# Relating 2 proofs of: If there are $m$ linearly independent vectors in $\mathbb{R}^n$, then $m\leq n$

I know of a proof using the exchange lemma, but I am trying to relate this approach to the approach using row reduction. The proof from my text (Linear Algebra Done Wrong) goes something like: since the vectors are linearly independent, the echelon form of the matrix with the $n$ vectors as columns has $n$ pivots. But there are only $m$ rows, so the number of pivots cannot exceed $m$. Hence $m\leq n$. However, I feel uneasy about the step, because it seems so much easier than the proof the exchange lemma. Where is the difficulty hidden in the proof using row reduction?

• The rows $r_1,..., r_n$ of a $n \times m$-matrix can be regarded as vectors in $\mathbb{R}^m$. They generate a subspace $V \subset \mathbb{R}^m$. In a row operation a row $r_k$ is replaced by a linear combination $\Sigma_{i=1}^n a_i r_i$ with $a_k \ne 0$ (a row exchange of $r_k$ and $r_l$ is the combinaton of three such operations: $r'_l = r_l + r_k$, $r'_k = -r_k + r'_l = r_l$, $r''_l = r'_l - r'_k = r_k$). Via row operations $\{ r_1,..., r_n \}$ is transformed into a certain basis $\{ b_1,...,b_k \}$ of $V$. – Paul Frost Jul 15 '18 at 12:47
• The same idea (replacement of a vector $v_k$ by a linear combination $\Sigma_{i=1}^n a_i v_i$ with $a_k \ne 0$ ) is used in the exchange lemma. – Paul Frost Jul 15 '18 at 12:47