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I know what linear independence is and that if the solution set is non-zero, then the system is not linearly independent. But I am a bit confused with this particular question,

If I have 3 vectors, $X_1, X_2$ and $X_3$ that are linearly independent, and I have 3 more, $Y_1= X_1+X_2, Y_2= 2X_2-X_3$, and $Y_3= X_1+X_2-2X_3$, how do I show that the $Y$ vectors are also linearly independent? I know that the determinant has to be non-zero in order for the system to be linearly independent, but I don't think you have to find the determinant here? I am a bit confused. If anyone would help, it would be really appreciated. Thank you.

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  • $\begingroup$ Can you write the Show that any of the vectors in question can be written as a linear combination of the others? $\endgroup$ – Rumplestillskin Dec 5 '17 at 7:32
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Since $$\begin{bmatrix}Y_1 & Y_2 & Y_3\end{bmatrix} = \begin{bmatrix}X_1 & X_2 & X_3\end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 1 & 2 & 1 \\ 0 & -1 & -2 \end{bmatrix} $$ you can check if the determinant of the right-most matrix is zero or not.

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There will likely be solutions here based on determinants, I will post one with a more pedestrian approach.

$Y_1, Y_2, Y_3$ will be linearly independent if and only if $\alpha_1Y_1+\alpha_2Y_2+\alpha_3Y_3=0$ implies $\alpha_1=\alpha_2=\alpha_3=0$. So, let's solve it for $\alpha_1, \alpha_2, \alpha_3$.

Substitute $Y_1, Y_2, Y_3$ and you get:

$$\alpha_1(X_1+X_2)+\alpha_2(2X_2-X_3)+\alpha_3(X_1+X_2-2X_3)=0$$

i.e. $$(\alpha_1+\alpha_3)X_1+(\alpha_1+2\alpha_2+\alpha_3)X_2+(-\alpha_2-2\alpha_3)X_3=0$$

Because $X_1, X_2, X_3$ are linearly independent, all those coefficients are zeros, i.e.:

$$\begin{align} \alpha_1 & & +\alpha_3 & =0 \\ \alpha_1 & +2\alpha_2 & +\alpha_3 & = 0 \\ & -\alpha_2 & -2\alpha_3 & = 0 \end{align}$$

which you can solve as a system of linear equations to obtain $\alpha_1=\alpha_2=\alpha_3=0$.

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Consider $c_1Y_1 + c_2Y_2 +c_3Y_3 = 0$ $$c_1(X_1+X_2) + c_2(2X_2 - X_3) + c_3(X_1+X_2 - 2X_3) = 0$$ $$(c_1 + c_3)X_1 + (c_1 +2c_2 +c_3)X_2 +(-c_2 -2c_3)X_3 = 0 $$

By linear independency of $\{X_1, X_2, X_3\}$, we get

$$c_1 +c_3 = 0, c_1 +2c_2 +c_3 = 0, -c_2 - 2c_3 =0$$ Solving last two equations gives us $c_1 - 3c_3 = 0$. Therefore $$c_1 + c_3 = 0, c_1 -3c_3 = 0$$ Hence $c_1= c_2= c_3 = 0$ that is $\{Y_1, Y_2, Y_3\}$ are linearly independent.

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Hint: Suppose that $c_1(x_1+x_2)+c_2(2x_2-x_3)+c_3(x_1+x_2-2x_3)=0$.
You must show that $c_1=c_2=c_3=0$.
Above equation can be formed following: $$c_1(x_1+x_2)+c_2(2x_2-x_3)+c_3(x_1+x_2-2x_3)$$ $$=(c_1+c_3)x_1+(c_1+2c_2+c_3)x_2+(-c_2-2c_3)x_3=0$$ What this is the meaning of this?
$c_1+c_3=0$
$c_1+2c_2+c_3=0$
$-c_2-2c_3=0$
Solve this homogeneous linear equations.So you check that determinant of $\begin{bmatrix} 1 & 0 & 1 \\ 1 & 2 & 1 \\ 0 & -1 & -2 \end{bmatrix}.$

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  • $\begingroup$ where did u get the numbers in the matrix from? $\endgroup$ – Lame Person Dec 5 '17 at 9:48
  • $\begingroup$ Look at coefficients of linear homogeneous equations. $\endgroup$ – 1ENİGMA1 Dec 5 '17 at 10:08

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