Linear Algebra 
Test #2

May 5, 2003

Name____________________ 
R. Hammack

Score ______

(5) Consider the matrix
(a) Find the eigenvalues of A.
Eigenvlaues are 1 and 1.
(b) Find the eigenspaces of A.
To find the eigenvectors for 1: Compute Null
x = z
y = 2z
z = free
Eigenspace:
To find the eigenvectors for 1: Compute Null(A + I)
x = free
y = 0
z = free
Eigenspace:
(c) Diagonalize the matrix A, that is find an invertible matrix P and a diagonal matrix D with .
(6) Suppose is
an orthogonal basis for having
the property that ,
,
and .
Suppose also that satisfies
, ,
and .
Find .
Notice that:
,
so
,
so
,
so
Now since the basis is orthogonal,
(7) Suppose is
a basis for , and
is a linear transformation for which , ,
and .
Moreover, suppose v is a vector in for
which . Find
T(v).
(8) Suppose A is a matrix which has 5 rows, and the rows are linearly
independent. Suppose also that Null(A) is twodimensional. How many columns
does A have?
Since the rows are linearly independent, the 5 rows form a basis for the row
space.
Thus rank(A) = dim(Row(A)) = 5.
By Rank Theorem: Rank(A) + Nullity(A) = n = (number of
columns)
Thus (number of columns) = 5 + 2 = 7
(9) Suppose A is a 5×5 matrix with a onedimensional null
space. Find det(A).
Since the null space is one dimensional, it follows that Null(A) = span(x)
for some nonzero vector x.
This means that Nullity(A) = 1. Since Nullity(A) is not zero,
A is not invertible, by the invertible matrix theorem. Hence det(A)
= 0.