Question

Let V ={(a,b,c,d) ϵ R^4! a+b+2c+2d = 0} and W ={(a,b,c,d) ϵ R4! a = -b;c = -d}

Find the dimensions of V and W.
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Answer 1

Answer on Question #44928 – Math – Linear Algebra

Question. Let $V = \{(a, b, c, d) \in \mathbb{R}^4 : a + b + 2c + 2d = 0\}$ and $W = \{(a, b, c, d) \in \mathbb{R}^4 : a = -b, c = -d\}$. Find the dimensions of $V$ and $W$.

Solution. Rewrite the $V$ in the next form: $V = \{(-b - 2c - 2d, b, c, d)\}$. Find the basis of $V$: $v_1 = (-1, 1, 0, 0)$, $v_2 = (-2, 0, 1, 0)$, $v_3 = (-2, 0, 0, 1)$. Check that $v_1, v_2, v_3$ are linearly independent.

Let $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = \bar{0} \Leftrightarrow (-\lambda_1, \lambda_1, 0, 0) + (-2\lambda_2, 0, \lambda_2, 0) + (-2\lambda_3, 0, 0, \lambda_3) = \bar{0} \Leftrightarrow$

$$\Leftrightarrow (- \lambda_ {1} - 2 \lambda_ {2} - 2 \lambda_ {3}, \lambda_ {1}, \lambda_ {2}, \lambda_ {3}) = \bar {0} \Leftrightarrow \left\{ \begin{array}{l} \lambda_ {1} = 0 \\ \lambda_ {2} = 0 \Rightarrow v _ {1}, v _ {2}, v _ {3} \text{ are linearly independent.} \\ \lambda_ {3} = 0 \end{array} \right.$$

Check that $\forall v \in V$ can be represented in the next form: $v = \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3$. Let $v = (-b - 2c - 2d, b, c, d) \in V$. Then $\begin{cases} \lambda_1 = b \\ \lambda_2 = c \\ \lambda_3 = d \end{cases}$, $b v_1 + c v_2 + d v_3 = (-b - 2c - 2d, b, c, d) == v$.

So dimension of $V$ is equal to 3.

Rewrite the $W$ in the next form: $W = \{(a, -a, c, -c)\}$. Find the basis of $W$: $w_1 = (1, -1, 0, 0)$, $w_2 = (0, 0, 1, -1)$. Check that $w_1, w_2$ are linearly independent.

Let $\lambda_1 w_1 + \lambda_2 w_2 = \bar{0} \Leftrightarrow (\lambda_1, -\lambda_1, 0, 0) + (0, 0, \lambda_2, -\lambda_2) = \bar{0} \Leftrightarrow (\lambda_1, -\lambda_1, \lambda_2, -\lambda_2) = \bar{0} \Leftrightarrow$

$$\Leftrightarrow \left\{ \begin{array}{l} \lambda_ {1} = 0 \\ \lambda_ {2} = 0 \end{array} \right. \Rightarrow w _ {1}, w _ {2} \text{ are linearly independent}.$$

Check that $\forall w \in W$ can be represented in the next form: $w = \lambda_1 w_1 + \lambda_2 w_2$. Let $w = (a, -a, c, -c) \in W$. Then $\begin{cases} \lambda_1 = a \\ \lambda_2 = c \end{cases}$, $a w_1 + c w_2 = (a, -a, c, -c) = w$. So dimension of $W$ is equal to 2.

Answer. $dimV = 3, dimW = 2$.

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