## 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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