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Show that the vectors $$\begin{bmatrix} 1 \\ 2 \\ 0 \\ 2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \end{bmatrix}\begin{bmatrix} 2 \\ 0 \\ 1 \\ 3 \end{bmatrix}$$ are linearly indepedent. Find a fourth vector so that the set is linearly independent.

Solution:

I row reduced a matrix composed of these vectors and found the columns were linearly independent.

To find a 4th vector so that the set is linearly independent - I picked a random vector, say, $ V= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

I then applied the Gram Schmidt process - I subtracted from V the projection of V onto each of the original vectors. Is this the correct way to find a linearly independent 4th vector?

My final answer was $\begin{bmatrix} \frac{17}{63} \\ \frac{-35}{63} \\ \frac{-30}{63} \\ \frac{-41}{63} \end{bmatrix}$

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There are simpler ways to obtain an $n$-th vector of a basis of $\Bbb R^n$ given the first $n-1$. For instance, in your case, the vector $(4,3,2,0)$ whose first 3 components are the sum of the corresponding components in the given 3 vectors but the last isn't is obviously linear independent from the others. –  Andrea Mori Apr 20 '12 at 8:18
    
But how is it obviously linearly independent? just because $1v_1 + 1v_2+ 1v_3 = (4,3,2,0)$ is false doesnt obviously rule out other possible combinations of the vectors making the statement true? –  Jim_CS Apr 20 '12 at 10:08
2  
forget the last component. Then the vectors $v_1, v_2, v_3$ form a basis of $\Bbb R^3$. Then the vector $(4,3,2)$ is a linear combination of $v_1,v_2,v_3$ in a unique way. When you include back the $4$-th coordinates, the relation either remains valid (and the vectors are linearly dependent), or it doesn't, as in the example. If the relation doesn't remain valid, no other relation can be obtained, as this "new" relation would give a second relation between the vectors without the last coordinates. –  Andrea Mori Apr 20 '12 at 16:52

1 Answer 1

Gram-Schmidt process is for orthogonalization (so that, given a set of already linearly independent vectors, you can construct an orthonormal basis).

If the set of original three vectors with $V$ you've chosen would have been linearly dependent, by substracting the projection of $V$ onto each of the original vectors you would get $0$.

The easiest way to find a fourth vector is to try all the $e_i$ vectors. As the original set is linearly independent, and its size is less than space dimension, at least one of $e_i$ is linearly independent of the original set.

You already found that $e_1$ is linearly independent of the original set, so you may use it as an answer.

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