I was reading a book which says that in case we are dealing with homogeneous linear equations,if A matrix has size of $m\times n$ and if $m<n$, then it will have infinite solutions other than trivial solution (where $x_1=x_2=x_3=\dots=0$), where $x_1,x_2,x_3$ are the variables in equation.

But i think the trivial solution will always possible even in case my matrix rank $<n$ ?Regardless of my rank trivial solution is always possible and in case of infinite solution one of the solution is trivial solution only? Am i correct?

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    $\begingroup$ Can you please adjust your question? It seems like something is missing. $\endgroup$ – Alberto Andrenucci Jun 22 '17 at 18:28
  • $\begingroup$ Yes.Doing it.It is showing half the question in preview itself. $\endgroup$ – rahul sharma Jun 22 '17 at 18:29

Trivial solution is always possible. Imagine that you take a standard matrix and multiply it by the $0$ vector. It will always give you the $0$ vector.

In fact, in a linear application, you have that $f(0)=0$.

What the book is trying to say is: if you have $n$ incognitas but just $m$ equations, with $m<n$, you will have $n-m$ degrees of freedom.

In fact, if you take $m$ equations, they give you conditions only on $m$ components of your vector. The other ones, $n-m$, are at your choice.

Let's do an example: you have just one equation, saying you that $x_1=4$. Considering that you are in $\mathbb{R}^3$ you have a lot of vectors satisfying this condition. In fact, you just have to put the first coordinate equal to $4$, for the other two you have free choice. In fact, having $2$ degrees of freedom, the $\text{Span}$ generated by vectors satisfying this condition has dimension $2$:

$$\text{Span}\left(\begin{pmatrix}4\\0\\1 \end{pmatrix}, \begin{pmatrix}4\\1\\0 \end{pmatrix} \right)$$

  • $\begingroup$ Got it. I took the wrong interpretation from the book. Thanks a lot. $\endgroup$ – rahul sharma Jun 22 '17 at 20:33
  • $\begingroup$ No problem, glad to help! $\endgroup$ – Alberto Andrenucci Jun 22 '17 at 20:33

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