## Linear Algebra and Its Applications, Exercise 2.3.19

Exercise 2.3.19. Suppose $A$ is an $m$ by $n$ matrix, with $n$ columns taken from $\mathbf{R}^m$. What is the rank of $A$ if its column vectors are linearly independent? What is the rank of $A$ if its column vectors span $\mathbf{R}^m$? What is the rank of $A$ if its column vectors are a basis for $\mathbf{R}^m$?

Answer: If the $n$ column vectors of $A$ are linearly independent then there must be a pivot in every one of the $n$ columns, so that the rank $r = n$.

If the $n$ columns of $A$ span $\mathbf{R}^m$ then we must have $n \ge m$. There can be no more than $m$ linearly independent vectors in $\mathbf{R}^m$ so out of the $n$ columns of $A$ only $m$ columns can have pivots. Therefore the rank $r = m$.

If the $n$ columns of $A$ are a basis for $\mathbf{R}^m$ then they are linearly independent, which means the rank $r = n$, and they also span $\mathbf{R}^m$ so we must also have $r = m$. We thus have $r = m = n$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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