For which values of x do the vectors $(x,0,x + 1),(1,x,1),(0,x,−x)$ form a basis for R3?

Question

For which values of x do the vectors $(x,0,x + 1),(1,x,1),(0,x,−x)$ form a basis for R3?

Is 1 a correct answer to this. The vectors would each be linearly independent, however this seems too simple and I'm guessing theres an error somewhere.

If 1 works then it would seem any real number would work since the location of the 0 in the first and last vector will always mean the three vectors are linearly independent, right?

Answer 1

Guide:

• Put those vectors as rows of a matrix, compute the determinant.

• If they form a basis, the determinant would not take value zero, hence you just have to find the root of the charactheristic polynomial and avoid those root to make it a basis.

• Remark: Find all values rather than find one that works.

Edit:

$$\begin{bmatrix} 1 & x & 1 \\ 0 & x & - x \\ x & 0 & x+ 1\end{bmatrix}$$

Perform $R_3-xR_1$:

$$\begin{bmatrix} 1 & x & 1 \\ 0 & x & - x \\ 0 & -x^2 & 1\end{bmatrix}$$

Perform $R_3+xR_2$:

$$\begin{bmatrix} 1 & x & 1 \\ 0 & x & - x \\ 0 & 0 & 1-x^2\end{bmatrix}$$

To make it nonsingular, each column must have a pivot point, hence we require $x \neq 0$ and $1-x^2 \neq 0$.

Hence the conditions that you need are $x \neq 0$ and $1-x^2 \neq 0$.

Answer 2

How did you find this value?

You can also "glue" the vectors together to form a matrix (dimension 3x3), and try to find its determinant. You will obtain a polynom in $x$ and the zeros of this polynom are the values for which the three vectors are linearly dependant.

Answer 3

A basis of $\mathbb{R}^3$ must have three linear independent vectors, so the equation $$ \lambda_1 (x,0,x+1)^t + \lambda_2 (1,x,1)^t + \lambda_3 (0,x,−x)^t = (0,0,0)^t $$ is allowed to have only the solution $(\lambda_1, \lambda_2, \lambda_3) = (0,0,0)$. In matrix form $$ A \lambda = 0 \iff \\ \begin{pmatrix} x & 1 & 0 \\ 0 & x & x \\ x+1 & 1 & -x \end{pmatrix} \begin{pmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ We can now go for the determinant of the matrix, which must not vanish for the solution to be unique, or we solve the homogeneous linear system and check what influence $x$ has on the number of solutions.

Method A: Solving the linear system

Still open are the possibilities one solution or infinite many solutions, as we know the null vector is a solution the case no solution is not applying here.

In augmented matrix form we have: $$ A \lambda = 0 \iff [A|0] \iff \\ \left[ \begin{array}{ccc|c} x & 1 & 0 & 0 \\ 0 & x & x & 0 \\ x+1 & 1 & -x & 0 \end{array} \right] \to \left[ \begin{array}{ccc|c} x & 1 & 0 & 0 \\ 0 & x & x & 0 \\ 1 & 0 & -x & 0 \end{array} \right] $$ Case 1: For $x\ne 0$ we can divide by $x$ and have $$ \left[ \begin{array}{ccc|c} 1 & 1/x & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & -x & 0 \end{array} \right] \to \left[ \begin{array}{ccc|c} 1 & 1/x & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & -1/x & -x & 0 \end{array} \right] \to \\ \left[ \begin{array}{ccc|c} 1 & 0 & -1/x & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1/x-x & 0 \end{array} \right] $$ Case 1.1: For $1/x-x \ne 0$ we can divide by $1/x-x$ and get $$ \left[ \begin{array}{ccc|c} 1 & 0 & -1/x & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] \to \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] $$ so for $$ x\ne 0 \wedge 1/x-x \ne 0 \iff \\ x\ne 0 \wedge 1/x \ne x \iff \\ x\ne 0 \wedge 1 \ne x^2 \iff \\ x \not\in \{-1, 0, 1 \} $$ the vectors are linear independent. As three linear independent vectors form a basis of $\mathbb{R}^3$ we are done with this case.

Case 1.2: For $x \ne 0$ and $x \in \{ -1, 1 \}$, which means $x=\pm 1$ we have $$ \left[ \begin{array}{ccc|c} 1 & 0 & \mp 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$ which gives the infinite many solutions $\lambda = (0, 0, \lambda_3)$. So in this case the three vectors are linear dependent, we have no basis for $\mathbb{R}^3$.

Case 2: For $x=0$ we have $$ \left[ \begin{array}{ccc|c} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array} \right] \to \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] $$ which again gives the infinite many solutions $\lambda = (0, 0, \lambda_3)$. So in this case we have no basis for $\mathbb{R}^3$.

Answer:

For $x \in \mathbb{R} \setminus \{-1, 0, 1\}$ the three vectors form a basis of $\mathbb{R}^3$.

Method B: Checking the determinant

$$ \det A = \\ \left\vert \begin{array}{ccc} x & 1 & 0 \\ 0 & x & x \\ x+1 & 1 & -x \end{array} \right\vert = \left\vert \begin{array}{ccc} x & 1 & 0 \\ 0 & x & x \\ 1 & 0 & -x \end{array} \right\vert = x \left\vert \begin{array}{cc} x & x \\ 0 & -x \end{array} \right\vert - \left\vert \begin{array}{ccc} 0 & x \\ 1 & -x \end{array} \right\vert = -x^3 + x = -x(x^2 - 1) $$ which vanishes for $x \in \{ -1, 0 , 1 \}$. For these $x$ values we have no unique solution to $A\lambda = 0$, which means the vectors are linear dependent and form no basis of $\mathbb{R}^3$. Any other real value will lead to a basis.