Linear Algebra

90 Chapter Two. Vector Spacescombination such as 2⃗t 1 − 3⃗t 2 and be sure the result is an element of T, that is,T doesn’t satisfy statement (2).Lemma 2.9 suggests that a good way to think of a vector space is as acollection of unrestricted linear combinations. The next two examples take somespaces and recasts their descriptions to be in that form.2.11 Example We can show that this plane through the origin subset of R 3⎛ ⎞x⎜ ⎟S = { ⎝y⎠ ∣ x − 2y + z = 0}zis a subspace under the usual addition and scalar multiplication operationsof column vectors by checking that it is nonempty and closed under linearcombinations of two vectors as in Example 2.2. But there is another way. Thinkof x−2y+z = 0 as a one-equation linear system and paramatrize it by expressingthe leading variable in terms of the free variables x = 2y − z.⎛⎜2y − z⎞⎛⎟S = { ⎝ y ⎠ ∣ ⎜2⎞ ⎛ ⎞−1⎟ ⎜ ⎟y, z ∈ R} = {y ⎝1⎠ + z ⎝ 0⎠ ∣ y, z ∈ R} (*)z0 1Now, to show that this is a subspace consider r 1 ⃗s 1 + r 2 ⃗s 2 . Each ⃗s i is a linearcombination of the two vectors in (∗) so this is a linear combination of linearcombinations.⎛⎜2⎞ ⎛⎟ ⎜−1⎞ ⎛⎟ ⎜2⎞ ⎛⎟ ⎜−1⎞⎟r 1 (y 1 ⎝1⎠ + z 1 ⎝ 0⎠) + r 2 (y 2 ⎝1⎠ + z 2 ⎝ 0⎠)0 10 1The Linear Combination Lemma, Lemma One.III.2.3, shows that this is a linearcombination of the two vectors and so Theorem 2.9’s statement (2) is satisified.2.12 Example This is a subspace of the 2×2 matrices M 2×2 .( )a 0 ∣∣L = { a + b + c = 0}b cTo parametrize, express the condition as a = −b − c.( ) ( ) ( )−b − c 0 ∣∣ −1 0 −1 0 ∣∣L = {b, c ∈ R} = {b + cb, c ∈ R}b c1 0 0 1As above, we’ve described the subspace as a collection of unrestricted linearcombinations. To show it is a subspace, note that a linear combination of vectorsfrom L is a linear combination of linear combinations and so statement (2) istrue.