Question

Find the dimension of the subspace spanned by the vectors e1,e2,e3 in v4(R)

Answer 1

Question 1.

Find the dimension of the subspace spanned by the vectors $e_{1},e_{2},e_{3}$ in $V_{4}(\mathbb{R})$.

Solution. Recall that

$e_{1}=(1,0,0,0),$

$e_{2}=(0,1,0,0),$

$e_{3}=(0,0,1,0).$

Prove that $e_{1},e_{2},e_{3}$ are linearly independent in over $\mathbb{R}$. Indeed, for any $\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{R}$ we have

$\alpha_{1}e_{1}+\alpha_{2}e_{2}+\alpha_{3}e_{3}=(\alpha_{1},\alpha_{2},\alpha_{3},0).$

So, if $\alpha_{1}e_{1}+\alpha_{2}e_{2}+\alpha_{3}e_{3}=(0,0,0,0)$, then immediately $\alpha_{1}=\alpha_{2}=\alpha_{3}=0$. Since these vectors are linearly independent, they form a basis of the space spanned by them. Thus, the dimension of this space equals the number of the vectors $e_{1},e_{2},e_{3}$, i. e. it is 3.

Answer: 3. $\square$