Check if polynomials are linearly independent.

Question

I would like to check if polynomials $1, 1+t^2, 1+t+t^2$ are linearly independent.
My idea is:
$1 \to [1,0,0]$
$1+t^2\to [1,1,0]$
$1+t^2+t^3 \to [1,1,1]$
And now $\left( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right)$
I would like to find rank of this array. Rank of this array is $3$ so columns are linearly independent.

Is it correct reasoning ?

You could also just compute the determinant of the - afterall hopefully correct - matrix — user190080, Sep 16 '16 at 20:47

score 11 · Answer 1 · answered Sep 16 '16 at 17:36

Just for fun you could use the definition of linear independence. That is, say that $$ a(1) + b(1 + t^2) + c(1 + t + t^2) = 0. $$ Then $$ (a + b + c)1 + (c)t + (b + c)t^2 = 0. $$ Now you probably know that $\{1,t,t^2\}$ is linearly independent. So $$ \begin{align} c &= 0 \\ b + c &= 0 \Rightarrow b = 0 \\ a + b + c &= 0 \Rightarrow a = 0. \end{align} $$

score 8 · Accepted Answer · answered Sep 16 '16 at 17:28

8

The coordinate vectors with respect to the basis $\{1,t,t^2\}$ are $$ \begin{bmatrix}1\\0\\0\end{bmatrix} \quad \begin{bmatrix}1\\0\\1\end{bmatrix} \quad \begin{bmatrix}1\\1\\1\end{bmatrix} $$ so the matrix you have to compute the rank of is $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} $$ Swapping rows 2 and 3 brings it into unitriangular form, so the rank is $3$.

Alternatively you can observe that $$ t=(1+t+t^2)-(1+t^2)\\ t^2=(1+t^2)-1 $$

answered Sep 16 '16 at 17:28

egreg

238,574

So checking rank is ok ? – Happy man Sep 16 '16 at 17:29
@Happyman Surely so! Provided you get the right matrix. ;-) – egreg Sep 16 '16 at 17:30

Check if polynomials are linearly independent.

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