 # Use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.

$1 + 2t^3, 2 + t – 3t^2, -t + 2t^2 – t^3$

This problem aims to familiarize us with vector equations, linear independence of a vector, and echelon form. The concepts required to solve this problem are related to basic matrices, which include linear independence, augmented vectors, and row-reduced forms.

To define linear independency or dependency, let’s say we have a set of vectors:

$\{ v_1 , v_2 ,…, v_k \}$

For these vectors to be linearly dependent, the following vector equation:

$x_1v_1 + x_2v_2 + ··· + x_kv_k = 0$

should only have the trivial solution $x_1 = x_2 = … = x_k = 0$ .

Hence, the vectors in the set $\{ v_1 , v_2 ,…, v_k \}$ are linearly dependent.

The first step is to write the polynomials in the standard vector form:

$1 + 2t^3 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 2 \end{pmatrix}$

$2 + t – 3t^2 = \begin{pmatrix} 2 \\ 1 \\ -3 \\ 0 \end{pmatrix}$

$-t + 2t^2 – t^3 = \begin{pmatrix} 0 \\ -1 \\ 2 \\ -1 \end{pmatrix}$

The next step is to form an augmented matrix $M$:

$M = \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & -3 & 2 & 0 \\ 2 & 0 & -1 & 0 \end{bmatrix}$

Performing a row operation on $R_4$, $\{ R_4 = R_4\space -\space 2R_1 \}$:

$M = \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & -3 & 2 & 0 \\ 0 & -4 & -1 & 0 \end{bmatrix}$

Next, $\{ R_3 = R_3 + 3R_2 \}$:

$M = \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & -4 & -1 & 0 \end{bmatrix}$

Next, $\{ R_4 = R_4 + 4R_2 \}$:

$M = \begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & -5 & 0 \end{bmatrix}$

Finally, $\{ -1R_3 \}$ and $\{R_4 = R_4 + 5R_3 \}$:

$M=\begin{bmatrix}1&2&0&0\\0&1&-1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

From the above matrix $M$, we can see that there are $3$ variables and $3$ equations. Hence, $1 + 2t^3, 2 + t – 3t^2, -t + 2t^2 – t^3$ are linearly independent.

## Numerical Result

The vector set $1 + 2t^3, 2 + t – 3t^2, -t + 2t^2 – t^3$ is linearly independent.

## Example

Is the set:

$\begin{Bmatrix} \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} & \begin{pmatrix}1 \\-1\\2\end{pmatrix}&\begin{pmatrix}3\\1\\4\end{pmatrix}\end{Bmatrix}$

linearly independent?

The augmented matrix of the above set is:

$M=\begin{bmatrix}1&1&3\\1&-1 &1\\-2& 2 &4\end{bmatrix}$

Row reducing the matrix gives us:

$M=\begin{bmatrix}1&0 &0\\0&1 &0\\0&0 &1\end{bmatrix}$

Hence, the set is linearly independent.

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