Answer to Question #158373 in Linear Algebra for Suchita

Question

Let W be a subspace of R^5, which is spanned by the vectors u₁ = (1, 2, 1, 0, 0) u₂=(0, 1, 3, 3, 1) u₃= (1, 4, 6, 4, 1)

Find a basis for W₀

Answer 1

Solution:

 Step 1: Set up a homogeneous system of equations

The set S = {v₁, v₂, v₃} of vectors in R⁵ is linearly independent if the only solution of (*) c₁v₁ + c₂v₂ + c₃v₃ = 0 is c₁, c₂, c₃ = 0.

In this case, the set S forms a basis for span S. Otherwise (i.e., if a solution with at least some nonzero values exists), S is linearly dependent. If this is the case, a subset of S can be found that forms a basis for span S.

With our vectors v₁, v₂, v₃, (*) becomes:

$c_1\begin{bmatrix} 1 \\ 2 \\ 1\\ 0\\ 0 \end{bmatrix}+c_2\begin{bmatrix} 0 \\ 1 \\ 3\\ 3\\ 1 \end{bmatrix}+c_3\begin{bmatrix} 1 \\ 4 \\ 6\\ 4\\ 1 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0\\ 0\\ 0 \end{bmatrix}$

Rearranging the left-hand side yields

$\begin{bmatrix} 1c_1 + 0c_2+1c_3 \\ 2c_1+1c_2+4c_3\\ 1c_1+3c_2+6c_3\\ 0c_1+3c_2+4c_3\\ 0c_1+1c_2+1c_3 \end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\\0 \end{bmatrix}$

The matrix equation above is equivalent to the following homogeneous system of equations

(**) 

$1c_1+0c_2+1c_3=0$

$2c_1+1c_2+4c_3=0$

$1c_1+3c_2+6c_3=0$

$0c_1+3c_2+4c_3=0$

$0c_1+1c_2+1c_3=0$

Step 2: Transform the coefficient matrix of the system to the reduced row echelon form

We now transform the coefficient matrix of the homogeneous system above to the reduced row echelon form to determine whether the system has

• the trivial solution only (meaning that S is linearly independent), or
• the trivial solution as well as nontrivial ones (S is linearly dependent).

$1\;\;0\;\;1\\2\;\;1\;\;4\\1\;\;3\;\;6\\0\;\;3\;\;4\\0\;\;1\;\;1$

can be transformed by a sequence of elementary row operations to matrix

$1\;\;0\;\;0\\0\;\;1\;\;0\\0\;\;0\;\;1\\0\;\;0\;\;0\\0\;\;0\;\;0$

 Step 3: Interpret the reduced row echelon form

The reduced row echelon form of the coefficient matrix of the homogeneous system (**) is

$1\;\;0\;\;0\\0\;\;1\;\;0\\0\;\;0\;\;1\\0\;\;0\;\;0\\0\;\;0\;\;0$

which corresponds to the system

$1c_1 \;\;\;\;\;\;\;\;\;\;\;=0\\\;\;\;\;\;\;1c_2\;\;\;\;\;=0\\\;\;\;\;\;\;\;\;\;\;\;1c_3=0\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;0=0\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;0=0$

Since each column contains a leading entry (highlighted in yellow), then the system has only the trivial solution, so that the only solution of (*) is c₁, c₂, c₃ = 0.

Therefore the set S = {v₁, v₂, v₃} is linearly independent.

Consequently, the set S forms a basis for span S.