# Transcribed Image Text: [1 1 1 0 0 0 [1 1 0 0 0 0 1 1 01 X1 X2 X3 X4 X5 X6 1

Transcribed Image Text: [1
1 1
0 0 0
[1
1 0 0 0
0 1 1
01
X1 X2 X3 X4 X5 X6
1
1
2
2
2
2 2
Problem 3: The 4 x 6 matrix A and its row-reduced echelon form are: A =
3
3
3
01
4
4]
1
1.
X1 X2 X3 X4 X5 X6
a. Fill out the following table of properties of the matrix:
Dimension of the
Dimension of
Dimension of
Is the transformation
Is the transformation
Rank
Domain of A
Codomain of A
null space
onto?
one-to-one?
X2
-X4
b. In parametric form, the solutions to the homogeneous equation AX =
are: x
X4
X4 (free)
X5
-X6
X6 (free)]
In the box, write a basis for the null space of matrix A.
Call the two vectors in the basis n, and n2.
Remember! Rows are rows and columns are columns.
Note that your two basis vectors just happen to be orthogonal!
(This is not always true, but the problem was designed this way.)
c. Normalize your two vectors so they form an orthonormal basis uj
and
for the null space N. Divide both vectors by their mutual 2-norm of
-1
d. The vector ỹ
= ŷ +z can be written (uniquely) as the sum of a vector ý in the null space of A, and a vector z orthogonal
to the null space. Find the component ý of which lies in the null space.
Hint: ŷ =
e. Find the component z of y which is orthogonal to the null space.
i = ỹ – ŷ
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to