Consider the set of vectors in :
Find a vector in which can be expressed as a linear combination of the other vectors in .
Any vector is a fine answer here. For example,
Make a new set of vectors by removing your vector from part (a) from . Is linearly independent? Explain.
Based on our answer to (a), we’d get
This set isn’t linearly dependent, since,
Find a vector in which can be expressed as a linear combination of the other vectors in .
From our discussion in (b)
Make a new set of vectors by removing your vector from part (c) from . Is linearly independent? Explain.
Based on our answer to (c), we’d get
which is a linearly independent set of vectors since the matrix,
has reduced row echelon form
that has a pivot in every column.
A unit vector is a vector whose magnitude is 1.
Describe all unit vectors in .
There are a few different ways to express this answer. One way is any vector of the form,
for any .
For which unit vectors is
a basis for ?
Any answer to part (a) except
works, since the determinant of the matrix
is . That is, it is a basis so long as .
Find a basis for the vector space of diagonal matrices.
There are many correct answers. One standard one is,
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 matrices.
There are many correct answers. One standard one is,
Consider the two bases for ;
Find a matrix that changes column vectors for basis into column vectors for basis .
The matrix,
changes -vectors into standard vectors, and the matrix
changes -vectors into standard vectors. On the other hand, changes standard vectors into -vectors.
So the matrix changes -vectors into -vectors.
Find a matrix that changes column vectors for basis into column vectors for basis .
We want to do the opposite of what we did in (a). This means, we want the inverse of our answer: is the matrix that changes -vectors into -vectors.
Is 5 an eigenvalue of the matrix:
Explain your answer.
Yes, it is. This is because, if we plug in into the characteristic polynomial of the matrix, we get the determinant of the matrix,
which is zero.