Class 25: Question 1
Which of the following vectors is
orthogonal to the row space of A?
Class 25: Answer 1: A
The row space is a subspace of Rn, as is the nullspace.
The two are orthogonal complements of each other which
means any vector in one is orthogonal to vectors in the
other.
Thus one way to solve the problem is to find the
nullspace of A, which ends up being span{(-1,-1,1)}.
Thus the only vector in the 4 given which fits is (1,1,-1)
Class 25: Question 2
Which of the following vectors is
orthogonal to the column space of A?
Class 25: Answer 2: C
The column space is a subspace of Rm, as is the left
nullspace. The two are orthogonal complements of each
other which means any vector in one is orthogonal to
vectors in the other. Since it’s easier to deal with the
column space, let’s do that.
The column space of A ends up being
span{(1,0,0),(0,2,1)}. Thus we need to see which of the
vectors is orthogonal to these two. One way would be to
take the cross product, but it’s easier just to check the
vectors we have
Thus the only vector in the 4 given which fits is (0,1,-2)
Class 25: Question 3
Which of the following vectors is
orthogonal to the nullspace of A?
Class 25: Answer 3: D
This question is asking about vectors which are
orthogonal to the nullspace, thus they must lie in the row
space.
The row space of A is clearly span{(1,0,1),(0,1,1)}.
Thus we need to see which of the vectors is in this vector
space.
Thus the only vector in the 4 given which fits is (2,0,2)
Class 25: Question 4
TRUE or FALSE: Any set of non-zero orthogonal
vectors must be linearly independent.
1. TRUE.
2. FALSE.
Class 25: Answer 4: A
In order to check whether a set of vectors is linearly
independent we need to solve the homogeneous linear
system Ax=0 where the columns of A are the given
vectors and x and 0 are vectors.
In other words:
If the set of vectors is orthogonal to each other then if we
take the dot product of both sides
Thus c1=0 since these are nonzero vectors.
The process can be repeated for all n vectors to show that
all the constants must be zero and thus the set of orthogonal
vectors is linearly independent.
Class 25: Question 5
1. TRUE.
2. FALSE.
Class 25: Answer 5 (B)
There are multiple bases for R2 so there must be multiple
orthogonal bases for R2. All you need to do is pick two
vectors that are not scalar multiples of each other, are
orthogonal and then normalize them to produce multiple
orthonormal bases in R2. Here are some more
Class 25: Question 6
Let Q be a square matrix with orthonormal columns.
TRUE or FALSE: Q-1=QT.
1. TRUE.
2. FALSE.
Class 25: Answer 6: A
If Q is a square matrix with orthonormal columns then QTQ
must equal the identity matrix.
This is clear because if you multiply a matrix by it’s transpose
then each element of the product consists of dot products of the
rows of the matrix with it’s own columns.
Since the matrix has orthonormal columns those dot products
will produce either 1 or 0, which will result in the identity
matrix. The product has 1 along the diagonal because that is
when a particular column vector is being dotted with itself.
Thus since QTQ=I that must mean that Q-1=QT.
Class 25: Question 7
Class 25: Answer 7 (A)
Class 25: Question 8
Class 25: Answer 8: A
The key thing to understand about this question is that it is the
same question as the one before!
In other words, to make an orthogonal projection onto a
subspace one needs to have a basis for that subspace.
What is a basis for the subspace corresponding to the line
y=x/2? The line corresponds to span{(2,1)}.
Thus (2,1) is a basis for the subspace.
Thus the orthogonal projection of (-3,1) onto (2,1) is (2,1).
Class 25: Question 9
Class 25: Answer 9: D
The line named l is y=3x. This is the subspace span{(1,3)}.
If z is the projection of b onto this subspace it will be some
scalar multiple of (1,3), i.e. (c,3c).
However, the more important part of the problem is the
interpretation of b-z which equals b-projl(b).
Recall that this new vector b-z will be orthogonal to l by
definition.
There’s no way b-z could be a point on l and orthogonal to it.
Thus only two of the statements are true.