bases is given by the formula \(M_{E \to B} = (m^i_j = \langle \vec b_i, \vec e_j \rangle)\)
So,
\[M_{E \to B} = \begin{pmatrix} \frac{-1}{\sqrt 2} & \frac{-1}{\sqrt 3} & \frac{-1}{\sqrt 6} \\ \frac{1}{\sqrt 2} & \frac{-1}{\sqrt 3} & \frac{-1}{\sqrt 6} \\ 0 & \frac{-1}{\sqrt 3} & \frac{2}{\sqrt 6} \end{pmatrix}\]
Find the explicit change of basis matrix from \(B\) to \(E\).
Change of basis matrices between orthonormal bases are orthogonal, so
\[M_{B\to E} = (M_{E\to B})^T = \begin{pmatrix} \frac{-1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\ \frac{-1}{\sqrt 3} & \frac{-1}{\sqrt 3} & \frac{-1}{\sqrt 3} \\ \frac{-1}{\sqrt 6} & \frac{-1}{\sqrt 6} & \frac{2}{\sqrt 6} \end{pmatrix}\]
Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of \(\mathbb R^4\) spanned by the set of vectors,
\[\left\{ \begin{pmatrix} -3 \\ -2 \\ 4 \\ 0 \end{pmatrix}, \begin{pmatrix} 8.5 \\ 3 \\ -3 \\ 1 \end{pmatrix}, \begin{pmatrix} -1.5 \\ -12 \\ -3.5 \\ 3.5 \end{pmatrix} \right\}\]Applying Gram-Schmidt yields the vectors,
\[\vec v_1^{\perp} = \begin{pmatrix}-3 \\ -2 \\ 4 \\ 0\end{pmatrix}\] \[\vec v_2^{\perp} = \begin{pmatrix}8.5 \\ 3 \\ -3 \\ 1\end{pmatrix} - \frac{-43.5}{29}\begin{pmatrix}-3 \\ -2 \\ 4 \\ 0\end{pmatrix} = \begin{pmatrix}4 \\ 0 \\ 3 \\ 1\end{pmatrix}\] \[\vec v_3^{\perp} = \begin{pmatrix} -1.5 \\ -12 \\ -3.5 \\ 3.5 \end{pmatrix} - \frac{14.5}{29} \begin{pmatrix}-3 \\ -2 \\ 4 \\ 0\end{pmatrix} - \frac{-13}{26} \begin{pmatrix}4 \\ 0 \\ 3 \\ 1\end{pmatrix} = \begin{pmatrix}2\\ -11\\ -4\\ 4\end{pmatrix}\]which are all orthogonal. To obtain an orthonormal basis, we divide each vector by its magnitude, and obtain
\[\left\{ \frac{1}{\sqrt{29}}\begin{pmatrix}-3 \\ -2 \\ 4 \\ 0\end{pmatrix}, \frac{1}{\sqrt{26}}\begin{pmatrix}4 \\ 0 \\ 3 \\ 1\end{pmatrix}, \frac{1}{\sqrt{157}}\begin{pmatrix}2 \\ -11\\ -4\\ 4\end{pmatrix} \right\}\]
Let
\[\vec v = \begin{pmatrix}9 \\ -1 \\ -3 \\ 9\end{pmatrix}.\]Find a basis of the subspace of \(\mathbb R^4\) of all vectors perpendicular to \(\vec v\)
Vectors perpendicular (orthogonal) to \(\vec v\) are all vectors \(\vec w\) so that
\[\vec v \cdot \vec w = 0\]This equation is the same as,
\[\vec v^T \vec w = 0\]Which is a matrix equation (there’s only one row, but it’s still a matrix equation):
\[\begin {pmatrix}9 & -1 & -3 & 9\end{pmatrix} \vec w = 0.\]We’re already in row echelon form, and there are three columns without pivots, with dependent variable,
\[a = \frac{b + 3c - 9d}{9}\]We play our usual game to find a basis (make all free variables zero except one, repeat, repeat):
\[\left\{ \begin{pmatrix} 1/9 \\ 1 \\ 0 \\ 0 \end{pmatrix} \begin{pmatrix} 3/9 \\ 0 \\ 1 \\ 0 \end{pmatrix} \begin{pmatrix} -9/9 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}.\]
If \(U\) is a subspace of a vector space \(W\), prove that \(U^{\perp}\) is a subspace of \(W\).
We follow the same recipe as we always do to prove something is a subspace:
Let \(\vec v\), \(\vec w\) be vectors in \(U^{\perp}\). Then we just need to show that \(a \vec v + b \vec w\) is, too. Let \(\vec u\) be a vector in \(U\). Then \(\langle a \vec v + b \vec w, \vec u \rangle = a\langle \vec v, \vec u \rangle + b \langle w, \vec u \rangle = 0\), meaning that it is indeed in \(U^\perp\) (why?). So \(U^{\perp}\) is a subspace of \(W\).
Find a basis for the kernel of the matrix \(A\),
\[A = \begin{pmatrix} 6 & -3 & 6 & 3 \\ -4 & 2 & -4 & -2 \end{pmatrix}.\]To find a basis for the kernel, we first row reduce the matrix \(A\), to obtain,
\[A = \begin{pmatrix} 1 & -1/2 & 1 & 1/2 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\]We then read off the solutions to the matrix equation \(A\vec x = \vec 0\) (just like finding bases for eigenspaces); there is one pivot and three free variables, so a basis would be,
\[\left\{ \begin{pmatrix}1/2 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix}-1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix}-1/2 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}\]
Find a basis for the column space of the matrix \(B\),
\[B = \begin{pmatrix} 3 & 5 & 1 \\ 5 & -4 & 14 \\ 5 & -4 & 14 \\ -2 & -4 & 0 \\ 4 & 9 & -1 \end{pmatrix}.\]To find a basis for the column space, we column reduce the matrix \(B\), to obtain
\[\begin{pmatrix} 1 & 0 & 0 \\ 14 & -74 & 0 \\ 14 & -74 & 0 \\ 0 & -4 & 0 \\ -1 & 14 & 0 \end{pmatrix}\]We can then take the nonzero columns as a basis for the column space:
\[\left\{ \begin{pmatrix}1 \\ 14 \\ 14 \\ 0 \\ -1\end{pmatrix}, \begin{pmatrix}0 \\ -74 \\ -74 \\ -4 \\ 14\end{pmatrix} \right\}\]