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Thanks for clarifications. Now i am posting the question in a different way.

Suppose a vector $V$ is orthogonal to vectors $X1$ and $X2$.

$X1$ and $X2$ are linearly independent.

Now if $V$ is also orthogonal to vectors $Y1$ and $Y2$ or in other words the dot product is zero, can we say the all vectors i.e., $X1\; X2 \;Y1 \;Y2$ are linearly dependent, since all vectors share the same orthoganal vectors.

Now let dot product of $V$ is nonzero with $Y3$, can we say $X1\; X2 \;Y3$ are linearly independent.

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    $\begingroup$ Certainly not, consider the case $X_1 = Y_1$ and $X_2 = Y_2$. $\endgroup$ – Travis Willse Jul 10 '15 at 15:08
  • $\begingroup$ For your edit, please see my updated answer. $\endgroup$ – hexaflexagonal Jul 10 '15 at 15:55
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No.

$$X_1=(1,0,0)$$ $$X_2=(0,1,0)$$ $$Y_1=(-1,0,0)$$ $$Y_2=(0,-1,0)$$ $$V=(0,0,1)$$

Plot [Plot via Wolfram Alpha]


Edit for the reposed question: Answer to the first part is still no. $$X_1=(1,0,0,0,0)$$ $$X_2=(0,1,0,0,0)$$ $$Y_1=(0,0,1,0,0)$$ $$Y_2=(0,0,0,1,0)$$ $$V=(0,0,0,0,1)$$ All vectors are linearly independent (and orthogonal).

As for $Y_3$, yes, we can say that it is linearly independent of $X_1$ and $X_2$. Put loosely, we know this because $Y_3$ is non-orthogonal to $V$, and any linear combination of $X_1$ and $X_2$ will be orthogonal to $V$ (since they themselves are orthogonal to $V$).

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  • $\begingroup$ No problem! Glad to help. $\endgroup$ – hexaflexagonal Jul 10 '15 at 18:41
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The vector, v= <0, 0, 1> is perpendicular to X1= <1, 0, 0> and X2= <0, 1, 0> as well as to Y1= <1, 1, 0> and Y2= <1, -1, 0> but those four vectors are not independent. I don't know what you mean by "uncorrelated" vectors.

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