MHB Using the given basis of a vector to prove other basis

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Hi Guys having a bit of trouble understanding vector basis.

If $$\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$$ is a basis for vector space $V$ over the field $F$

and $${f}_{1}=-{e}_{1}, {f}_{2}={e}_{1}-{e}_{2}, {f}_{3}={e}_{1}-{e}_{3}$$

how can I go about proving that $$\left\{{f}_{1},{f}_{2},{f}_{3}\right\}$$ is also a basis for $V$?
 
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Kronos said:
Hi Guys having a bit of trouble understanding vector basis.

If $$\left\{{e}_{1},{e}_{2},{e}_{3}\right\}$$ is a basis for vector space $V$ over the field $F$

and $${f}_{1}=-{e}_{1}, {f}_{2}={e}_{1}-{e}_{2}, {f}_{3}={e}_{1}-{e}_{3}$$

how can I go about proving that $$\left\{{f}_{1},{f}_{2},{f}_{3}\right\}$$ is also a basis for $V$?

Hi Kronos,

Solve the system of equations for $e_1$, $e_2$, and $e_3$. Your resulting equations will show that $e_1$, $e_2$, and $e_3$ all belong to the linear span of $f_1$, $f_2$, and $f_3$. Thus Span($e_1$, $e_2$, $e_3$) $\subset$ Span($f_1$, $f_2$, $f_3$). Since $e_1$, $e_2$, and $e_3$ form a basis for $V$, they span $V$ and thus $V = \text{Span}(f_1, f_2, f_3)$. This means that $f_1$, $f_2$, and $f_3$ span $V$. The $f$-vectors also linearly independent since $V$ is three-dimensional (as $e_1$, $e_2$, and $e_3$ form a basis for $V$).
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

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