Let
A basis for the row space of is
You should be able to explain and justify your answer. Enter a coordinate vector, such as \( \verb+<1,2,3,4>+ \) or \( \verb+<1,2,3,4,5>+ \), or a comma separated list of coordinate vectors, such as \( \verb+<1,2,3,4>,<5,6,7,8>+ \) or \( \verb+<1,2,3,4,5>,<6,7,8,9,10>+ \).
The dimension of the row space of is
because (select all correct answers -- there may be more than one correct answer):
A.
is the identity matrix.
B.
has a pivot in every row.
C.
The basis we found for the row space of has three vectors.
D.
One of the five columns in is a free variable column.
E.
Three of the four rows in have pivots.
F.
Two of the four rows in do not have a pivot.
The row space of is a subspace of
because
choose
each row of A is a vector in R^5
A has 4 rows
.
The geometry of the row space of is
choose
the origin inside R^4
a 1-dimensional line through the origin inside R^4
a 2-dimensional plane through the origin inside R^4
a 3-dimensional subspace of R^4
R^4
the origin inside R^5
a 1-dimensional line through the origin inside R^5
a 2-dimensional plane through the origin inside R^5
a 3-dimensional subspace of R^5
a 4-dimensional subspace of R^5
R^5
.