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

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!

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    $\begingroup$ "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 $a \in \mathbb R \setminus \{-2,1\}$. $\endgroup$ Commented Oct 29, 2023 at 11:45
  • $\begingroup$ 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. $\endgroup$ Commented Oct 29, 2023 at 12:00
  • $\begingroup$ @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 $a$!). $\endgroup$ Commented Oct 29, 2023 at 12:11
  • $\begingroup$ @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? $\endgroup$
    – J__n
    Commented Oct 29, 2023 at 12:24
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    $\begingroup$ @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. $\endgroup$ Commented Oct 29, 2023 at 12:45

2 Answers 2

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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.

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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$}$

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