### Subspace

Exercise
Which of the following are subspaces of some euclidean spaces?

1) **H** = {(`a` + `b`, 3`a` - `b`, 2`a` + `b`): `a`, `b` ∈ **R**}

Answer **H** is a subspace.

`u`, `v` ∈ `H`

⇒ **u** = (`a` + `b`, 3`a` - `b`, 2`a` + `b`), **v** = (`a'` + `b'`, 3`a'` - `b'`, 2`a'` + `b'`)

⇒ **u** + **v** = (`a` + `b` + `a'` + `b'`, 3`a` - `b` + 3`a'` - `b'`, 2`a` + `b` + 2`a'` + `b'`) = (`a''` + `b''`, 3`a''` - `b''`, 2`a''` + `b''`) ∈ `H`,

where `a''` = `a` + `a'`, `b''` = `b` + `b'`.

`u` ∈ `H`, `c` ∈ **R**

⇒ **u** = (`a` + `b`, 3`a` - `b`, 2`a` + `b`)

⇒ `c`**u** = (`ca` + `cb`, 3`ca` - `cb`, 2`ca` + `cb`) = (`a'` + `b'`, 3`a'` - `b'`, 2`a'` + `b'`) ∈ `H`

where `a'` = `ca`, `b'` = `cb`.

2) **H** = {(`a` + `b`, 3`a` - `b`, 2`a` + `b` + 1): `a`, `b` ∈ **R**}

Answer If **H** is a subspace, then it must contain the zero vector. In other words, we can find `a` and `b`, such that

`a` + `b` = 0, 3`a` - `b` = 0, 2`a` + `b` + 1 = 0.

However, the first two equations imply `a` = `b` = 0, which does not satisfy the third equation. We conclude that **H** is not a subspace.

3) **H** = {(`a` + `b`, 3`a` - `b`, 2`a` + `b`): `a`, `b` ∈ **Z**}

Answer By taking `a` = 1, `b` = 0, we see that (1, 2, 3) ∈ **H**. If **H** is a subspace, then we should also have 1/2(1, 2, 3) = (1/2, 1, 3/2) ∈ **H**. On the other hand, we cannot find two *integers* `a` and `b` such that `a` + `b` = 1/2. Therefore **H** is not a subspace.

4) **H** = {(`x`, `y`, `z`): `x` + `y` - `z` = 0, 2`x` - 3`y` + `z` = 0}

Answer **H** is a subspace.

`u`, `v` ∈ `H`

⇒ **u** = (`x`, `y`, `z`), `x` + `y` - `z` = 0, 2`x` - 3`y` + `z` = 0; **v** = (`x'`, `y'`, `z'`), `x'` + `y'` - `z'` = 0, 2`x'` - 3`y'` + `z'` = 0

⇒ **u** + **v** = (`x` + `x'`, `y` + `y'`, `z` + `z'`), (`x` + `x'`) + (`y` + `y'`) + (`z` + `z'`) = `x` + `y` - `z` + `x'` + `y'` - `z'` = 0, 2(`x` + `x'`) - 3(`y` + `y'`) + (`z` + `z'`) = 2`x` - 3`y` + `z` + 2`x'` - 3`y'` + `z'` = 0

⇒ `u` + `v` ∈ `H`.

The verification of `u` ∈ `H`, `c` ∈ **R** ⇒ `c`**u** ∈ `H` is similar.

5) **H** = {(`x`, `y`, `z`): `x` + `y` - `z` = 0, 2`x` - 3`y` + `z` = 1}

Answer Since `x` = `y` = `z` = 0 does not satisfy the condition 2`x` - 3`y` + `z` = 1, **H** does not contain the zero vector. Therefore **H** is not a subspace.

6) **H** = {(`x`, `y`, `z`): `x`^{2} + `y`^{2} - `z`^{2} = 0}

Answer Since (1, 0, 1), (0, 1, 1) ∈ `H` and (1, 0, 1) + (0, 1, 1) = (1,1,2) ∉ `H`, we conclude that `H` is not a subspace.

7) **H** = {(`x`, `y`): `xy` ≥ 0}

Answer Since (2, 0), (-1, -1) ∈ `H` and (2, 0) + (-1, -1) = (1, -1) ∉ `H`, we conclude that `H` is not a subspace.

8) **H** = {(`x`, `y`, 0): `x`, `y` ∈ **R**}

Answer **H** is a subspace.

`u`, `v` ∈ `H`

⇒ **u** = (`x`, `y`, 0), **v** = (`x'`, `y'`, 0)

⇒ **u** + **v** = (`x` + `x'`, `y` + `y'`, 0 + 0) = (`x` + `x'`, `y` + `y'`, 0) ∈ `H`.

The verification of `u` ∈ `H`, `c` ∈ **R** ⇒ `c`**u** ∈ `H` is similar.