[Math] non trivial solution of Homogenous equation

linear algebra

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:

  1. The determinant of the Matrix $\mathbf{A}=\left[\mathbf{a_1}\space\mathbf{a_2}\space\dots\space\mathbf{a_n} \right]$ will be non-zero because the Row reduced Echelon form has all pivots as non-zeros.
  2. Independence doesn't mean orthogonality. The vectors $\left[1 \space1\right]$ and $\left[1 \space 2\right]$ are independent but they are not orthogonal. On the other hand, $\left[1 \space1\right]$ and $\left[1 \space -1\right]$ are independent and orthogonal. It can be proven that Orthogonality implies independence.
  3. The equation $\mathbf{A}\mathbf{x}=\mathbf{0}$ will have no non-trivial solution because of point 1; the value of $x_1, x_2,\dots,x_n$ will automatically equate to zero since there are no free-variables. All of them become Pivot variables.
  4. For the 3-dimensional case as posed in your question, it means that the vectors $\mathbf{a_1},\mathbf{a_2},\mathbf{a_3}$ cannot be coplanar for the equation to hold a non-zero solution. If they are coplanar, it means that they are "linearly" dependent.
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