Suppose the non zero $n \times 1$ column vector $x$ solves the system of equations $Ax=b$ where $A$ is $m \times n$ matrix whose columns are the vectors $a_1,a_2,\ldots,a_n$ and $b$ is a $m \times 1$ column vector. then the set of the vectors $\{a_1, a_2, \ldots, a_n, b\}$ is Linearly dependent . Please explain how? Shouldn't there be conditon the set of vectors in "$A$" matrix ?


Note that $$Ax = b \iff \begin{bmatrix} a_1 & ... & a_n\end{bmatrix}\begin{bmatrix}x_1\\ \vdots \\ x_n\end{bmatrix} - b = 0 \iff \sum_{i=1}^n x_i a_i - b = \sum_{i=1}^n x_i a_i + (-1)\times b =0.$$

Since $-1\ne 0$, this proves that $(a_1,...,a_n,b)$ are linearly dependent.

  • $\begingroup$ Please explain the last step, Since -1 not equal to zero, this gives desired relation $\endgroup$ – Anonymous May 10 '18 at 17:36
  • $\begingroup$ I edited my answer to be clearer. $\endgroup$ – paf May 10 '18 at 17:37
  • $\begingroup$ No, what I want to ask is how (-1) came ? $\endgroup$ – Anonymous May 10 '18 at 17:43
  • $\begingroup$ It's the coefficient of b, see my edit. $\endgroup$ – paf May 10 '18 at 17:44
  • $\begingroup$ Now it's clear ,Upvoted!! $\endgroup$ – Anonymous May 10 '18 at 17:48

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.