2212. Solutions to linear algebra exercises

Vector spaces and subspaces

On this sheet, all exercises are for assessment except for those marked (P),which will be worked in the problem session. Marking scheme; 1 mark

per part

13. Which of the following subsets of Rn are vector spaces over R under ordi-nary addition and scalar multiplication of vectors? For each which is not

a vector space over R, explain why not. For each which is a vector space,either write down two non-zero elements of the set or show that it containsjust one element, the ‘zero’ vector.

(a)

(x1, x2, x3, x4, x5) : x1 + x2 + x3 + x4 + x5 = 0,x3 + x4 + x5 = 0,

x4 − x5 = 0

This is a vector space (as the set of solutions to a homogeneous systemof linear equations). Because it’s in echelon form it’s easy to findelements. We can set x5, x2 to have any value, and then computex4, x3, x1. e.g. if x2 = 0, x5 = 1, we have x4 = 1, x3 = −2, x1 = 0,giving (0, 0,−2, 1, 1), and if x2 = 1, x5 = 0, we have x4 = 0, x3 =0, x1 = −1, giving (−1, 1, 0, 0, 0).

(b) (P)

(x1, x2, x3, x4, x5) : x1 + x2 + x3 + x4 + x5 = 5x3 + x4 + x5 = 3

x4 − x5 = 0

This is not a vector space. We see that (0, 0, 0, 0, 0) is not in the set.(The problem here is that the system of equations is not homogeneous,that is the right hand side is non-zero.)

(c)

(x1, x2, x3) : x1 + 2x2 + 3x3 = 6x2 + x3 = 2

x3 = 1

This is not a vector space. We see that (0, 0, 0) is not in the set. (Theproblem here is that the system of equations is not homogeneous, thatis the right hand side is non-zero.)

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(d)

(x1, x2, x3) : x1 + 2x2 + 3x3 = 0x2 + x3 = 0

x3 = 0

This is a vector space (as the set of solutions to a homogeneous systemof linear equations). The coefficient matrix has full rank 3, and so thesystem of equations has a unique solution. That means the vectorspace has just one element, (0, 0, 0).

(e) (P) {(x, y, z) : x, y, z,∈ R, x2 − y2 − z2 = 0}This isn’t a vector space. It contains (5, 4, 3) and (13, 12, 5) but nottheir sum (18, 16, 8), since

182 = 324 6= 256 + 64 = 162 + 82.

(The problem here is that it is the solution set of a non-linear equa-tion.)

(f) {(x, y, z) : x, y, z,∈ R, x = 3y − 4z}This is a vector space. It is the solution space of a linear, homogeneousequation. We find solutions by giving y, z any values we choose, andthen computing x in terms of them, e.g. if z = 1, y = 0, we get x = −4,and if y = 1, z = 0 we get x = 3. So (3, 1, 0), (−4, 0, 1) are in the space.

(g) {(x, y, z) : x, y, z,∈ R, x = yz}This is not a vector space. It contains (1, 1, 1) but not 2(1, 1, 1) =(2, 2, 2). It is the solution set of a non-linear equation.

(h) {(x, y, z) : x, y, z,∈ R, x2 + y2 + z2 ≤ 1}This is not a vector space. It contains (1, 0, 0) but not 2(1, 0, 0) =(2, 0, 0). We wouldn’t expect the solution set of an inequality to be avector space; if such a set contains a non-zero vector, it’s hard for allmultiples of it also to satisfy the inequality.

(i) {(a, b, 2a + 3b) : a, b ∈ R}This is a vector space. Setting a = 1, b = 0 and then a = 0, b = 1 weget two elements (1, 0, 2) and (0, 1, 3).

(j) {(a, b, 2a + 3b + 4) : a, b ∈ R}This is not a vector space. It doesn’t contain (0, 0, 0). Because for anelement to have its first two coordinates zero we need a = 0, b = 0,but then the last coordinate is 4.

(k) (P) {(a − 1, b + 1, 2a + 3b + 1) : a, b ∈ R} This is a vector space. Wenotice that 2a + 3b + 1 = 2(a− 1) + 3(b + 1) Setting first a = 0, b = 0and then a = 0, b = 1 we get the elements (−1, 1, 1) and (−1, 2, 4).

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(l) {(a2, b2, a2 + b2) : a, b ∈ R}This is not a vector space. It contains (1, 1, 2) but not (−1,−1,−2) =−(1, 1, 2).

(m) {(a2, b2, c2) : a, b, c ∈ R}This is not a vector space. It contains (1, 1, 1) but not (−1,−1,−1) =−(1, 1, 1).

14. Which of the following sets of functions f : R → R are vector spaces overR under ordinary addition and scalar multiplication of functions (where(f + g)(x) = f(x) + g(x) and (λf)(x) = λ(f(x)))? For each which is not

a vector space over R, explain why not. For each which is a vector space,either write down two non-zero elements of the set or show that it containsjust one element, the ‘zero’ vector.

(a) {f : R → R : f ′′(x) − 3f ′(x) + 2f(x) = ex}This is not a vector space. It contains −xex but not xex. (The differ-ential equation is linear but not homogeneous.)

(b) {f : R → R : f ′′(x) − 3f ′(x) + 2f(x) = 0}This is a vector space. It’s general solution is Aex + Be2x. So, e.g. ex

and e2x are both in the set. (It’s the solution set of a linear, homoge-neous equation.)

(c) (P) {f : R → R : f ′(x) − x[f(x)] = 0} This is a vector space. It’s thesolution set of a linear homogeneous differential equation. We solve

dy

dx− xy = 0

by separating the variables. We have

dy

dx= xy

∫

dy

y=

∫

xdx,

log(|y|) =1

2x2 + C,

y = kex2

We see thatf(x) = ex

2

, f(x) = 2ex2

are both in the set.

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(d) {f : R → R : f ′′(x) + f(x) = 0}This is a vector space. It’s the solution space of a linear, homoge-neous differential equation. The general solution is f(x) = A cos(x) +B sin(x), and in particular f(x) = cos(x) and f(x) = sin(x) are in theset.

(e) (P) {f : R → R : f ′(x) − [f(x)]2 = 0} This is not a vector space. It’sthe solution set of a non-linear differential equation. We can solve theequation by separating the variables.

dy

dx= y2,

∫ dy

y2=

∫

dx,

−1

y= x + c,

y =−1

x + c.

So we see that f(x) = − 1

xis in the set but f(x) = 1

xis not.

(f) {f : R → R : f ′(x) −√

f(x) = 0}This is not a vector space. It is the solution set of a non-linear differ-ential equation. We can solve the equation by separating the variables.

dy

dx=

√y,

∫

dy√y

=∫

dx,

2√

y = x + c,

y2 =1

2x + b.

So we see that f(x) =√

x/2 is in the set but f(x) =√

x, which is ascalar multiple of that, is not.

15. Which of the following are subspaces of R3? For each which is not, explainwhy not. For each which is, either write down two distinct non-zero vectorsv1, v2, of the subspace or show that the subspace is just the zero vector.

(a) {(x, y, z) ∈ R3 : z = 0}This is a subspace. It contains (1, 0, 0) and (0, 1, 0).

(b) (P) {(x, y, z) ∈ Z3} This is not a subspace. It contains (1, 1, 1) butnot (1

2, 1

2, 1

2).

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(c) {(x, y, z) ∈ R3 : x ≥ 0, y ≥ 0, z ≥ 0}This is not a subspace. It contains (1, 1, 1) but not (−1,−1,−1).

(d) {(x, y, z) ∈ R3 : x − 2y + 3z = 0}This is a subspace. It contains (2, 1, 0) and 3, 0,−3).

(e) (P) {(x, y, z) ∈ R3 : x − 2y + 3z + 4 = 0} This is not a subspace. Itdoesn’t contain (0, 0, 0).

(f) {(x, y, z) ∈ R3 : x2 − 2y + 3z = 0}This is not a subspace. (The equation is non-linear.) It contains(2, 1, 0) but not 2(2, 1, 0) = (4, 2, 0).

(g) {(x, y, z) ∈ R3 : x − 2y + 3z > 0}This is not a subspace. ) It contains (1, 0, 0) but not −1(1, 0, 0) =(−1, 0, 0).

(h) {(x, y, z) ∈ R3 : x − 2y + 3z = 0, 3x + 4y − z = 0}This is a subspace. It;s the solution set of a system of homoge-neous linear equations. It’s equivalent to the system x − 2y + 3z =0, 10y − 8z = 0,, which is easier to solve. We can find solutions(−7, 4, 5), (−14, 8, 10).

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