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Problem 191

# Let $$U$$ and $$W$$ be the following subspaces of $$R^{4}$$. \begin{aligned} &\mathrm{U}=\\{(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}): \mathrm{b}+\mathrm{c}+\mathrm{d}=0\\} \\ &\mathrm{W}=\\{(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}): \mathrm{a}+\mathrm{b}=0, \mathrm{c}=2 \mathrm{~d}\\} \end{aligned} Find the dimension and a basis of: (i) $$\mathrm{U}_{2}$$ (ii) $$\mathrm{W}$$, (iii) $$\mathrm{U} \cap \mathrm{W}$$.

Expert verified
In summary, for the given subspaces U and W of $$R^4$$, we have: (i) The dimension of U_2 is 3, and its basis is $$\{(1,0,0,0),(0,1,-1,0),(0,0,-1,1)\}$$. (ii) The dimension of W is 2, and its basis is $$\{(-1,1,0,0),(0,0,2,1)\}$$. (iii) The dimension of U ∩ W is 2, and its basis is $$\{(-1,1,0,0),(0,0,2,1)\}$$.
See the step by step solution

## Step 1: General form for vectors in U_2

Since U_2 is not given explicitly in the exercise, we assume that U_2 refers to U. In this case, vectors in U have the property b + c + d = 0, so we can rewrite any vector in U as (a, b, -b-d, d).

## Step 2: Finding a basis for U

A basis for U can be found by taking the linearly independent vectors formed from the general form of a vector in U. Let the basis vectors be u_1, u_2, and u_3: $$u_1 = (1, 0, 0, 0)$$, $$u_2 = (0, 1, -1, 0)$$, $$u_3 = (0, 0, -1, 1)$$. These vectors are linearly independent and span U. Therefore, the basis for U (or U_2) is {u_1, u_2, u_3}.

## Step 3: Dimension of U_2

Since U (or U_2) has a basis consisting of three vectors, the dimension of U_2 is 3. (ii) Finding the basis and dimension of W

## Step 4: General form for vectors in W

Vectors in W have the properties a + b = 0 and c = 2d, so we can rewrite any vector in W as (-b, b, 2d, d).

## Step 5: Finding a basis for W

A basis for W can be found by taking the linearly independent vectors formed from the general form of a vector in W. Let the basis vectors be w_1 and w_2: $$w_1 = (-1, 1, 0, 0)$$, $$w_2 = (0, 0, 2, 1)$$. These vectors are linearly independent and span W. Therefore, the basis for W is {w_1, w_2}.

## Step 6: Dimension of W

Since W has a basis consisting of two vectors, the dimension of W is 2. (iii) Finding the basis and dimension of U ∩ W

## Step 7: Intersection of U and W

In the intersection of U and W, we will have vectors (a, b, c, d) that satisfy both the properties of U and W, i.e., b + c + d = 0 and a + b = 0, c = 2d.

## Step 8: Finding a basis for U ∩ W

From the general form of vectors in U and W, we have U: (a, b, -b-d, d), W: (-b, b, 2d, d). Set the general forms equal to each other and solve for a, b, c, and d: $$(-b, b, 2d, d) = (a, b, -b-d, d)$$ We get a = -b, c = 2d. So, the vector in U ∩ W is in the form (-b, b, 2d, d). We can rewrite it as v = b(-1, 1, 0, 0) + d(0, 0, 2, 1) = b w_1 + d w_2, where w_1 and w_2 are already the basis vectors for W. Since the vector v can be written as a linear combination of the basis vectors of W, the intersection U ∩ W has the same basis as W, which is {w_1, w_2}.

## Step 9: Dimension of U ∩ W

Since U ∩ W has a basis consisting of two vectors, the dimension of U ∩ W is 2.

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