If $y$ is linearly dependent, then $Bx=y$

Question

I saw a proof of $(AB=I) \Rightarrow (BA=I)$ here.

And I'm a little confused by following statement:

Since $B x_1,Bx_2 \ldots Bx_n$ is a basis, every vector y can be represented as a linear combination of those vectors. This means that for any vector y there exists some vector x such that Bx=y.

I don't understand the last sentence of the statement. I get, that when $y$ is linearly dependent, it can be expressed like the linear combination of basis vectors, e.g. $Bx_1 + 2Bx_2 = y$. The problem is, the statement above tells, that every $y$ can be expressed using only one basis vector. How is that possible? Or am I understanding it wrong?

Answer 1

Because, if there exist $n$ coefficients $c_i$ such that $\boldsymbol y=c_1\left(\boldsymbol B \boldsymbol x_1\right) + ... c_n\left(\boldsymbol B \boldsymbol x_n\right)$ then,

exploiting the linearity of the matrix product $\boldsymbol y=\boldsymbol B(c_1 \boldsymbol x_1+ ... + c_n \boldsymbol x_n )=\boldsymbol B\boldsymbol x$

where $(c_1 \boldsymbol x_1+ ... + c_n \boldsymbol x_n )=\boldsymbol x$ is the $\boldsymbol x$ the proof talks about.

Answer 2

Note that in $Bx=y$, $x$ doesn't have a subscript and and hence (most likely) isn't a basis vector. In fact if $a_j$ are appropriate scalars and $$ \sum_j{a_j}Bx_j=y $$ then $$ x=\sum_j a_j x_j $$ by virtue of linearity of matrix multiplication.