I have the matrix:

$$ \begin{pmatrix} 3 & 2 & -1 & 4 \\ 1 & 0 & 2 & 3 \\ -2 & -2 & 3 & -1 \\ \end{pmatrix} $$

I have 2 questions to answer:

  1. Consider the columns of the matrix as vectors in $R^3$. How many of these vectors are linearly independent?
  2. Consider for $R^4$. How many vectors are linearly independent?

Is the answer in both cases 2 or am I totally wrong about how to solve this.


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    $\begingroup$ I believe the answer is 2 in both cases since the rank is 2. Is this correct? $\endgroup$ – zeeks Feb 26 '18 at 3:44
  • $\begingroup$ or for $R^3$, there are 0 linearly independent vectors since there are 4 vectors and we have R^3? $\endgroup$ – zeeks Feb 26 '18 at 3:46
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    $\begingroup$ Correct!${}{}{}$ $\endgroup$ – FYY Feb 26 '18 at 3:49
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    $\begingroup$ Hint: the middle row is the sum of the other two. $\endgroup$ – dxiv Feb 26 '18 at 3:51

I interpret the first part of the question as asking for the column rank of the matrix and the second part of the question as asking for the row rank of the matrix.

Yes, compute its RREF and you can see that there are two pivot columns, hence the rank is $2$.

It is known that the row rank is equal to the column rank and the answer is $2$ for both.

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  • $\begingroup$ among the notes the professor gave us, he said if the number of vectors is higher than n in $R^n$, which in the first case is $R^3$, then the vectors are linearly dependent. does that mean, for the first question, the answer is 0? or i misunderstood his point. $\endgroup$ – zeeks Feb 26 '18 at 3:51
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    $\begingroup$ No, it is still $2$. The answer is $2$ in the sense that out of all these vectors, you can find two of them such that they form a linearly independent set. Linearly dependent doesn't mean we can't find any such vector at all, the answer is not zero unless you have the zero matrix. $\endgroup$ – Siong Thye Goh Feb 26 '18 at 3:59

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