The set of vectors {a, b, c} is a basis of R³. Determine if another set, {3a-2b, a+4b+5c, a-2c}, is also a basis of R³.

Question

Since the set of vectors $\{\vec{a},\vec{b},\vec{c}\}$ are a basis of $\mathbb{R^3}$, we can understand that they are linearly independent. This means that:

• $\alpha_1\cdotp\vec{a}+\alpha_2\cdotp\vec{b}+\alpha_3\cdotp\vec{c}=\vec{0}$ , or
• $\det\{\vec{a},\vec{b},\vec{c}\}\neq0$ .

I tried to solve the second option by assigning values to the $\{\vec{a},\vec{b},\vec{c}\}$. I concluded that $\{3\vec{a}-2\vec{b},\vec{a}+4\vec{b}+5\vec{c},\vec{a}-2\vec{c}\}$ is also linearly independent, therefore, is a basis of $\mathbb{R^3}$:

$$\vec{a}=(1,0,1)\\\vec{b}=(-2,1,4)\\\vec{c}=(0,3,1)$$ $$\det\begin{vmatrix}1&-2&0\\0&1&3\\1&4&1\end{vmatrix}=-17$$
$$3\vec{a}-2\vec{b}=(7,-2,-5)\\\vec{a}+4\vec{b}+5\vec{c}=(-7,19,20)\\\vec{a}-2\vec{c}=(1,-6,-1)$$ $$\det\begin{vmatrix}7&-7&1\\-2&19&-6\\-5&20&-1\end{vmatrix}=\underline{566}$$

So, $\{3\vec{a}-2\vec{b},\vec{a}+4\vec{b}+5\vec{c},\vec{a}-2\vec{c}\}$ is also linearly independent, therefore, is a basis of $\mathbb{R^3}$.
The problem is, the question asks for a general statement, which I cannot prove by giving an example. So how could I determine whether the set of $\{3\vec{a}-2\vec{b},\vec{a}+4\vec{b}+5\vec{c},\vec{a}-2\vec{c}\}$ is also a basis of $\mathbb{R^3}$ or not?

Answer 1

The matrix of the second set is

$[ 3 a - 2 b , a + 4 b + 5 c, a - 2 c] = [a,b,c] \begin{bmatrix} 3 && 1 && 1 \\ -2 && 4 && 0 \\ 0 && 5 && -2 \end{bmatrix} $

Now, $\begin{vmatrix} 3 && 1 && 1 \\ -2 && 4 && 0 \\ 0 && 5 && -2 \end{vmatrix} = 3(-8) + 2 (-7) = -38 \ne 0 $

And since the determinant of a product of two matrices is the product of the determinants of the two matrices, we conclude that the determinant of $[3 a - 2 b , a + b + 5c, a -2 c]$ is non-zero, and this means its columns are linearly independent, and thus they form a basis for $\mathbb{R}^3$.

Answer 2

$$\alpha (3a-2b)+\beta (a+4b+5c)+\gamma (a-2c)=(3\alpha +\beta +\gamma)a+(-2\alpha +4\beta)b+(5\beta -2\gamma)c$$ Now, since $a,b,c$ are linearly independent, the RHS is non-zero, and hence so must be the LHS. So, any linear combination of $3a-2b,a+4b+5c,a-2c$ is non-zero (since $\alpha, \beta, \gamma$ were chosen arbitrarily), and so the vectors are linearly independent.

Since dimension of $\mathbb R^3$ is $3$, any three linearly independent vectors must span $\mathbb R^3$ and so $\{3a-2b,a+4b+5c,a-2c\}$ is indeed a basis of $\mathbb R^3$.