# Linear combination

Let x=$$\begin{array}{cc|l} 1 \\ 1 \\ 7 \ \end{array}$$
write x as a linear combination of u using theorem.

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

u2=$$\begin{array}{cc|l} 2/3 \\ 2/3 \\ 1/3 \ \end{array}$$

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

v=$$\sum^n_{i=1} ciui$$

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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## Answers and Replies

Mark44
Mentor
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.

I did that but it didn't turn out to be the same as using the sum definition.

If you use rref, x=$$\frac{-13\sqrt{2}}{3}$$u1+$$\frac{11}{3}$$u2.

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

How can I use the sum definition to achieve that answer?

vela
Staff Emeritus
Homework Helper
Hint: The u vectors aren't linearly independent, so....

Error in entry u3 second row should be negative

vela
Staff Emeritus
Homework Helper
Oh, then you're making an arithmetic error somewhere. The answer is unique, so either method should give the same result.

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.

Mark44
Mentor
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
$$\left[ \begin{array}{c c c} u_{11} & u_{21} & u_{31} \\ u_{12} & u_{22} & u_{32} \\ u_{13} & u_{23} & u_{33} \end{array} \right] \left[ \begin{array} {c} c_1 \\ c_2 \\ c_3 \end{array} \right] = \left[ \begin{array} {c} 1 \\ 1 \\ 7 \end{array} \right]$$

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

Mark44
Mentor
v=$$\sum^n_{i=1} ciui$$
BTW, don't use the sub or sup tags inside tex tags. For subscripts, use underscore, e.g., c_i

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

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

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)

Mark44
Mentor
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.

$$\frac{-\sqrt{2}}{3}$$u1+$$\frac{5}{3}$$u2

This is the answer and none of the methods posted obtain this.

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vela
Staff Emeritus
Homework Helper
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).

x=$$\frac{-13\sqrt{2}}{3}$$u1+$$\frac{11}{3}$$u2.

This is what the answer comes out to be in using all mentions suggested.

Mark44
Mentor
$$\frac{-\sqrt{2}}{3}$$u1+$$\frac{5}{3}$$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.

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.