Question

[10 points] Let v1 = (−2, −7, 2), v2 = (5, 1, −5), v3 = (−4, −9, 4), v4 = (−3, −6, 3), v5 = (1, 2, −1). Find a subset of {v1, v2, v3, v4, v5} that forms a basis for the subspace of R 3 spanned by those five vectors. Express each non-basis vector as a linear combination of basis vectors

Answer 1

Answer:

A basis is {v1,v4}

v2 = 3 v1 -11/3 v4

v3 = 1/3 v1 + 10/9 v4

v5 = -1/3 v4

Step-by-step explanation:

Lets triangulate the matrix that has the elements v1, v2, v3, v4 and v5 as their rows

$\left[\begin{array}{ccc}-2&-7&2\\5&1&-5\\-4&-9&4\\-3&-6&3\\1&2&-1 \end{array}\right]$

Lets divide each row by its first element, which is possible because it is different than 0

$\left[\begin{array}{ccc}1&3.5&-1\\1&0.2&-1\\1&2.25&-1\\1&2&-1\\1&2&-1 \end{array}\right]$

Thus v5 = v4 / -3 = -1/3 * v4. We can take it out. Also, we will take each row except the first one and substract the value of the first row from them, obtaining

$\left[\begin{array}{ccc}1&3.5&-1\\0&3.3&0\\0&1.25&0\\0&1.5&0\end{array}\right]$

Note that now, from the second row onwards, each row is a multiple of the other. So we can remove every row except 2 of them, for example the first one and the fourth one

We can form a base of the subspace of R³ spanned by v1,v2,v3,v4,v5 with {v1,v4}. Also, we have

v5 = -1/3*v4

(v1/-2-v3/-4)*1.5 = (v1/-2-v4/-3)*1.25, hence

0.25v1/-2 + 1.25 v4/-3 = 1.5 v3/-4

Thus

v3 = -8/3 *0.25 v1/-2 - 8/3*1.25 v4/-3 = 1/3 v1 + 10/9 v4

In a similar way,

(v1/-2-v2/5)*1.5 = (v1/-2-v4/-3)*3.3

Thus

v2 = 10/3 * (-1.8v1/-2 + 3.3v4/-3) = 3 v1 -11/3 v4