The vectors must be pairwise orthogonal, i.e. they must be perpendicular to each other (now, how is it you test this?)
hint
Two vectors are orthogonal, exactly when their scalar product is 0, but more is needed.
hint
A basis is orthonormal, if the vectors are mutually orthogonal and each has the length (norm) 1.
answer
$\mv_1\cdot\mv_2=\mv_1\cdot\mv_3=\mv_2\cdot\mv_3=0$ and all three vectors have the norm 1, so the three vectors constitute an orthonormal basis for $\reel^3$.
D
State an orthonormal basis in $\,\reel^3\,$ where $\,\,\displaystyle{(\frac {\sqrt 2}2,\frac {\sqrt 2}2,0)}\,\,$ is the first basis vector.
hint
E.g. you can guess a unit vector that is perpendicular to the given vector, and then find the third basis vector using the cross product. Check that the basis vectors found constitute an orthonormal basis.
Exercise 3: Orthonormalization. By Hand
E
Determine the solution set for the homogeneous equation
$$x_1+x_2+x_3=0$$
and explain that it (of course) is a subspace in $\,\reel^3\,.$ Find an orthonormal basis for this solution space.
hint
If we shall find an orthonormal basis for the solutions space, we must first find a basis for the solution space, i.e. we must solve the equation.
When two vectors are to be orthonormalized, we must on the one hand rotate the vectors within the space they are spanning, such that they become orthogonal (are mutually perpendicular) and continue to span the same space, and on the other hand they must be normalized, i.e. lengthened or shortened, so they all have the length 1.
hint
Luckily we can do all this by following the Gram-Schmidt method.
answer
An orthonormal basis for the solution space of the equation can be the vectors
$\,\mA\,$ will have two real roots (counted with multiplicity). Note, that the opposite is not true: Two real eigenvalues can also occur from a non-symmetric matrix, jf. the previous exercises.
D
Show as a consequence: If $\,\mA\,$ is not a diagonal matrix, then it has two different (real) eigenvalues.
answer
The discriminant is 0, exactly when $\,c=0\,$ and $\,a=b\,.$
Exercise 7: Diagonalization by Orthogonal Substitution
$\mV$ can be orthogonalized by using Gram-Schmidt on the columns.
hint
Since $\mA$ is symmetric, the corresponding eigenspaces are orthogonal. So, it is only the two-dimensional eigenvector space, that you in this exercise need to expose to Gram-Schmidt. The one-dimensional eigenvector space only needs to be normalized.
Show that the characteristic polynomial for $\,\mA\,$ has a simple root -4 and a double root.
answer
The characteristic polynomial has the simple root -4 and the double root 2.
G
Determine a proper eigenvector $\,\mv_1\,$ of $\,\mA\,$ corresponding to the simple root -4.
answer
$\,\mv_1=\begin{matr}{rrr} -1 \\ -1 \\ 2 \end{matr}\,$ is a proper eigenvector for $\,\mA\,$ corresponding to the eigenvalue -4.
H
Show that the characteristic polynomial for $\,\mB\,$ also has a simple root 0 and a double root.
answer
The characteristic polynomial has the simple root 0 and the double root 3.
I
Determine a proper eigenvector $\,\mv_2\,$ for $\,\mB\,$ corresponding to the simple root 0.
answer
$\,\mv_2=\begin{matr}{rrr} 1 \\ 1 \\ 1 \end{matr}\,$ is a proper eigen vector for $\,\mB\,$ corresponding to the eigenvalue 0.
J
Show that $\,\mv_1\cdot\mv_2=\mnul\,.$
answer
When $\mv_1\cdot\mv_2=\mnul,$$\mv_1$ and $\mv_2$ are orthogonal.
K
Determine by using the results above an orthogonal matrix $\,\mathbf Q\,$ that can diagonalize both $\,\mA\,$ and $\,\mB\,$ by orthogonal substitution. State the results of both $\,\mathbf Q\transp\cdot\mA\cdot\mathbf Q\,$ and $\,\mathbf Q\transp\cdot\mB\cdot\mathbf Q\,.$
hint
$\mA$ and $\mB$ are symmetric. What is special about symmetrix matrices in this context?
hint
For $\mA$, $\mv_1$ is orthogonal to the eigenvector space $E_2$. In the same way $\mv_2$ is orthogonal to the eigenvector space $E_3$ for $\mB$.
hint
We can find the last, common eigenvector $\mv_3$ in the intersecting line between the two planes, $E_2$ for $\mA$ and $E_3$ for $\mB$, respectively, but can we do it more easily?
hint
$\mv_3$ is perpendicular both to $\mv_1$ and $\mv_2.$
hint
Therefore we can find $\mv_3$ as the cross product of $\mv_1$ and $\mv_2.$
hint
Then the three vectors only need to be normalized.
constitutes the coordinate vector with respect to the $v$-basis.
M
State an orthonormal basis for $\,\mathbf U\,.$
hint
The vector $\,\mv_1\,$, $\,\mv_2\,$ and $\,\mv_3\,$ constitute a basis for $\,\mathbf U\,,$ so they only need to be orthonormalized using Gram-Schmidt.
