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$B = (v_1 , v_2,\ldots, v_n)$ form a basis of a general vector space. Given $P$, an invertible matrix, prove that BP is a basis.

Here's my thought - we know that an invertible matrix is a product of multiplying elementary matrices. Meaning, for each $v_i$, multiplying it with $P$ will result in some elementary actions done to the vector. These actions are simply linear combinations of them - so, BP will be a basis with vectors that are all linear combinations of vectors in $B$, meaning, they will still span and be linear independent.

Is this direction correct? I'll formalize it, ofcourse, but wanted opinions. Thanks!

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  • $\begingroup$ In “Linear combinations of them” what does “them” refer to? If it’s the rows of $B$, then how do you know that all of these linear combinations of the rows are linearly independent? $\endgroup$
    – amd
    Commented Jan 3, 2020 at 21:21

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HINT: Consider a linear combination of $BP$ that is zero:

$$\sum_{k=1}^n \lambda_kPv_k = 0.$$

Because $P$ is invertible, you may apply $P^{-1}$ to both sides. Do you think you can finish?

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    $\begingroup$ Thanks for the hint! I think I got it. $\endgroup$
    – Omri. B
    Commented Jan 3, 2020 at 21:37
  • $\begingroup$ You're welcome! Glad to have helped. $\endgroup$
    – Fimpellizzeri
    Commented Jan 3, 2020 at 21:39

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