## Linear Algebra and Its Applications, Exercise 2.3.15

Exercise 2.3.15. If the vector space $V$ has dimension $k$ show that

a) if a set of $k$ vectors in $V$ is linearly independent then that set forms a basis

b) if a set of $k$ vectors in $V$ spans $V$ then that set forms a basis

Answer: a) Assume that we have a set of linearly independent vectors $v_1, \dotsc, v_k$ in $V$. Suppose that this set does not span $V$. By theorem 2L (page 86) we can extend this set to form a basis by adding additional vectors $v_{k+1}, \dotsc, v_{k+p}$. But if this expanded set of vectors is a basis for $V$ then the dimension of $V$ is (by definition) $k+p > k$ which contradicts the assumption that the dimension of $V$ is $k$. Therefore the linearly independent set $v_1, \dotsc, v_k$ must span $V$ and be a basis for it.

b) Assume that we have a set of vectors $w_1, \dotsc, w_k$ in $V$ that span $V$. Suppose that this set is not linearly independent and thus not a basis. By theorem 2L (page 86) we can reduce this set to form a basis by removing one or more vectors so that the new set has $l$ vectors where $l < k$. But since the dimension of $V$ is $k$ then there must exist some set of $k$ vectors that is a basis for $V$. But if the reduced set of $l$ vectors is a basis for $V$ and the other set of $k$ vectors is also a basis then by theorem 2K we must have $l = k$ not $l < k$. We therefore conclude that the spanning set $w_1, \dotsc, w_k$ is in fact linearly independent and is thus a basis for $V$.

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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