# Can a matrix whose rows are linear combinations of rows of a full-rank matrix be guaranteed to be full rank?

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?

• 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? Aug 23 '17 at 6:39
• Your notation is ambiguous. Do you mean $c_3 m_2+c_4 m_3$ or $c_3 m_1 + c_4 m_3$? Aug 23 '17 at 7:23
• 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.
– ArKi
Aug 23 '17 at 15:05
• 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. 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 $n\ge4$.