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?

Thank you in advance!

  • $\begingroup$ $v_{4}$? You only have $v_{1}, v_{2}, v_{3}$! $\endgroup$
    – Elmex80s
    Dec 14, 2020 at 17:27

1 Answer 1


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 $\mathbb R^4$.

  • $\begingroup$ Thank you for the answer! My attempt to find the right $h$ to determine $dim(S)$ is right, isn't it? $\endgroup$
    – C. Bishop
    Dec 14, 2020 at 18:04
  • $\begingroup$ Yes it is correct. $\endgroup$ Dec 14, 2020 at 18:23
  • $\begingroup$ Sorry, could you give me a reference for the theorem you mentioned above? Thank you in advance! $\endgroup$
    – C. Bishop
    Dec 30, 2020 at 8:43
  • 1
    $\begingroup$ @C.Bishop Steinitz exchange lemma $\endgroup$ Dec 30, 2020 at 12:35

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