Question

Let S={v1,v2,v3} be a linearly independent set of vectors in

. Determine whether T={2 v1+ v2 - 2 v3, 4v1+ 3v2 ,3 v1+ 2 v2- v3} is linearly independent.

Answer 1

S = {v₁, v₂, v₃} is linearly independent, which means

c₁v₁ + c₂v₂ + c₃v₃ = 0

if and only if c₁ = c₂ = c₃ = 0.

T = {w₁, w₂, w₃} = {2v₁ + v₂ - 2v₃, 4v₁ + 3v₂, 3v₁ + 2v₂ - v₃}

is linearly independent if the same is true for the vectors w₁, w₂, w₃, i.e.

c₁w₁ + c₂w₂ + c₃w₃ = 0

iff c₁ = c₂ = c₃ = 0.

Expanding this equation, we get

c₁ (2v₁ + v₂ - 2v₃) + c₂ (4v₁ + 3v₂) + c₃ (3v₁ + 2v₂ - v₃) = 0

(2c₁ + 4c₂ + 3c₃) v₁ + (c₁ + 3c₂ + 2c₃) v₂ + (-2c₁ - c₃) v₃ = 0

Put the coefficients of v₁, v₂, v₃ in this expansion into an augmented matrix and solve the linear system:

`\left[\begin{array}c2&4&3&0\\1&3&2&0\\-2&0&-1&0\end{array}\right]`

Row-reduce the matrix:

• Replace row 2 with [row 1 + (-2) row 2]:

`\left[\begin{array}c2&4&3&0\\0&-2&-1&0\\-2&0&-1&0\end{array}\right]`

• Replace row 3 with [row 1 + row 3]:

`\left[\begin{array}c2&4&3&0\\0&-2&-1&0\\0&4&2&0\end{array}\right]`

Right away, we see that row 3 is a scalar multiple of row 2, since row 3 = -2 (row 2), so T is not linearly independent.

Continuing with the row reduction, we'd end up with

`\left[\begin{array}c2&0&1&0\\0&2&1&0\\0&0&0&0\end{array}\right]`

so if we choose, for instance, c₃ = 1, so that c₁ = c₂ = -1/2, we get

-1/2 (2v₁ + v₂ - 2v₃) - 1/2 (4v₁ + 3v₂) + (3v₁ + 2v₂ - v₃)

= (-v₁ - 1/2 v₂ + v₃) + (-2v₁ - 3/2v₂) + (3v₁ + 2v₂ - v₃)

= (-1 - 2 + 3) v₁ + (-1/2 - 3/2 + 2) v₂ + (1 - 1) v₃) = 0