Say that the vectors $\vec{v_1}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} , \vec{v_2}=\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}$. The two vectors are linearly independent.

I want to try to find all vectors $\vec{y}$ that will make the $span \left \{ \vec{v_1}, \vec{v_2}, \vec{y} \right \} = \mathbb{R^3}$. There will be many such vectors $\vec{y}$. At first, I thought I could do a projection onto the column space of $\vec{v_1}$ and $\vec{v_2}$ and get a perpendicular vector. But even before I could do such a projection, I need a vector that is not in the column space of $\vec{v_1}$ and $\vec{v_2}$ to project from. And if I could find one such vector, then I wouldn't even need to do a projection because that will just fit in well as a vector for $\vec{y}$ to span the whole of $\mathbb{R^3}$.

Subsequently, base on the above thought, I begin to think that $\vec{y}$ not necessarily has to be perpendicular to $\vec{v_1}$ and $\vec{v_2}$. As long as $\vec{y}$ is not in the column space of $\vec{v_1}$ and $\vec{v_2}$, it will work. So as long as $\vec{y} \neq c_1\vec{v_1} + c_2\vec{v_2}$, I got all all the vectors that can make the span equals to the whole of $\mathbb{R^3}$. But expressing it this way isn't very right. How can I express this in the more usual explicit form from this idea?

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    $\begingroup$ I think all you can do is what you've done: the span is ${\bf R}^3$ provided $y$ isn't a linear combination of $v_1$ and $v_2$. $\endgroup$ – Gerry Myerson Nov 20 '11 at 11:30
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    $\begingroup$ Which is iff det $\{v_1,v_2,y\} \neq 0$. $\endgroup$ – user38268 Nov 20 '11 at 11:56

Since either $\vec{v_i}\,,\vec{y}\in\mathbb{R}^3$ and $\vec{v_1}\neq k\vec{v_2}, k\neq0$ ,you need to find a vector orthogonal to both $\vec{v_i}$, that is $\vec{x}=\vec{v_1}\times\vec{v_2}$ , the outer product of the vectors, and $\vec{y}$ thus could be expressed like this: $$ \vec{y}=k_1\vec{v_1}+k_2\vec{v_2}+k_3\vec{x} $$ Where $k_3\neq 0$.

  • $\begingroup$ Even so that I want to find an orthogonal vector, I need to do a projection. And to do so, I need a vector not the column space of $\vec{v_1}$ and $\vec{v_2}$ to do a projection. But if I had this, I wouldn't even have to do a projection. $\endgroup$ – xenon Nov 20 '11 at 12:35
  • $\begingroup$ I don't see any reason you should do a projection, notice that outer the product of $\vec{v_i}$ is orthogonal to span($\vec{v_i}$) as you can always prove it. And you can calculate the outer product by calculating the determinant of the following matrix: $$\begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 3 & 4 \end{pmatrix}$$ $\endgroup$ – 7O'clock Nov 20 '11 at 12:54

Just find one vector $\vec{y}_0$ that is not of the form $c_1 \vec{v}_1 + c_2 \vec{v}_2$ and then your set of all $\vec y$ is the linear span of $\vec y_0$, $\vec v_1$ and $\vec v_2$ with a non-vanishing coefficient of $\vec y_0$. You could write this as $\vec y \in \{ a \vec y_0 + c_1 \vec{v}_1 + c_2 \vec{v}_2 \ |\ a \neq 0\}$.

That is, your set of $\vec y$ can be chosen from the entirety of $\mathbb{R}^3$ excluding the $\vec v_1$-$\vec v_2$ plane.

  • $\begingroup$ I like your explanation and solution better. Basically, we need to find the condition for linearly independent rows (or columns) of a matrix formed by these vectors. For a matrix A (with $\vec{v_1}$, $\vec{v_2}$ and $\vec{y}$ being the rows) with rank 3, we can find the condition that requires the vector to stay outside $y_1 -2y_2+y_3= 0$. which is exactly the plane spanned by $\vec{v_1}$ and $\vec{v_2}$. $\endgroup$ – C. Y. Cheng Jan 21 '14 at 4:38

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