# How to find $v_4$ such that $span\{v_1, v_2, v_3, v_4\}=\mathbb{R}^4$?

I am in trouble with these exercise: Consider the vectors $$v_1=(1, 0, 0, 1), v_2=(2, h, h, 2), v_3=(1, 1+h, 2h +1, 1)$$ with $$h\in\mathbb{R}$$. Determine the dimension of the subspace $$S$$ generated by $$v_1, v_2, v_3, v_4$$ by varying $$h\in\mathbb{R}$$. Thus, choosen $$h\in\mathbb{R}$$ such that $$dim(S) =3$$, find $$v_4$$ such that $$span\{v_1, v_2, v_3, v_4\} = \mathbb{R}^4$$.

In order to find the dimension of the generated subspace $$S$$ by varying $$h\in\mathbb{R}$$, I evaluate

$$\begin{vmatrix} 1 &2&1\\ 0&h &1+h\\ 0& h&2h+1\\ 1&2&1 \end{vmatrix}$$

I observe that $$rank(A) = 3$$ for all $$h\neq 0$$ and if $$h=0, rank(A) = 2$$. So (I guess) the answer to the first question is that $$dim(S) =3$$ for all $$h\neq 0$$.

How to proceed to find $$v_4$$ such that $$span\{v_1, v_2, v_3, v_4\}=\mathbb{R}^4$$? Could anyone please help me?

• $v_{4}$? You only have $v_{1}, v_{2}, v_{3}$! Dec 14, 2020 at 17:27

Let $$W$$ be linear space of finite dimension $$n$$ with $$\{w_1, \dots , w_n\}$$ a basis of $$W$$.

It is a theorem that for a linear subspace $$V$$ spanned by $$\{v_1, \dots, v_m\}$$ with $$m \lt n$$, you can pickup $$w \in \{w_1, \dots , w_n\}$$ such that $$\{v_1, \dots, v_m, w\}$$ is linearly independent.

So in your case, take the canonical basis $$\{e_1, \dots e_4\}$$. If $$h \neq 0$$, $$\{v_1, v_2, v_3\}$$ are linearly independent. You can pick up one of the vectors $$e_i$$ of the canonical basis in a way to get a basis $$\{v_1,v_2,v_3,e_i\}$$ of $$\mathbb R^4$$. Just test them one after the other... one will work!

In fact,

$$\begin{vmatrix} 1 & 1 &2&1\\ 0 & 0&h &1+h\\ 0 & 0& h&2h+1\\ 0& 1&2&1 \end{vmatrix} = \begin{vmatrix} 0&h &1+h\\ 0& h&2h+1\\ 1&2&1 \end{vmatrix} = \begin{vmatrix} h &1+h\\ h&2h+1 \end{vmatrix} = h^2 \neq 0$$

So you already win with $$e_1$$ and $$\{v_1,v_2,v_3,e_1\}$$ is a basis of $$\mathbb R^4$$.

• Thank you for the answer! My attempt to find the right $h$ to determine $dim(S)$ is right, isn't it? Dec 14, 2020 at 18:04
• Yes it is correct. Dec 14, 2020 at 18:23
• Sorry, could you give me a reference for the theorem you mentioned above? Thank you in advance! Dec 30, 2020 at 8:43
• @C.Bishop Steinitz exchange lemma Dec 30, 2020 at 12:35