## 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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