Proof for the number of basic vectors are the same for any basis If $v_1, ... , v_m$ and $w_1, ... , w_n$ are both bases for the same vector space, then $m
=n$

Proof
Assume that there are more $w$'s than $v$'s, ie., $n&gt;m$
$$
w_1=a_{11}v_1+....+a_{m1}v_m
$$
Similarly, each $w$ is a linear combination of the $v$'s
$$
\begin{bmatrix}w_1&amp;w_2&amp;\dots&amp;w_n\end{bmatrix}=\begin{bmatrix}v_1&amp;v_2&amp;\dots&amp;v_m\end{bmatrix}\begin{bmatrix}a_{11}&amp;\dots&amp;a_{1n}\\
\vdots&amp;&amp;\vdots\\
a_{m1}&amp;\cdots&amp;a_{mn}\end{bmatrix}\iff W=VA
$$
Then it says &quot;$A$ is a short-wide matrix, since we assumed $n&gt;m$. So $Ax= 0$ has a nonzero solution&quot;
How come $n&gt;m$ alone says anything about the system ?
I understand that if $rank(A)&lt;n$(not Full rank), then $Ax=0$ must have nonzero solutions, but how can I know about the rank of the matrix $A$ from the given details ?
Reference: Page 170, Dimension of Vector Spaces, Introduction to Linear Algebra - Ed 5, Gilbert Strang
When row reducing $A$ to row echelon form, the number of pivots (which is the rank of $A$) can never exceed the number of columns or the number of rows of $A$. Hence, when $A$ is an $m\times n$ matrix with $m&lt;n$, you can have at most $m$ pivot variables in $Ax=0$, leaving you at least $n-m&gt;0$ free variables. Hence there is a non-zero solution to $Ax=0$ that can be found by choosing non-zero values for the free variables and solving for the pivot variables.
Hint: $Ax=0$ has a unique solution iff $A$ is square and has full rank. Can you proceed now?