Let $v_1$, $v_2$, $v_3$ be a basis of the $\mathbb{R}$-vector space $\mathbb{R}^3$

Question

I'm not that good at math and would be very happy if you could give me some hints and so on

So my task is:

Let $\{v_1, v_2, v_3\}$ be a basis of the $\mathbb{R}$-vector space $\mathbb{R}^3$

• show that $\{v_1 + v_2, v_1 + v_3, v_2 + v_3\}$ is also a base of $\mathbb{R}^3$

• let $a, b, c \in \mathbb{R}$, so that $a ≠ b ≠ c$; show that $\{v_1 + v_2 + v_3, av_1 + bv_2 + cv_3, a^2v_1 + b^2v_2 + c^2v_3\}$ is also a base of $\mathbb{R}^3$

I will be gratefully for every help!

Answer 1

A systematic way to solve it would be to arrange the given vectors in a matrix, with respect to the given basis $(v_1,v_2,v_3)$.
It's usually done by writing the coordinates of the vectors (w.r.t the given basis!) in the columns of the matrix.
Since we have $v_1+v_2={\bf1}\cdot v_1+{\bf1}\cdot v_2+{\bf0}\cdot v_3$, its coordinates w.r.t basis $(v_1,v_2,v_3)$ are $(1,1,0)$, and we will use it as a column. Similarly taken coordinates for the other two vectors, for part 1, we get the matrix $$\pmatrix{1&1&0\\1&0&1\\0&1&1}$$ For part 2., the matrix is $$\pmatrix{1&a&a^2\\1&b&b^2\\1&c&c^2}$$ To conclude that the columns of a square matrix form a basis, all we have to check is that its determinant is nonzero.