Worked examples · Sheet 3

Bases and dimension of subspaces defined by equations

Theory used: Cartesian form From Cartesian to parametric form Basis Dimension Homogeneous and non-homogeneous equations

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Find a basis and the dimension of the subspace

W = {(x, y, z) \in R^{3} : x - 2 y + z = 0} .

x = 2 y - z

(x, y, z) = (2 y - z, y, z) = y (2, 1, 0) + z (- 1, 0, 1)

W = ⟨(2, 1, 0), (- 1, 0, 1)⟩, dim W = 2

There is only one equation and it has rank $1$ , so by the formula $dim W = n - rk A$ of the Cartesian form we have $dim W = 3 - 1 = 2$ . To find a basis we solve for $x$ in terms of the free variables $y$ and $z$ and write the generic vector of $W$ factoring out $y$ and $z$ . The two vectors $(2, 1, 0)$ and $(- 1, 0, 1)$ span $W$ and are not proportional, so they are independent: they form a basis of $W$ , which is a plane through the origin. The check is done by substituting into the equation: $2 - 2 + 0 = 0$ and $- 1 - 0 + 1 = 0$ .

Exercise 2

Find a basis and the dimension of the subspace of $R^{4}$

W : {x_{1} + x_{2} - x_{3} = 0 x_{2} + x_{3} - x_{4} = 0

x_{2} = x_{4} - x_{3}, x_{1} = x_{3} - x_{2} = x_{3} - (x_{4} - x_{3}) = 2 x_{3} - x_{4}

(x_{1}, x_{2}, x_{3}, x_{4}) = (2 x_{3} - x_{4}, x_{4} - x_{3}, x_{3}, x_{4}) = x_{3} (2, - 1, 1, 0) + x_{4} (- 1, 1, 0, 1)

W = ⟨(2, - 1, 1, 0), (- 1, 1, 0, 1)⟩, dim W = 2

The coefficient matrix has rows $(1, 1, - 1, 0)$ and $(0, 1, 1, - 1)$ , which are not proportional, so it has rank $2$ and $dim W = 4 - 2 = 2$ . The system is already in echelon form, with pivots on the variables $x_{1}$ and $x_{2}$ , therefore $x_{3}$ and $x_{4}$ are chosen as free variables. From the second equation we obtain $x_{2}$ , then we substitute into the first and obtain $x_{1}$ . Factoring out $x_{3}$ and $x_{4}$ in the generic vector we obtain two vectors that span $W$ ; they are independent, because their last two components are $(1, 0)$ and $(0, 1)$ . Hence they form a basis of $W$ .

Exercise 3

Find a basis and the dimension of the subspace of $R^{4}$

W : ⎩ ⎨ ⎧ x_{1} + x_{2} + x_{3} + x_{4} = 0 x_{1} - x_{2} + x_{3} - x_{4} = 0 3 x_{1} + x_{2} + 3 x_{3} + x_{4} = 0

113 1 - 1 1 113 1 - 1 1 R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - 3 R_{1} 100 1 - 2 - 2 100 1 - 2 - 2 R_{3} \to R_{3} - R_{2} 100 1 - 2 0 100 1 - 2 0

rk A = 2, dim W = 4 - 2 = 2

- 2 x_{2} - 2 x_{4} = 0 ⟹ x_{2} = - x_{4}, x_{1} = - x_{2} - x_{3} - x_{4} = - x_{3}

(x_{1}, x_{2}, x_{3}, x_{4}) = (- x_{3}, - x_{4}, x_{3}, x_{4}) = x_{3} (- 1, 0, 1, 0) + x_{4} (0, - 1, 0, 1)

W = ⟨(- 1, 0, 1, 0), (0, - 1, 0, 1)⟩

There are three equations, but row reduction makes the third row vanish: the third equation is the sum of twice the first and the second, so it adds no conditions. The rank is $2$ , therefore $dim W = 4 - 2 = 2$ , and counting the equations without computing the rank would give a wrong dimension. From the echelon system we obtain $x_{2} = - x_{4}$ and then $x_{1} = - x_{3}$ , with $x_{3}$ and $x_{4}$ free. The two vectors obtained by factoring out the free variables are independent, so they form a basis of $W$ .

Exercise 4

Let $B = (1011)$ . Find a basis and the dimension of the subspace

W = {A \in M_{2} (R) : A B = B A} .

A = (a c b d), A B = (a c a + b c + d), B A = (a + c c b + d d)

A B = B A ⟺ ⎩ ⎨ ⎧ a = a + c a + b = b + d c + d = d ⟺ {c = 0 d = a

A = (a 0 b a) = a (1001) + b (0010)

W = ⟨ I_{2}, E_{12} ⟩, dim W = 2

We write the generic matrix $A$ with four unknowns and impose the equality $A B = B A$ entry by entry. The entries in position $(2, 1)$ always coincide, so three equations remain, which reduce to $c = 0$ and $d = a$ . The matrices of $W$ therefore depend on the two free parameters $a$ and $b$ , and factoring them out we obtain $A = a I_{2} + b E_{12}$ . The matrices $I_{2}$ and $E_{12}$ span $W$ and are not proportional, therefore they form a basis of $W$ and $dim W = 2$ . In coordinates with respect to the standard basis of $M_{2} (R)$ , $W$ is the subspace of $R^{4}$ defined by the equations $c = 0$ and $d - a = 0$ , and the dimension $4 - 2 = 2$ agrees with the formula of the Cartesian form.