i would like to recall why is that if
$A*x=0$
where $A$ is some square matrix has non trivial solution if matrix is singular?i was trying to recall it,we can represent $A*x=0$ as for example in case of three by three matrix we have
$a(1,1)*x_1+a(1,2)*x_2+a(1,3)*x_3=0$
$a(2,1)*x_1+a(2,2)*x_2+a(2,3)*x_3=0$
$a(3,1)*x_1+a(3,2)*x_2+a(3,3)*x_3=0$
we can represent each equation as dot product of each vector $a$ and $x$,from which it is clear they are orthogonal to each other,let me know if i am wrong,does determinant is zero means that all these vectors that are orthogonal to $x$ vector,must be located on plane?as i know determination of vectors are co planar is that determinant of matrix creates by these vectors as a column elements must be zero?thanks in advance
Best Answer
For an $n$-dimensional space, Independence of vectors implies a lot. Let $\mathbf{a_1},\mathbf{a_2},\mathbf{a_3},\dots,\mathbf{a_n}\in\mathbb{R}^n$. The vectors are independent if and only if $\alpha_1\mathbf{a_1}+\alpha_2\mathbf{a_2}+\dots+\alpha_n\mathbf{a_n}=0$ has no non-zero solution to it and the only solution is $\alpha_i=0,\forall i=1,2,3,\dots,n$. Now the condition can be rephrased in several different ways: