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$$1 + 2t+ t^2, 3-9t^2,1 + 4t + 5t^2$$

(A) Linearly dependent or (B) Linearly independent

The answer is A from the answer key.

This is a test review.

I don't see that either polynomial is a scalar multiple of any of the other polynomials. How can I test for linear dep|indep with polynomials? If these were vectors I would just put them in a matrix and row reduce (or if a square matrix, check the determinant), but I'm not sure what to do here.

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    $\begingroup$ The mathematical fact is that, as you see, polynomials in $x$ do not form a finite dimensional vector space, but polynomials of degree 2 (or less) do form a 3 dimensional vector space with basis $1,x,x^2$. Linear independence of such polynomials can be judged inside that vector space. So you can treat these as (three dimensional) vectors. $\endgroup$ – Colin McLarty Jul 6 '15 at 7:52
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    $\begingroup$ The key in your question is that you looked for one polynomial to be a scalar multiple of another, whereas you should have looked instead for a polynomial to be a linear combination (with scalar multipliers) of the others. $\endgroup$ – Laurent LA RIZZA Jul 6 '15 at 14:26
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Hint

You represent $1 + 2t + t^2$ as a column vector \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} and you represent $3 -9t^2$ as a column vector \begin{bmatrix} 3 \\ 0 \\ -9 \end{bmatrix} and you represent $1 + 4t + 5t^2$ as a column vector \begin{bmatrix} 1 \\ 4 \\ 5 \end{bmatrix}

Now you have a matrix

\begin{bmatrix} 1 & 3 & 1 \\ 2 & 0 &4 \\1 & -9 & 5 \end{bmatrix}

Now I believe You got it from here

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$$6\cdot (1+2t+t^2) - 3\cdot (1 + 4t + 5t^2) = 3 - 9t^2.$$ We can write one of the polynomials as a linear combination of the two other polynomials and therefore they are linearly dependent.

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  • $\begingroup$ I denote $c_1$ as the coefficient of the first polynomial and $c_2$ as the coefficient of the second polynomial (so in this case $c_1 = 6$ and $c_2 = -3$ but we don't know that yet). For the $t$'s to cancel out, we know that $c_1 = -2c_2$. Furthermore, since there is a 3 in the final polynomial, $c_1+c_2=3$. Substituting $c_1 = -2c_2$ into the second equation yields $c_2 = -3$ and subsequently $c_1 = 6$. $\endgroup$ – molarmass Jul 5 '15 at 23:04
  • $\begingroup$ Thanks; I didn't think you saw my Q so I deleted it. Wow, that's great and I'm not sure if I'll be able to do something like this in my head during an exam. $\endgroup$ – asdf Jul 5 '15 at 23:15
  • $\begingroup$ You can actually write it all out: \begin{align}c_1 + 2c_1t + c_1t^2 + c_2 + 4c_2t + 5c_2t^2 &= 3-9t^2\\ (c_1 + c_2) + (2c_1 + 4c_2)t + (c_1 + 5c_2)t^2 &= (3) + (0)t + (-9)t^2\end{align} $\endgroup$ – molarmass Jul 5 '15 at 23:16
  • $\begingroup$ I did as you suggested and got it. Thanks. :)) $\endgroup$ – asdf Jul 5 '15 at 23:21
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An alternative way to solve this problem is by using the Wronskian. Put \begin{align*} f(t) &= 1 + 2\,t+ t^2 & g(t) &= 3-9\,t^2 & h(t) &= 1 + 4\,t + 5\,t^2 \end{align*} and define $$ W(t)= \begin{bmatrix} f(t) & g(t) & h(t) \\ f^\prime(t) &g^\prime(t) & h^\prime(t) \\ f^{\prime\prime}(t) & g^{\prime\prime}(t) & h^{\prime\prime}(t) \end{bmatrix} = \begin{bmatrix} 1+2\,t+t^2 & 3-9\,t^2 & 1+4\,t+5\,t^2 \\ 2+2\,t & -18\,t & 4+10\,t \\ 2 & -18 & 10 \end{bmatrix} $$ If there exists a $t_0$ such that $\det W(t_0)\neq 0$, then $\{f,g,h\}$ is linearly independent. Since each of $f$, $g$, and $h$ is analytic, if $\det W(t)=0$ for all $t\in\Bbb R$, then $\{f,g,h\}$ is linearly dependent.

Now, note that adding $\DeclareMathOperator{Col}{Col}9\cdot\Col_1$ to $\Col_2$ and subtracting $5\cdot\Col_1$ from $\Col_3$ gives $$ \det W(t) = \begin{vmatrix} 1+2\,t+t^2 &12+18\,t& -4-6\,t \\ 2+2\,t&18&-6 \\ 2&0&0 \end{vmatrix} $$ Then adding $3\cdot\Col_3$ to $\Col_2$ gives $$ \det W(t) = \begin{vmatrix} 1+2\,t+t^2 & 0 & -4-6\,t \\ 2+2\,t & 0 & -6 \\ 2 & 0 & 0 \end{vmatrix} = 0 $$ This implies that $\{f,g,h\}$ is linearly dependent.

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This family is linearly dependent. In fact, we have the following $$2(1+2t+t^2)-\frac{1}{3}(3-9t^2)=1+4t+5t^2.$$ Khadija

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You still can just put them in a matrix! Just pretend $1,t,t^2$ are the standard coordinate vectors.

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Briefly,

You can think about a space of polynomials of maximum degree 3. This is a linear space on real numbers. (Prove it). Then what is it's basic? Obviously {$1$,$t$,$t^2$} gives you three linear independent vectors and they span all vector space. So, what we have is actually that you can see that this problem is not different from saying that $1=e_1$, $t=e_2$, $t^2=e_3$ where we just renamed those vectors.

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You can build the matrix with coefficients so that each polynomial is a column and each monomial is a row, then you show that this matrix $M$ does not have full rank. One way to do this is to show that the equation $Mv = 0$ has a solutions sub-space of positive dimensions.

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    $\begingroup$ Because the non-trivial solutions in null-space of M imply linear dependence, correct? $\endgroup$ – asdf Jul 5 '15 at 23:05
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You can think at the polyonomial $a+bx+cx^2$ as the vector $(a,b,c)\in\Bbb R^3$ (because of the obvious isomorphism $\Bbb R[X]_{\le2}\simeq\Bbb R^3$).

Thus your three vectors are $(1,2,1),(3,0,-9),(1,4,5)$. If you consider the $3\times3$ matrix having these thee vectors as colums, its determinant is $3(10-4)-9(2)=0$ thus they are linearly dependent.

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  • $\begingroup$ I like the co-factor expansion determinant check, Thanks! $\endgroup$ – asdf Jul 5 '15 at 22:51
  • $\begingroup$ You're welcome!:-) $\endgroup$ – Joe Jul 5 '15 at 22:53
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Finishing Alakbary's answer, I believe you calculate the determinant of his 3x3 matrix and it comes to zero if and only if they are linearly dependent.

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