For a set of vectors to span R4 question

[Samuelb88]

Homework Statement

The set of vectors u = {1,-2,2,1}, v = {1,3,1,1}, w = {3,4,4,3} cannot span R4. Complete this set to create a set of vectors that will span R4. Show that your set of vectors spans R4.

The Attempt at a Solution

Let $$y = {y_1,y_2,y_3,y_4}$$. I write span{u,v,w,y} as the coefficient matrix:

[1,1,3,y_1
-2,3,4,y_2
2,1,4,y_3
1,1,3,y_4]

Using the first row to produce zeros in each row below yields:

[1,1,3,y_1
0,5,10,y_2+2y_1
0,-1,-2,y_3-2y_1
0,0,0,y_4-y_1]

Using the second row to produce zeros in the row below yields:

[1,1,3,y_1
0,5,10,y_2+2y_1
0,0,0,5y_3-9y_2+2y_1
0,0,0,y_4-y_1]

So to my understanding, it would seem given the set {u,v,w}, a fourth vector y cannot be chosen so that the set {u,v,w,y} spans R4 since not every row can contain a pivot position in this case. Please correct me if I am wrong.

- Sam

PS sorry for the messy work - I don't know how to write matrices in latex.

 

[Mark44]

The problem is that u, v, and w are linearly dependent. Have you learned about the Gram-Schmidt process for constructing a basis yet?

 

[Samuelb88]

The problem is that u, v, and w are linearly dependent. Have you learned about the Gram-Schmidt process for constructing a basis yet?

No I haven't. To my understanding though, the Gram-Schmidt process creates a set (for this case) of vectors {u',v',w'} so that v' is orthogonal to u', w' is orthogonal to v' and I suppose u', v', and w' are linearly independent so I may choose a fourth vector y' to "complete" this set so that this completed set spans R4?

 

[Mark44]

You can use G-S to construct all four vectors of a basis for R⁴.

 

[vela]

So to my understanding, it would seem given the set {u,v,w}, a fourth vector y cannot be chosen so that the set {u,v,w,y} spans R4 since not every row can contain a pivot position in this case. Please correct me if I am wrong.

You're right. You can't add just one more vector and span R⁴. You need to add two or more because, as Mark noted, the three you have are linearly dependent.

PS sorry for the messy work - I don't know how to write matrices in latex.

You can use the bmatrix environment to produce matrices in LaTeX. Click on the one below to see what the code is.

$$\begin{bmatrix}<br /> 1 & 1 & 3 & y_1 \\<br /> -2 & 3 & 4 & y_2 \\<br /> 2 & 1 & 4 & y_3 \\<br /> 1 & 1 & 3 & y_4<br /> \end{bmatrix}$$

 

[Mark44]

Don't include 0. Your u, v, and w are linearly dependent, which can be seen from the first 3 columns of your last matrix (but taken one step further).

$$\begin{bmatrix}1 & 0 & 1 \\0 & 1 & 2 \\0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}$$

This says c₁ + c₃ = 0, and that c₂ + 2c₃= 0, where c₁, c_w, and c₃ are the coefficients of u, v, and w. If you let c₃ = 1, then c₁ = -1, c₂ = -2, and c₃ = 1.

In other words -u -2v + w = 0, so it can be seen that anyone of these vectors can be solved for in terms of the other two.

Without knowing G-S, I suppose the best way is to pick two of the three vectors, say u and v, and then try to find two more vectors out of the set of standard basis vectors for R⁴ - {<1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, <0, 0, 0, 1>} - so that you have four linearly independent vectors.

For your spanning set you can throw in w.