0
$\begingroup$

$V$ be a n-dimensional vector space over the field $F$, with fixed Basis $\{\ \alpha_1, ...\alpha_n\}$ . A system of linear equitation - $$a_{11}x_1+a_{12}x_1+a_{11}x_2 \cdots +a_{1n}x_n=0$$ $$a_{21}x_1+a_{22}x_1+a_{11}x_2 \cdots +a_{2n}x_n=0$$ $$\cdots$$ $$a_{k1}x_1+a_{k2}x_1+a_{11}x_2 \cdots +a_{kn}x_n=0$$

is independent if and only if the collection of vectors $$v_1=\sum a_{1j}\alpha_j, v_2=\sum a_{2j}\alpha_j, \cdots v_k=\sum a_{kj}\alpha_j$$

in $V$ are independent.

So,if I am not wrong, elements of basis is solution to the linear system, since $v_i$ is the solution and it is the linear combination of coefficient $a_{ij}$ and basis $\alpha_j$.

Can any one explain how the set of fixed basis become the solution of linear system?

I am new so please teach me. Thanks.

I got this from a lecture.Please see below-

$\endgroup$
0
$\begingroup$

Your understanding is incorrect, but that's not too surprising. That definition of independence of systems of linear equations seems needlessly confusing, IMHO. There's no need for an abstract $n$-dimensional vector space $V$, nor is there a need for an abstract basis $(\alpha_1, \ldots, \alpha_n)$.

I think a better (and equivalent) definition would be to say, the system $$a_{11}x_1+a_{12}x_1+a_{13}x_2 \cdots +a_{1n}x_n=0$$ $$a_{21}x_1+a_{22}x_1+a_{23}x_2 \cdots +a_{2n}x_n=0$$ $$\vdots$$ $$a_{k1}x_1+a_{k2}x_1+a_{k3}x_2 \cdots +a_{kn}x_n=0$$ is independent if the row vectors of the coefficient matrix $$\begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{k1} & a_{k2} & a_{k3} & \cdots & a_{kn} \end{pmatrix}$$ are linearly independent in $F^n$.

Why are they equivalent? Fix the standard basis $(e_1, \ldots, e_n)$ in $F^n$. We can uniquely define a linear transformation $T : V \to F^n$ by making $T\alpha_i = e_i$ for all $i$. Because this linear transformation maps a basis to a basis, it is a bijective linear transformation (as a matter of fact, it is easily seen to be the coordinate vector map for the basis of $\alpha$s). As such, $T$ preserves the property of linear independence.

Tracking the logic, the system of equations is independent if and only if the vectors $$v_l = a_{l1}\alpha_1 + a_{l2}\alpha_2 + \ldots + a_{ln} \alpha_n, \quad l=1, \ldots, k$$ is linear independent. Since $T$ preserves linear independence, and is linear, this set is linearly independent if and only if $$T(v_l) = a_{l1}T(\alpha_1) + a_{l2}T(\alpha_2) + \ldots + a_{ln} T(\alpha_n), \quad l=1, \ldots, k$$ is linear independent. But, $T(\alpha_j) = e_j$, so \begin{align*} T(v_l) &= a_{l1}(1, 0, \ldots, 0) + a_{l2}(0, 1, \ldots, 0) + \ldots + a_{ln} (0, 0, \ldots, 1) \\ &= (a_{l1}, a_{l2}, \ldots, a_{ln}), \end{align*} which is the $l$th row of the coefficient matrix. Hence, the system is independent if and only if the row vectors of the coefficient matrix are linearly independent in $F^n$.

Note how much simpler this definition is! There's really no need for an abstract $n$-dimensional vector space $V$ over $F$, or an abstract basis. Any set of $n$ linearly independent vectors will do, from any space!

So, to address your actual question, the basis of $\alpha$s will not be a solution to the system, simply because they may be really abstract vectors over $F$. The actual elements of $V$ (including the basis of $\alpha$s) matter very little to this definition.

$\endgroup$
  • $\begingroup$ can you recommend any lecture or book or pdf that gives a good, easy explanation and proof related to the results? $\endgroup$ – Andrew May 7 at 6:18
0
$\begingroup$

The system of linear equations are independent if it has exactly one solution. Now treating $a_{i}$ as vectors we can say if these vectors are linearly independent then we have exactly one solution.

Now considering this for three dimension, we see

$a_1 = (a_{11},a_{12},a_{13}),a_2 = (a_{21},a_{22},a_{23}), a_3 = (a_{31},a_{32},a_{33})$

We have the basis here $\{\ \alpha_1,\alpha_2,\alpha_3\}$, we write each of this vectors as

$\alpha_1 = (\alpha_{11},\alpha_{12},\alpha_{13}),\alpha_2 = (\alpha_{21},\alpha_{22},\alpha_{23}), \alpha_3 = (\alpha_{31},\alpha_{32},\alpha_{33})$

(Above all vectors are broken down in their Cartesian components)

$v_1=\sum a_{1j}\alpha_j, v_2=\sum a_{2j}\alpha_j, v_3=\sum a_{3j}\alpha_j$, so

$v_1=((a_{11}\alpha_{11}+a_{12}\alpha_{21}+a_{13}\alpha_{31}), (a_{11}\alpha_{12}+a_{12}\alpha_{22}+a_{13}\alpha_{32}),(a_{11}\alpha_{13}+a_{12}\alpha_{23}+a_{13}\alpha_{33}))$

$v_2=((a_{21}\alpha_{11}+a_{22}\alpha_{21}+a_{23}\alpha_{31}), (a_{21}\alpha_{12}+a_{22}\alpha_{22}+a_{23}\alpha_{32}),(a_{21}\alpha_{13}+a_{22}\alpha_{23}+a_{23}\alpha_{33}))$

$v_3=((a_{31}\alpha_{11}+a_{32}\alpha_{21}+a_{33}\alpha_{31}), (a_{31}\alpha_{12}+a_{32}\alpha_{22}+a_{33}\alpha_{32}),(a_{31}\alpha_{13}+a_{32}\alpha_{23}+a_{33}\alpha_{33}))$

Now when $v_1, v_2,v_3$ are independent, $k_1\overrightarrow{v_1}+k_2\overrightarrow{ v_2}+k_3\overrightarrow{v_3} = 0$, $\textbf{only when}$, $k_1=0,k_2=0,k_3=0$, now using above equations, we get after simplification

$k_1\overrightarrow{ v_1}+k_2\overrightarrow{ v_2}+k_3\overrightarrow{v_3} = \\(k_1a_{11}+k_2a_{21}+k_3a_{31})\overrightarrow{\alpha_1}+(k_1a_{12}+k_2a_{22}+k_3a_{32})\overrightarrow{\alpha_2}+(k_1a_{13}+k_2a_{23}+k_3a_{33})\overrightarrow{\alpha_3}$

Now again repeating the above can be zero $\textbf{only when}$, $k_1=0,k_2=0,k_3=0$, otherwise not. But we also see that as $\alpha_1,\alpha_2,\alpha_3$ are basis vectors we also have (for the above to be true)

$k_1a_{11}+k_2a_{21}+k_3a_{31} = 0$

$k_1a_{12}+k_2a_{22}+k_3a_{32} = 0$

$k_1a_{13}+k_2a_{23}+k_3a_{33} = 0$

Now the above set of equations are true only when $k_1=0,k_2=0,k_3=0$, otherwise not.

We also see that above three equations can be summed up as $k_1\overrightarrow{ a_1}+k_2\overrightarrow{a_2}+k_3\overrightarrow{a_3}=0$

That indicates $a_1,a_2,a_3$ are independent vectors. So as mentioned at the begining when $a_1,a_2,a_3$ are independent vectors we have one unique solution(and you can actually see the solution), so the system of equations are independent.

$\endgroup$
  • $\begingroup$ You wrote, $a_1 = (a_{11},a_{12},a_{13}),a_2 = (a_{21},a_{22},a_{23}), a_3 = (a_{31},a_{32},a_{33})$, I assume, $a_{ij}$ is the coordinate of vector $a_j$, but then how can you write $\alpha_1 = (\alpha_{11},\alpha_{12},\alpha_{13}),\alpha_2 = (\alpha_{21},\alpha_{22},\alpha_{23}), \alpha_3 = (\alpha_{31},\alpha_{32},\alpha_{33})$ using basis $\{\ \alpha_1,\alpha_2,\alpha_3\}$,are they same as $a_1, a_2, a_3$ ? can you elaborate ? $\endgroup$ – Andrew May 7 at 6:01
  • $\begingroup$ Any vector (including any basis vector) can be expressed in terms of basis vectors of original Cartesian coordinate system. In this case $\alpha_1 = \alpha_{11}i+\alpha_{12}j+\alpha_{13}k$ as $a_1 = a_{11}i+a_{12}j+a_{13}k$, $\alpha_{11},\alpha_{12},\alpha_{13}$ have some values which we do not know and also we do not need to know. What is important I guess is to fix the same original Cartesian coordinate system for every vector in this system (including $\alpha_1$ and $a_1$), which is done here. i,j,k are the basis vectors in Cartesian coordinate system. $\endgroup$ – amitava May 7 at 6:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.