The
vector space
of subtractive color mixing is created from real multiples of vectors
red
,
yellow
, and
blue
with the understanding that
red
+
yellow
+
blue
=
0
. (When mixing colors, red, yellow, and blue combine to make black.)
For instance, some elements of the vector space are
2
red
+
π
yellow
and
−
3
red
+
2
9
yellow
+
e
3
blue
.
However,
3
red
+
2
blue
and
red
−
2
yellow
are the same vector
,
since they differ by a multiple of
0
.
(
They differ by
2
⋅
(
red
+
yellow
+
blue
)
=
0
.
)
What is true of the following sets of vectors in this vector space?
{
red
+
yellow
+
blue
,
red
+
2
yellow
+
4
blue
}
{
red
,
yellow
,
blue
}
{
red
+
yellow
,
yellow
+
blue
}
Only 2 spans the space
Only 3 is linearly independent
1 and 3 are linearly independent
1 and 3 are linearly independent; 2 spans the space
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Option 1 contains
red
+
yellow
+
blue
=
0
, so it is not linearly independent. Since there is no way to express
red
as a multiple of the other elements, it does not span the space.
Option 2 spans the space, since the space was defined in terms of those elements. It is not linearly independent though, because the sum of its basis elements is
0
.
Option 3 is linearly independent, since there are no real numbers
a
and
b
, aside from
a
=
0
and
b
=
0
, such that
a
⋅
(
red
+
yellow
)
+
b
⋅
(
yellow
+
blue
)
=
0
.
Furthermore, it spans the space, since any element of the space can be written in the form
x
red
+
y
yellow
by subtracting the appropriate number of copies of
red
+
yellow
+
blue
. And any element of that form is expressible as a linear combination of
red
+
yellow
and
red
=
0
−
(
yellow
+
blue
)
.
So option 3 is a basis and neither other option is.
How do we get the vector red+blue from the third option? I am confused..
Option 1 contains red + yellow + blue = 0 , so it is not linearly independent. Since there is no way to express red as a multiple of the other elements, it does not span the space.
Option 2 spans the space, since the space was defined in terms of those elements. It is not linearly independent though, because the sum of its basis elements is 0 .
Option 3 is linearly independent, since there are no real numbers a and b , aside from a = 0 and b = 0 , such that a ⋅ ( red + yellow ) + b ⋅ ( yellow + blue ) = 0 . Furthermore, it spans the space, since any element of the space can be written in the form x red + y yellow by subtracting the appropriate number of copies of red + yellow + blue . And any element of that form is expressible as a linear combination of red + yellow and red = 0 − ( yellow + blue ) .
So option 3 is a basis and neither other option is.