Basis and solution of System of equation

$$V$$ 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-

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.

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

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.

• 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 ? – Andrew May 7 at 6:01
• 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. – amitava May 7 at 6:41