Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $\{v_{1},v_{2},\cdots,v_{n}\}$ is $n$ linearly independent vectors in $\mathbb{R}^{n}$, what would be necessary and sufficient condition of $A$ ($n\times n$ matrix) $A$ so that the vectors $Av_{1}$, $Av_{2}$, $\cdots$, $Av_{n}$ are linearly independent?

share|cite|improve this question
up vote 1 down vote accepted


The vectors $Av_1, \ldots ,Av_n$ are linearly independent if, and only if, $\det \left([Av_1 | \ldots |Av_n]_{n\times n}\right)\neq 0$.

Now note that $[Av_1 | \ldots |Av_n]_{n\times n}=A[v_1|\ldots |v_n]_{n\times n}$.

share|cite|improve this answer
why is ther vertical lines in the matrix, before and after the dots? – Mark Oct 9 '13 at 16:49
@Mark It's a notation which suggests that I'm 'writing the matrix by columns'. The $i^{\text{th}}$ column of the matrix $[Av_1 | \ldots |Av_n]_{n\times n}$ is $Av_i$. Is that clear now? – Git Gud Oct 9 '13 at 16:52
yes so you gave me a property of matrix with column vectors that are linearly independent. So I need to find the condition of A that makes the matrix product with nonzero determinant correct? – Mark Oct 9 '13 at 17:03
So, the matrix product $\left([Av_1 | \ldots |Av_n]_{n\times n}\right)$ has to be invertible here – Mark Oct 9 '13 at 17:05
@Mark Yes, but that's not what I'm hinting at. Think about the last line in my answer. – Git Gud Oct 9 '13 at 17:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.