Consider the set of vectors in \(\mathbb R^3\):
\[S = \left\{ \begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}2\\4\\6\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}1\\4\\5\end{pmatrix} \right\}\]Find a vector in \(S\) which can be expressed as a linear combination of the other vectors in \(S\).
Any vector is a fine answer here. For example,
\[\begin{pmatrix}2\\4\\6\end{pmatrix} = 2\begin{pmatrix}1\\2\\3\end{pmatrix}\]
Make a new set of vectors \(T\) by removing your vector from part (a) from \(S\). Is \(T\) linearly independent? Explain.
Based on our answer to (a), we’d get
\[T = \left\{ \begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}1\\4\\5\end{pmatrix} \right\}\]This set isn’t linearly dependent, since,
\[\begin{pmatrix}1\\4\\5\end{pmatrix} = 2\begin{pmatrix}1\\2\\3\end{pmatrix} - \begin{pmatrix}1\\0\\1\end{pmatrix}\]
Find a vector in \(T\) which can be expressed as a linear combination of the other vectors in \(T\).
From our discussion in (b)
\[\begin{pmatrix}1\\4\\5\end{pmatrix} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \begin{pmatrix}1\\0\\1\end{pmatrix}\]
Make a new set of vectors \(U\) by removing your vector from part (c) from \(T\). Is \(U\) linearly independent? Explain.
Based on our answer to (c), we’d get
\[U = \left\{ \begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix} \right\}\]which is a linearly independent set of vectors since the matrix,
\[\begin{pmatrix}1 & 1 \\ 2 & 0 \\ 3 & 1 \end{pmatrix}\]has reduced row echelon form
\[\begin{pmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}\]that has a pivot in every column.
A unit vector is a vector whose magnitude is 1.
Describe all unit vectors \(\vec x\) in \(\mathbb R^2\).
There are a few different ways to express this answer. One way is any vector of the form,
\[\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}\]for any \(\theta\).
For which unit vectors \(\vec x\) is
\[S = \left\{ \begin{pmatrix}1\\0\end{pmatrix}, \vec x \right\}\]a basis for \(\mathbb R^2\)?
Any answer to part (a) except
\[\begin{pmatrix}\pm 1 \\ 0\end{pmatrix}\]works, since the determinant of the matrix
\[\begin{pmatrix}1 & \cos \theta\\ 0 & \sin \theta\end{pmatrix}\]is \(\sin \theta\). That is, it is a basis so long as \(\theta \ne \pi n\).
Find a basis for the vector space of diagonal \(2 \times 2\) matrices.
There are many correct answers. One standard one is,
\[\left\{ \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} \right\}\]
An upper triangular matrix is a matrix whose entries below diagonal entries are all 0. Find a basis for the vector space of upper triangular \(2 \times 2\) matrices.
There are many correct answers. One standard one is,
\[\left\{ \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} \right\}\]
Consider the two bases for \(\mathbb R^2\);
\[B = \left\{ \begin{pmatrix}2\\2\end{pmatrix}, \begin{pmatrix}4\\-1\end{pmatrix} \right\} \qquad C = \left\{ \begin{pmatrix}1\\3\end{pmatrix}, \begin{pmatrix}-1\\-1\end{pmatrix} \right\}\]Find a matrix \(M\) that changes column vectors for basis \(B\) into column vectors for basis \(C\).
The matrix,
\[M_B = \begin{pmatrix}2 & 4 \\ 2 & -1\end{pmatrix}\]changes \(B\)-vectors into standard vectors, and the matrix
\[M_C = \begin{pmatrix}1 & -1 \\ 3 & -1\end{pmatrix}\]changes \(C\)-vectors into standard vectors. On the other hand, \(M_C^{-1}\) changes standard vectors into \(C\)-vectors.
So the matrix \(M_C^{-1} M_B\) changes \(B\)-vectors into \(C\)-vectors.
Find a matrix \(N\) that changes column vectors for basis \(C\) into column vectors for basis \(B\).
We want to do the opposite of what we did in (a). This means, we want the inverse of our answer: \((M_C^{-1} M_B)^{-1} = M_B^{-1}M_C\) is the matrix that changes \(C\)-vectors into \(B\)-vectors.
Is 5 an eigenvalue of the matrix:
\[\begin{pmatrix}1&4\\2&3\end{pmatrix}?\]Explain your answer.
Yes, it is. This is because, if we plug in \(5\) into the characteristic polynomial of the matrix, we get the determinant of the matrix,
\[\begin{pmatrix}1-5 & 4 \\ 2 & 3-5\end{pmatrix}\]which is zero.