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

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

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!

• Do you know the definition of basis? – Jerry Jan 13 at 10:53
• @Jerry yeah, let V be a K-vector space. Then it means U ⊆ V a base of V, If U is a generator of V and linearly independent. The empty set ∅ is the base of the zero space {0} – user634297 Jan 13 at 11:08
• Did you study determinants? – Berci Jan 13 at 11:17
• @Berci No, Not yet but I looked now a YouTube video how to calculate determinants, and the first matrix is then -2 correct ? – user634297 Jan 13 at 11:31
• Yes, it's $-2$ for that. The determinant of a 3x3 matrix is the 3d (signed) volume of the parallelepiped spanned by its column vectors, and if those really span 3d, then the volume is nonzero, while if the vectors lie in a plane, the volume is 0. – Berci Jan 13 at 11:34

A systematic way to solve it would be to arrange the given vectors in a matrix, with respect to the given basis $$(v_1,v_2,v_3)$$.
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.