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I have a question that asks for values of alpha so that the polynomials $\alpha t^2+t+1$, $ t^2+\alpha t+1$ and $ t^2+t+\alpha$ are linearly independent.

For that to be true, the answers for $c_1(\alpha t^2+t+1) + c_2(t^2+\alpha t+1) +c_3(t^2+t+\alpha) $ must be trivial, with $c_1=c_2=c_3 = 0$

$t^2(\alpha c_1+c_2+c_3)+t(c_1+\alpha c_2 +c_3)+1(c_1+c_2+\alpha c_3)=0 $

So the following system has non-trivial solution if the determinant is equal to $0$, which would be a linearly dependent system: $$ \left\{ \begin{array}{c} \alpha c_1+c_2+c_3=0\\ c_1+\alpha c_2 +c_3=0\\ c_1+c_2+\alpha c_3=0 \end{array} \right. $$

$$ \det \begin{vmatrix} \alpha & 1 & 1\\ 1 & \alpha & 1\\ 1 & 1 & \alpha \end{vmatrix} =0 $$

$Row_1 - \alpha Row_3$ and $ Row_2 - Row_3$ :

$$ =\det \begin{vmatrix} 0 & 1-\alpha & 1-\alpha^2\\ 0 & \alpha-1 & 1-\alpha\\ 1 & 1 & \alpha \end{vmatrix} =0 $$

In the $3\times 3$ matrix above I use the rule of Sarrus to get

$ ( 1-\alpha^2) - (1-\alpha^2)(\alpha-1)=0$

$= ( 1-\alpha^2)(2-\alpha)=0$

$\alpha =1$ or $\alpha=2$

This would mean that for all $\alpha \neq 1$ and $\alpha \neq 2 $, the polynomials are linearly independent. However the ansers are $\alpha \neq 1$ and $\alpha \neq -2 $. Where am I going wrong? Thanks everyone in advance.

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  • $\begingroup$ row(2)-row(3) is wrongly done. $\endgroup$ – Kavi Rama Murthy Sep 15 '18 at 0:27
  • $\begingroup$ thanks, I edited it here. On my paper calculations I had done this part correctly though $\endgroup$ – Lucas A Sep 15 '18 at 0:29
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    $\begingroup$ The first term in the computation of det. is $(1-\alpha)^{2}$, not $1-\alpha^{2}$. $\endgroup$ – Kavi Rama Murthy Sep 15 '18 at 0:31
  • $\begingroup$ Thanks! That's where my error came from! $\endgroup$ – Lucas A Sep 15 '18 at 0:47
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It might be easiest to just multiply it out without the row operations and find the roots.

$(a^3 -3a + 2) = 0\\ (a-1)(a^2+a-2)\\ (a-1)(a + 2)(a-1)$

But, there are some ways you might guess your way to an answer.

If $a = 1$ then all of the rows are identical and hence linearly independent.

If $a = -2$ all of the row sum to $0.$ And that means that the vector $(1,1,1)$ is in the kernel, and the matrix is singular.

If you know about eigenvalues....

What are the eignevalues of

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

The sum of the eigenvalues equal the trace of the matrix, and the product of the eigenvalues equals the determinant of the matrix.

We know that $-1, 2$ are eigenvalues, the trace is $0$ and the determinant is $2.$ The last eigenvalue must be $-1$

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  • $\begingroup$ Thanks! I understood the first two methods you used, but I will have to read further about eigenvalues $\endgroup$ – Lucas A Sep 15 '18 at 0:59
  • $\begingroup$ Eigenvalues will come along soon enough. $\endgroup$ – Doug M Sep 15 '18 at 1:01
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Set $f_1(t) = \alpha t^2 + t +1$, $f_2(t) = t^2+\alpha t + 1$, $f_3(t) = t^2 + t+1$.

If $\{f_1, f_2, f_3\}$ is linearly dependent, then $\exists \alpha_1, \alpha_2,\alpha_3$ scalars not all zero such that $\sum_{i=1}^3 \alpha_if_i = 0$. Taking the derivative we also get $\sum_{i=1}^3 \alpha_if_i' = 0$ and $\sum_{i=1}^3 \alpha_if_i'' = 0$ so the matrix

$$\begin{bmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t) \\ f_1''(t) & f_2''(t) & f_3''(t) \\\end{bmatrix}$$

has linearly dependent columns.

In particular the determinant (known as the Wronskian) is equal to $0$:

$$0= \begin{vmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t) \\ f_1''(t) & f_2''(t) & f_3''(t) \\\end{vmatrix} = \begin{vmatrix} \alpha t^2 + t +1 & t^2+\alpha t + 1 & t^2+t+\alpha \\ 2\alpha t + 1 & 2t+\alpha & 2t+1 \\ 2\alpha & 2 & 2 \\\end{vmatrix} = -2(\alpha-1)^2(\alpha+2)$$

Hence $\alpha = 1$ or $\alpha = -2$.

Therefore if $\alpha \notin \{1,-2\}$, the set $\{f_1, f_2, f_3\}$ is surely linearly independent.

If $\alpha = 1$ then $f_1 = f_2 = f_3$, and if $\alpha = -2$ then $\sum_{i=1}^3 f_i = 0$ so in those cases we clearly have linear dependence.

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