# Linear algebra proof, dependent or independent

Suppose that $p_1=4-3x+6x^2+2x^3$, $p_2=1+8x+3x^2+x^3$, and $p_3=3-2x-x^2$ are vectors in $P_3$. Determine if $p_1$,$p_2$,and $p_3$ are linearly independent or dependent. Justify your answer.

So far, what I did was say:

Suppose $k_1$, $k_2$ and $k_3$ are constants.

Then I said:

$$k_1 \begin{pmatrix} 4\\ -3\\ 2\\ 6 \end{pmatrix} + k_2 \begin{pmatrix} 1\\ 8\\ 3\\ 1 \end{pmatrix} + k_3 \begin{pmatrix} 3\\ -2\\ -1\\ 0 \end{pmatrix}$$

Then I made the augmented matrix $$\begin{pmatrix} 4 & 1 & 3 &0 \\ -3 & 8 & -2 & 0\\ 6 & 3 & -1 & 0\\ 2 & 1 & 0 & 0 \end{pmatrix}$$

and found that the RREF is

\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}

Therefore, $k_1 = k_2 = k_3 = 0$. Therefore, $p_1$, $p_2$ and $p_3$ are linearly independent.

So far this is what I have. If this is right then how to I go about justifying my answer?

Also, is there a better way of writing vectors on this website?

• What is the RREF? – Emanuele Paolini Mar 12 '13 at 19:23
• row reduced echelon form – nicholas Mar 12 '13 at 19:28
• Assuming your calculation is correct, it follows from the fact that $\{1,x,x^2,x^3\}$ span $P_3$. – Keaton Mar 12 '13 at 19:31
• @nicholas codecogs.com/latex/eqneditor.php – user4167 Mar 12 '13 at 19:35
• awesome thanks euler – nicholas Mar 12 '13 at 20:24

## 1 Answer

That absolutely justifies that $p_1,p_2,p_3$ are linearly independent.

• ok just wanted to check, thanks – nicholas Mar 12 '13 at 20:17
• what about the last row being all zeros? What does it tell us – Pirategull Mar 16 '20 at 21:08
• @Pirategull It tells us that the image of the map isn't the full space. This is a good thing, because if it were, then the nullspace of the map would have dimension $-1,$ whatever that would mean. – Cameron Buie Mar 16 '20 at 21:35