# Find $a\in\mathbb{R}$ such that the following three vectors from a basis of vector space $\mathbb{R}^3$

$$\text{Let } v_1=\begin{pmatrix}a \\ 1 \\ 1\end{pmatrix}, \ v_2 = \begin{pmatrix}1 \\ a \\ 1\end{pmatrix},\ v_3 = \begin{pmatrix}1 \\ 1 \\ a\end{pmatrix}.\ \text{Determine } a\in\mathbb{R} \text{ such that the previous vectors form a basis for }\mathbb{R}^3$$.

Though I have an idea for how to solve this, I am not sure if my logic is correct. Here's what I have thought of. For $$v_1, v_2,v_3$$ to be a basis, they need to satisfy two conditions: they are linearly independent and form a sistem of generators. And so I check both conditions: $$v_1,v_2,v_3 \text{ linearly independent} \Rightarrow (\alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 = 0,\ \alpha_1,\alpha_2,\alpha_3\in\mathbb{R} \Rightarrow \alpha_1 = \alpha_2 = \alpha_3 = 0)$$

The first relation creates the following system of equations: $$\begin{cases}a\alpha_1a+\alpha_2+\alpha_3 = 0 \\ \alpha_1 + a\alpha_2+\alpha_3=0 \\ \alpha_1 + \alpha_2 + a\alpha_3 = 0\end{cases}, \text{ where } \alpha_1,\alpha_2,\alpha_3 \text{ are the unkowns and } a \text{ is a parameter.}$$ This system, we know, has the solution $$\alpha_1=\alpha_2=\alpha_3=0$$. For the vectors to be linearly independent, this needs to be the only solution of the system, which means the determinant of the system needs to be not null: $$\begin{vmatrix}a & 1&1\\1 & a & 1 \\ 1&1&a\end{vmatrix} \neq 0 \iff (a+2)(a-1)^2\neq 0 \iff a \in\mathbb{R}\setminus\{-2, 1\}$$

We now need to check if the vectors form a system of generators. We need to check that $$\forall x,y,z\in\mathbb{R}, \exists \alpha_1,\alpha_2,\alpha_3\in\mathbb{R} \text{ such that } \begin{pmatrix}x\\y\\z\end{pmatrix} = \alpha_1 v_1 + \alpha_2 v_2 +\alpha_3 v_3$$

The following system is obtained: $$\begin{cases}a\alpha_1 + \alpha_2 + \alpha_3 = x \\ \alpha_1 + a\alpha_2 + \alpha_3 = y \\ \alpha_1 + \alpha_2 + a\alpha_3 = z\end{cases}$$

This system needs a solution for every $$x,y,z$$, so I attempt to solve using echelon form: $$\begin{pmatrix}a & 1 & 1 & | & x \\ 1 & a & 1 & | & y \\ 1 & 1 & a & | & z \end{pmatrix} \sim \begin{pmatrix}1 & 1 & a & | & z \\ 0 & a-1 & 1-a & | & y-z \\ 0 & 0 & 2-a-a^2 & | & x+y-2z\end{pmatrix} \Rightarrow (a+2)(a-1)\alpha_3 = x+y-2z$$

Here is where I get stuck. Let's pretend I started the problem by checking for the system of generators condition first (which means I wouldn't know that $$a$$ cannot be $$1$$) Is $$a\neq1$$ and $$a\neq2$$ a condition I should add now? Or does it not matter? By which I mean that, if, say, $$a=2$$, then $$\alpha_3$$ can be any number in $$\mathbb{R}$$, and we will find $$\alpha_1$$ and $$\alpha_2$$ depending on whatever value $$\alpha_3$$ is. I think that this step does not exclude $$a$$ from being neither $$1$$ nor $$2$$, but I am not sure. At the same time, I do not know if this part is supposed to bring along any new conditions or if it simply checks that the vectors form a system of generators. Any help is much appreciated!

• "For $v_1,v_2,v_3$ to be a basis, they need to satisfy two conditions: they are linearly independent and form a sistem of generators" Actually, because there are precisely three vectors $v_1,v_2,v_3$ that you're looking at in $\mathbb R^3$, you need to check only one of those conditions, and the other will follow automatically. You already checked the determinant condition : the answer is then precisely $a \in \mathbb R \setminus \{-2,1\}$. Commented Oct 29, 2023 at 11:45
• Your determinant is correctly computed, and your conclusion is correct as well. Check your echelon form, it is not correct. That leads to some confusion, for $a=2$ shouldn't play a role. Commented Oct 29, 2023 at 12:00
• @J__n well now that makes sense: since $a=1$ doesn't provide a basis, some conditions should be enforced on $(x,y,z)$ to be a combination of your three vectors (which are not a basis for that $a$!). Commented Oct 29, 2023 at 12:11
• @WishYouTheBest I was again mistaken. Apparently, I do not know how to subtract properly! I now have the correct echelon form, from which it results that $(2-a-a^2)\alpha_3 = (a+2)(a-1)\alpha_3=x+y-2z$. However, what I still do not fully understand is, if I were to first check for the vectors to form a system of generators and not for their linear independence (which definitely imposes that $a\neq -2, 1) as I did, why does it matter that$a\neq 2\$ to generate any vector from those 3?
– J__n
Commented Oct 29, 2023 at 12:24
• @J__n I did not see the rest of your work, unfortunately. I will take a look at this and then get back to you. While I'm at that, +1 and I hope your question is answered well. Commented Oct 29, 2023 at 12:45

This is focusing only on the "row echelon" part of your attempt. As remarked earlier, given three vectors $$v_1,v_2,v_3$$ on a $$\mathbf{3}$$-dimensional space $$\mathbb R^3$$, all the following are equivalent :

• $$v_1,v_2,v_3$$ form a basis for $$\mathbb R^3$$.

• $$v_1,v_2,v_3$$ are linearly independent over $$\mathbb R^3$$.

• $$v_1,v_2,v_3$$ form a system of generators over $$\mathbb R^3$$.

• For any basis, if one expresses $$v_1,v_2$$ and $$v_3$$ in that basis and forms a matrix whose columns are the coordinates of $$v_i$$ in this basis, then the determinant of that matrix is non-zero.

Using the equivalence of the first and fourth conditions, the vectors $$v_1,v_2,v_3$$ form a basis of $$\mathbb R^3$$ precisely when $$\begin{vmatrix} a&1&1\\ 1&a&1\\ 1&1&a \end{vmatrix} \neq 0 \iff a \neq -2,1.$$

Thus, the answer is in fact $$a \in \mathbb R \setminus \{-2,1\}$$.

Now, if we use the equivalence of the first and third conditions, then we need to check if $$v_1,v_2,v_3$$ generate all of $$\mathbb R^3$$. Now, you drew up the row-echelon form of $$\mathbb R^3$$ by doing : $$\begin{pmatrix}a & 1 & 1 & | & x \\ 1 & a & 1 & | & y \\ 1 & 1 & a & | & z \end{pmatrix} \sim \begin{pmatrix}1 & 1 & a & | & z \\ 0 & a-1 & 1-a & | & y-z \\ 0 & 0 & 2-a-a^2 & | & x+y-2z\end{pmatrix}$$

And by reading off the third row of the echelon form, you know that $$(2-a-a^2)\alpha_3 = x+y-2z$$.

At this point, you were confused. To resolve that, you need to go back to a statement you wrote yourself :

We now need to check if the vectors form a system of generators. We need to check that $$\forall x,y,z\in\mathbb{R}, \exists \alpha_1,\alpha_2,\alpha_3\in\mathbb{R} \text{ such that } \begin{pmatrix}x\\y\\z\end{pmatrix} = \alpha_1 v_1 + \alpha_2 v_2 +\alpha_3 v_3$$

We need to find all values of $$a$$ for which the above is true. The point is, that when $$a=-2$$ or when $$a=1$$, the above is NOT true. To see why, if $$a=-2$$ or $$a=1$$ ,then what you get is $$x+y-2z = 0$$ above.

This tells you the following :

If $$a=-2$$ or $$a=1$$, and I assume that there exist $$\alpha_1,\alpha_2,\alpha_3 \in \mathbb R$$ such that $$\alpha_1v_1+\alpha_2v_2+\alpha_3v_3 = (x,y,z)$$, then it must be the case that $$x+y-2z = 0.$$

Think about it. The very starting of the row echelon argument is the existence of these $$\alpha_i$$s such that $$\alpha_1v_1+\alpha_2v_2+\alpha_3v_3 = (x,y,z)$$. You started with that assumption, and then got to $$x+y-2z = 0$$. But then, obviously, if $$x,y,z$$ satisfy this condition, then they can't be arbitrary, because the value of any of $$x,y,z$$ is controlled by the other two.

This means that the statement that $$v_1,v_2,v_3$$ is a system of generators does not hold, and therefore that $$a=-2,1$$ can be ruled out. This resolves your confusion on how to proceed once you hit that particular equality when $$a=-2$$ or $$a=1$$.

However, what happens if $$a\neq -2,1$$? In that case, you can confidently divide by $$2-a-a^2$$ and get $$\alpha_3 = \frac{x}{2-a-a^2}.$$

Some simple algebra will then give you the values of $$\alpha_1$$ and $$\alpha_2$$ as well, which you can check are real numbers. Once you do that, because you did everything without assuming anything about $$x,y,z$$, you know that for any $$x,y,z$$ you can express $$(x,y,z)$$ as a linear combination of $$v_1,v_2,v_3$$. That shows that the very first block-quoted statement in this answer is true, and hence that $$v_1,v_2,v_3$$ is a system of generators in this case. That is how the case $$a \neq -2,1$$ is handled.

If it's a space in $$\mathbb{R}^3$$ then, like you said, $$\begin{vmatrix}a&1&1\\1&a&1\\1&1&a\end{vmatrix}\neq0.$$
Let's calculate:
$$a\ast\begin{vmatrix}a&1 \\ 1&a \end{vmatrix}-1\ast\begin{vmatrix}1&1\\1&a\end{vmatrix}+1\ast\begin{vmatrix}1&1\\a&1\end{vmatrix}=a(a^2-1)-1(a-1)+1(1-a)=a^3-a-a+1+1-a=a^3-3a+2\neq0.\Longrightarrow \fbox{a\neq -2,1}$$