Linear Combination of 3 Vectors equal to whole $\mathbb R^3$

Question

Let $\mathbf u=(1,2,-1), \mathbf v = (0,2,5), \mathbf w = (1,0,-2)$

Is every vector in $\mathbb R^3$ a linear combination of $\mathbf u, \mathbf v, \mathbf w$ ?

I decided to let $(C_1, C_2, C_3)$ be every vector in $\mathbb R^3$, if $\mathbb R^3$ is a linear combination of $\mathbf u, \mathbf v, \mathbf w$, then there exists $a, b, c \in \mathbb R$ such that:

$a\mathbf u + b\mathbf v + c\mathbf w = (C_1, C_2, C_3)$

I will then need to solve the following system of linear equations:

$$a+ c=C_1$$ $$2a+2b=C_2$$ $$-a+5b-2c=C_3$$

Solving the system using Gaussian Elimination gives me a identity matrix on the left side of the augmented matrix, the right side is some constants made up of $C_1, C_2, C_3$

Will this be sufficient to show that every vector in $\mathbb R^3$ a linear combination of $\mathbf u, \mathbf v, \mathbf w$ ?

Thanks for the help in advance!

Answer 1

You can further check that the three vectors form a basis for $\mathbb{R^3}$. If

$$A(1,2,-1)+B(0,2,5)+C(1,0,-2)=(0,0,0)$$

then $A+C=0, 2A+2B=0$ and $-A+5B-2C=0$ so that $A=B=C=0$, which shows that the vectors are linearly independent.

Answer 2

There are many ways to do it:

This question is equivalent to asking whether the list u,v,w is linearly independent. If it is linearly independent, then the answer to your question is yes. If it is not linearly independent (i.e. it's linearly dependent), then the answer is no.

The systematic way to solve this is to solve the homogenous equation $Ax = 0$ where the matrix A has the candidate basis vectors for columns. If the system of linear equations row reduces to the identity, then it is a basis.

Or, calculate the determinant to be 8, which is not 0. By the invertible matrix theorem, the answer to your question is yes.