- 1.
- The projection
matrix P onto the plane in
R3 spanned by the vectors
u =
1, 1, 0
and
v =
1, 1, 1
is the product
P = A
ATA
AT, where u and v are the columns of A.
BITMAPSETProbSolvHint0.2006in0.243in0inq1
Note that Pw is a linear combination of u and v for any vector
w =
x, y, z
in
R3, so P maps
R3 onto the
plane spanned by u and v.
To find a nonzero vector b that is projected to zero, leave the insertion
point in the matrix P, and from the Matrices submenu choose
Nullspace Basis.
- 2.
- The matrix


has characteristic polynomial
X - 2
, minimum
polynomial
4 - 4X + X2 =
X - 2
, and eigenvalue and
eigenvectors BITMAPSETProbSolvHint0.2006in0.243in0inq2
The minimal polynomial is a factor of the characteristic polynomial. The
eigenvalue 2 occurs with multiplicity 4 as a root of the characteristic
polynomial
X - 2
. The eigenvalue 2 has two linearly
independent eigenvectors. Note that
- 3.
- The solutions of this equation are in
R3,
and the column space of A is a subset of
R2, so these solutions
cannot be the column space of A. They do form the nullspace of A by the
definition of nullspace; consequently, this set is a subspace of
R3. The product of A with the first row of A is


![$\left.\vphantom{
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 2
\end{array}
}\right]$](img13.png)


= 

, which is not


, so the solution set is not the row space of A.BITMAPSETProbSolvHint0.2006in0.243in0inq3
To determine whether this subspace of
R3 is a point, line, or
plane, solve the system of equations: from the Solve submenu,
choose Exact
The subspace is a line. It is the line that passes through the origin and
the point


.
- 4.
- Apply the following operations in turn.BITMAPSETProbSolvHint0.2006in0.243in0inq4
This result proves the first identity. For the second part, carry out the
following steps.
- Evaluate
- Evaluate
- Leave the insertion point in the matrix and choose Simplify,
or from the Combine submenu choose Trig Functions.