- #1
sponsoredwalk
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Suppose we want to express the vector v = (1,-2,5) as a linear combination of the vectors
u = (1,1,1)
v = (1,2,3)
w = (2,-1,1)
We follow the method here:
[itex] x \ \begin{bmatrix}
1 \\
1 \\
1 \end{bmatrix} \ + \ y \
\begin{bmatrix}
1 \\
2 \\
3 \end{bmatrix} \ + \ z \
\begin{bmatrix}
2 \\
- 1 \\
1 \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix} \ =\ \begin{bmatrix}
x \\
x \\
x \end{bmatrix} \ + \
\begin{bmatrix}
y \\
2y \\
3y \end{bmatrix} \ + \
\begin{bmatrix}
2z \\
- z \\
z \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix}[/itex]
and continue, but I don't understand this fully.
The way I understand a 3-tuple is that (1,2,3) is
1 in the x axis, 2 in the y-axis & 3 in the z-axis.
I can't help but return to thinking this way & wanting
to write
u = (1,1,1)
v = (1,2,3)
w = (2,-1,1)
as
[itex]
x \ \begin{bmatrix}
1 \\
1 \\
2 \end{bmatrix} \ + \ y \
\begin{bmatrix}
1 \\
2 \\
-1 \end{bmatrix} \ + \ z \
\begin{bmatrix}
1 \\
3 \\
1 \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix} \ =\ \begin{bmatrix}
x \\
x \\
2x \end{bmatrix} \ + \
\begin{bmatrix}
y \\
2y \\
- y \end{bmatrix} \ + \
\begin{bmatrix}
z \\
3z \\
z \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix}[/itex]
Needless to say it's because I don't understand the reason why we do it
one way and not the other, I mean it doesn't make sense because in my
underdeveloped and confused understanding of linear algebra we can
transpose the vector and switch between the two ways I've done the
matrices here & none of it makes sense.
It would help a lot if anyone could clear this up with a good
explanation!
u = (1,1,1)
v = (1,2,3)
w = (2,-1,1)
We follow the method here:
[itex] x \ \begin{bmatrix}
1 \\
1 \\
1 \end{bmatrix} \ + \ y \
\begin{bmatrix}
1 \\
2 \\
3 \end{bmatrix} \ + \ z \
\begin{bmatrix}
2 \\
- 1 \\
1 \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix} \ =\ \begin{bmatrix}
x \\
x \\
x \end{bmatrix} \ + \
\begin{bmatrix}
y \\
2y \\
3y \end{bmatrix} \ + \
\begin{bmatrix}
2z \\
- z \\
z \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix}[/itex]
and continue, but I don't understand this fully.
The way I understand a 3-tuple is that (1,2,3) is
1 in the x axis, 2 in the y-axis & 3 in the z-axis.
I can't help but return to thinking this way & wanting
to write
u = (1,1,1)
v = (1,2,3)
w = (2,-1,1)
as
[itex]
x \ \begin{bmatrix}
1 \\
1 \\
2 \end{bmatrix} \ + \ y \
\begin{bmatrix}
1 \\
2 \\
-1 \end{bmatrix} \ + \ z \
\begin{bmatrix}
1 \\
3 \\
1 \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix} \ =\ \begin{bmatrix}
x \\
x \\
2x \end{bmatrix} \ + \
\begin{bmatrix}
y \\
2y \\
- y \end{bmatrix} \ + \
\begin{bmatrix}
z \\
3z \\
z \end{bmatrix} \ = \ \begin{bmatrix}
1 \\
-2 \\
5 \end{bmatrix}[/itex]
Needless to say it's because I don't understand the reason why we do it
one way and not the other, I mean it doesn't make sense because in my
underdeveloped and confused understanding of linear algebra we can
transpose the vector and switch between the two ways I've done the
matrices here & none of it makes sense.
It would help a lot if anyone could clear this up with a good
explanation!