Let's say I have a matrix $M$, whose rows are linearly independent and has full rank, that can be described as:

$$M=\begin{bmatrix} m_1 \\ m_2 \\ ... \\ m_n \\ \end{bmatrix}$$

Now, let's say I create a matrix $N$, whose rows are linear combinations of the exactly two rows in $M$, such that:

$$N=\begin{bmatrix} c_1m_1 + c_2m_2 \\ c_3m_1 + c_4m_3 \\ ... \\ c_{x-1}m_{n-1} + c_{x}m_n \\ \end{bmatrix}$$

Here, $c_1...c_x$ are non-zero constants that we can choose. Does there exist a set of values for these constants such that $N$'s rows are linearly independent?

  • 1
    $\begingroup$ Your numbering isn't clear. If you want $m_1$ and $m_3$ in row 2, then you don't want $m_{n-1}$ and $m_n$ in the bottom row. But anyway you could have every $c_{odd}$ be zero, the others be one, right? $\endgroup$ Aug 23 '17 at 6:39
  • $\begingroup$ Your notation is ambiguous. Do you mean $c_3 m_2+c_4 m_3$ or $c_3 m_1 + c_4 m_3$? $\endgroup$ Aug 23 '17 at 7:23
  • $\begingroup$ Sorry the notation is unclear. $N$ consists of every pair-wise linear combination of $M$. I also edited the question and specified that the constants must be non-zero. $\endgroup$
    – ArKi
    Aug 23 '17 at 15:05
  • $\begingroup$ If you use every pair of rows of $M$, you get order of $n^2$ elements of an $n$-dimensional vector space. They can't possibly be linearly independent. $\endgroup$ Aug 23 '17 at 23:22

Let $n=4$. Then $M$ has four rows, and they generate a vector space of dimension 4. $N$ has six rows, and they all live in that 4-dimensional vector space, so they can't be linearly independent.

This argument works for all $n\ge4$.


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.