Find Linear Combination of u to Represent x

In summary: I solved it the right way and now I know how to do it in the future.In summary, x is a linear combination of u using theorem.
  • #1
Dustinsfl
2,281
5
Let x=[tex]
\begin{array}{cc|l}
1 \\
1 \\
7 \
\end{array}
[/tex]
write x as a linear combination of u using theorem.

u1=[tex]
\begin{array}{cc|l}
1/{3\sqrt{2}} \\
1/{3\sqrt{2}} \\
-4/{3\sqrt{2}} \
\end{array}
[/tex]

u2=[tex]
\begin{array}{cc|l}
2/3 \\
2/3 \\
1/3 \
\end{array}
[/tex]

u3=[tex]
\begin{array}{cc|l}
1/\sqrt{2} \\
-1/\sqrt{2} \\
0 \
\end{array}
[/tex]

v=[tex]\sum^n_{i=1} ciui[/tex]

I first did the rref of u and then wrote x in terms of the linear combination but it isn't the same as using the sum which I am not sure how to do.
 
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  • #2
You want to solve the matrix equation Ac = x for c.

Here A is a 3x3 matrix whose columns are u1, u2, and u3, c is <c1, c2, c3>T, the coefficients that specify a particular linear combination of the u vectors.
 
  • #3
I did that but it didn't turn out to be the same as using the sum definition.
 
  • #4
If you use rref, x=[tex]\frac{-13\sqrt{2}}{3}[/tex]u1+[tex]\frac{11}{3}[/tex]u2.


However, if I use the sum definition, the answer should be [tex]\frac{-\sqrt{2}}{3}[/tex]u1+[tex]\frac{5}{3}[/tex]u2


How can I use the sum definition to achieve that answer?
 
  • #5
Hint: The u vectors aren't linearly independent, so...
 
  • #6
Error in entry u3 second row should be negative
 
  • #7
Oh, then you're making an arithmetic error somewhere. The answer is unique, so either method should give the same result.
 
  • #8
I have entered it into a ti89, maple, and double checked all the rows and columns before doing rref and everything yields the same result.

I think it has to do with using the summation. This problem has to do with orthonormal basis.

I have already verified that it is orthonormal.

This is the Kronecker delta summation if that helps.
 
  • #9
Dustinsfl said:
I did that but it didn't turn out to be the same as using the sum definition.
I don't know what you're talking about. The matrix equation and the sum definition are describing exactly the same thing.

Here's the sum, expanded:
c1 u1 + c2 u2 + c3 u3 = x

Compare that to
[tex]\left[ \begin{array}{c c c}
u_{11} & u_{21} & u_{31} \\
u_{12} & u_{22} & u_{32} \\
u_{13} & u_{23} & u_{33} \end{array} \right]
\left[ \begin{array} {c}
c_1 \\
c_2 \\
c_3 \end{array} \right] =
\left[ \begin{array} {c}
1 \\
1 \\
7 \end{array} \right][/tex]

Notice that c1 multiplies the first column of the matrix, and c2 and c3 multiply the 2nd and 3rd columns, respectively.
 
  • #10
Dustinsfl said:
v=[tex]\sum^n_{i=1} ciui[/tex]
BTW, don't use the sub or sup tags inside tex tags. For subscripts, use underscore, e.g., c_i
 
  • #11
Mark44 said:
Here's the sum, expanded:
c1 u1 + c2 u2 + c3 u3 = x

Compare that to
[tex]\left[ \begin{array}{c c c}
u_{11} & u_{21} & u_{31} \\
u_{12} & u_{22} & u_{32} \\
u_{13} & u_{23} & u_{33} \end{array} \right]
\left[ \begin{array} {c}
c_1 \\
c_2 \\
c_3 \end{array} \right] =
\left[ \begin{array} {c}
1 \\
1 \\
7 \end{array} \right][/tex]

Notice that c1 multiplies the first column of the matrix, and c2 and c3 multiply the 2nd and 3rd columns, respectively.

That obtains my original answer which isn't the one I am looking for. (I just did it again that way)
 
  • #12
The ci's in the vector sum and in the matrix equation are the same, so why are you looking for different values? With a given set of vectors, there should be only one linear combination of them that gets you your vector x.

If you changed the coordinates of any of your vectors (such as by rationalizing the denominator), then you'll get a different set of constants, because you are, after all, working with a different set of vectors.
 
  • #13
[tex]\frac{-\sqrt{2}}{3}[/tex]u1+[tex]\frac{5}{3}[/tex]u2


This is the answer and none of the methods posted obtain this.
 
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  • #14
Isn't that what you found earlier?

That linear combination is equal to (1,1,1); your row-reduction solution is for x=(1,1,7).
 
  • #15
x=[tex]\frac{-13\sqrt{2}}{3}[/tex]u1+[tex]\frac{11}{3}[/tex]u2.

This is what the answer comes out to be in using all mentions suggested.
 
  • #16
Dustinsfl said:
[tex]\frac{-\sqrt{2}}{3}[/tex]u1+[tex]\frac{5}{3}[/tex]u2


This is the answer and none of the methods posted obtain this.
Does it check? The two answers you posted are easy enough to check.
 
  • #17
Well that was good practice. I have seen the issue. I wrote the problem down wrong; therefore, we all had a fun practice that wasn't needed.

x should be (1,1,1) transpose.

That was fun.
 

FAQ: Find Linear Combination of u to Represent x

How do you find the linear combination of u to represent x?

The linear combination of u to represent x is found by solving a system of equations. This involves setting up the given vectors as columns in a matrix, and then using row operations to reduce the matrix to row-echelon form. The resulting equations will give the coefficients of u that can be combined to represent x.

What is a linear combination?

A linear combination is a mathematical expression that involves multiplying each element of a set of vectors by a corresponding scalar coefficient, and then adding the resulting products together. This results in a new vector that is a combination of the original vectors.

Why is it important to find the linear combination of u to represent x?

Finding the linear combination of u to represent x is important in solving systems of linear equations, as it allows us to express one vector as a combination of other vectors. This can also be useful in analyzing data and making predictions based on multiple variables.

Are there any specific methods for finding the linear combination of u to represent x?

Yes, there are several methods for finding the linear combination of u to represent x. These include the Gaussian elimination method, the method of substitution, and the method of determinants. The method used will depend on the specific system of equations and personal preference.

Can the linear combination of u to represent x be found for any set of vectors?

No, not all sets of vectors have a linear combination that can represent a given vector. For example, if the vectors are linearly dependent, meaning that one vector can be expressed as a scalar multiple of another, then there is no unique solution for the linear combination. In this case, the vectors are said to span a subspace rather than forming a basis.

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