Linear Algebra Done Right 2nd Edition – Chapter 1 Exercises 5(b), 5(c), 5(d), 6, 7

Chapter 1

5.(b)Determine if U = {(x_1,x_2,x_3) \in \textbf{F}^3:x_1 + 2x_2 + 3x_3 = 4} is a subspace of \textbf{F}^3 .

Solution: Since 0 + 0x_2 + 0x_3 \neq 4, (0,0,0) \notin U. Thus U is not a subspace of \textbf{F}^3 .

5.(c)Determine if U = {(x_1,x_2,x_3) \in \textbf{F}^3:x_1x_2x_3 = 0} is a subspace of \textbf{F}^3 .

Solution: Since x_1x_2x_3 = 0 at least one of x_i = 0. Consider (0,1,1) and (1,0,1). Both are in U but (0,1,1) + (1,0,1) = (1,1,2) \notin U. Thus, U is not a subspace of \textbf{F}^3 .

5.(d)Determine if U = {(x_1,x_2,x_3) \in \textbf{F}^3:x_1 = 5x_3} is a subspace of \textbf{F}^3 .

Solution:

Clearly (0,0,0) \in U.

Consider (a,b,c) and (x,y,z) both in U.
Then (5c,b,c) + (5z,y,z) =  (5c+5z, b+y, c+z) = ( 5(c+z), (b+y), (c+z) ) \in U

Let (x,y,z) = (5z, y, z) \in U and a \in \textbf{F}
Then a(x,y,z) = (a5z, ay, az) = (5az, ay, az) \in U.

U is a subspace of \textbf{F}^3

6. Give an example U of \mathbb{R}^2 such that U is closed under addition and under taking additive inverses (meaning -u \in U whenever u \in U), but U is not a subspace of \mathbb{R}^2.

Solution 1:

Let U = \{ (x,y) \in \mathbb{R}^2 : x,y \text{ are rational } \}

U \neq  \emptyset since (0,0) \in U.

If (a,b), (c,d) \in U then (a+c, b+d) \in U since the sum of rationals is rational. Thus, U is closed under addition.

Also, -a, -b are are rational since a, b, -1 are rational and the product of rationals is rational. Then (-a,-b) \in U. So U is closed under taking additive inverses.

Now, let a \in \mathbb{R}, a \text{ is irrational } . Consider (1,1) \in U. Then a(1,1) = (a,a) \notin U since the product of an irrational and a non-zero rational is an irrational. Thus, U is not closed under scalar multiplication.

U is not a subspace of \mathbb{R}^2

Solution 2: This is the one in the solution manual.

Let U = \{ (x,y) \in \mathbb{R}^2 : x,y \in \mathbb{Z} \}

U \neq  \emptyset since (0,0) \in U.

Given (a,b) \in U and (c,d) \in U then a+c \in \mathbb{Z} and b+d \in \mathbb{Z} and (a+c,b+d) \in U.
Thus, U is closed under addition.

Also, -a, -b are integers since a, b are integers. Then (-a,-b) \in U.
So U is closed under taking additive inverses.

Consider (1,1) \in U and \frac{1}{2} \in \mathbb{R}.
Then \frac{1}{2} (1,1) = (\frac{1}{2},  \frac{1}{2}) \notin U.
Thus, U is not closed under scalar multiplication.

U is not a subspace of \mathbb{R}^2

7. Give an example of a nonempty subset U of \mathbb{R}^2 such that U is closed under scalar multiplication, but U is not a subspace of \mathbb{R}^2.

Solution:

Let U = \{ (x,y) \in \mathbb{R}^2 : \text{ if } x = 0, y = 0 \text{ otherwise } y \neq 0 \}

Just for fun, this is a different subset than the one found in the solution manual.

U \neq  \emptyset since (0,0) \in U.

Let (x,y) \in U and a \in \mathbb{R}. If a = 0 or (x,y) = (0,0), then a(x,y) = (0,0) otherwise a(x,y) = (ax, ay) : ay \neq 0. So a(x,y) \in U. Thus, U is closed under scalar multiplication.

Consider (10, 1) and (-5, -1). Both are in U. But (10, 1) + (-5, -1) = (5, 0) \notin U.

U is not a subspace of \mathbb{R}^2.

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