FiessyFrimatsd0

2022-01-31

Can a matrix transformation ever make a linearly dependent matrix linearly independent?
For example, if A is a linearly dependent matrix, and B any matrix, could BA ever come out to be linearly independent?

Ronald Alvarez

Expert

Step 1
No it can not:
$A\in {K}^{m×n},B\in {K}^{p×m}$
A is linear dependent, so there exist ${\lambda }_{k}\in K$ not all equal to 0 with
$\sum _{k=1}^{n}{\lambda }_{k}{a}_{k}=0.$
where $A=\left({a}_{1},{a}_{2},\dots ,{a}_{n}\right)$ or: there is a non-zero vector ${x}_{0}=\left({\lambda }_{k}\right)\ne 0$ with
$A{x}_{0}=0.$
For BA we have
$\left(BA\right){x}_{0}=B\left(A{x}_{0}\right)=B0=0$
so regardless of the choice of B the vector ${x}_{0}$ is a non-zero kernel vector for BA, so BA can not be linear independent.

Lily Thornton

Expert

Step 1
Let the matrices A and B be given. Suppose the columns of A are the column vectors ${A}_{i}$ where i runs over the number of columns, let that number be k.
Now,
$BA=\left[B{A}_{1}|B{A}_{2}|\dots B{A}_{k}\right]$
So, your question boils down to asking can B act on the set of linearly independent vectors $\left\{{A}_{1},{A}_{2},\dots {A}_{k}\right\}$ such that the resulting set $\left\{B{A}_{1},B{A}_{2},\dots B{A}_{k}\right\}$ is linearly dependent.
The answer is yes, let B be the transformation that mixes the columns of A, for example, define $B{A}_{1}={A}_{1}+{A}_{2}$ and $B{A}_{i}={A}_{i}$ for $i\ge 2$

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