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Lesson /10
Subspaces

One of the most confusing concepts in linear algebra: Subspaces.
\textbf{Subspaces are certain space that is within another space that satisfies certain conditions.}
Space can be anything from a line, a plane, or a hyperplane.

The conditions that must be satisfied are:

  1. $\vec{0}$ must exist in the space
  2. When vectors in the space are added, then that resulting vector must still be in the original space (Closed Under Addition)
  3. When a vector in the space is multiplied by a scalar, then the resulting vector must still be in the original space (Closed Under Scalar Multiplication)
When you are proving that a space is a subspace, it must be done generally, but to disprove that it is a subspace, then a counterexample is completely fine.

I will do two examples, one that is a subspace, and other one that is not a subspace.

 

Subspace Example 1

Prove or disprove that $S=\{\begin{bmatrix}x_1\\x_2\end{bmatrix} | x_2 = 2x_1\}$ is a subspace of $\mathbb{R}^2$.

This is an equation of a line (you might recognize that if I expressed it as y=2x).

The general rule of thumb for approaching the subspace proof is to use LHS=RHS approach, where you go through LHS and RHS of a equation that you are trying to prove separately, and see if they are equal to each there at the end.

1.Zero Vector

$\vec{0} = \begin{bmatrix}0\\0\end{bmatrix}$
(LHS)

0
(RHS)
2(0)
=0
Since LHS = RHS, $\vec{0}$ exists in the space.

2. C.U.S.
We define two vectors \textbf{that are part of the space} $\vec{x}$ and $\vec{y}$.

$$\vec{x} = \begin{bmatrix}x_1\\x_2\end{bmatrix}, \vec{y} = \begin{bmatrix}y_1\\y_2\end{bmatrix}$$

We already established that the two vectors are already part of the space, therefore it is possible to rewrite the vectors as following:

$$\vec{x} = \begin{bmatrix}x_1\\2x_1\end{bmatrix}, \vec{y} = \begin{bmatrix}y_1\\2y_1\end{bmatrix}$$

Now, we add the two vectors, to produce a new vector, $\vec{x}+\vec{y}=\begin{bmatrix}x_1+y_1\\2x_1+2y_1\end{bmatrix}$.
To prove that the new vector is part of the space agin, we need to see if it still satisfies the condition, $x_2=2x_1$

(LHS)
$2x_1+2y_1$
(RHS)
$2(x_1+y_1)$
$=2x_1+2y_1$
Since second coordinate of the new vector is 2 times the first coordinate, we proved that the vector formed by addition is still part of the space.

3. C.U.S.M.
Again, we define a vector that is part of the space,

$$\vec{x} = \begin{bmatrix}x_1\\x_2\end{bmatrix}$$

Again, we already defined that $\vec{x}$ is part of the space, so we can rewrite the vector like before.

$$\vec{x} = \begin{bmatrix}x_1\\2x_1\end{bmatrix}$$

Additionally, we define a scalar value, $t, \in \mathbb{R}$.
The resulting vector formed after scalar multiplying the vector is

$$t\vec{x} = \begin{bmatrix}tx_1\\t2x_1\end{bmatrix}$$

Again, we have to see if this resulting vector is still part of the space that satisfies that the condition, $x_2=2x_1$.
(LHS)
$t2x_1$
$=2tx_1$
(RHS)
$2(tx_1)$
$=2tx_1$

Since $LHS = RHS$, we know that the space is closed under scalar multiplication.
In conclusion, since the space has zero vector included, closed under addition, and closed under scalar multiplication, we know that space S is a subspace of $\mathbb{R}^2$.

 

Subspace Example 2

Prove or disprove that $S=\{\begin{bmatrix}x_1\\x_2\end{bmatrix} | x_2 = {x_1}^2\}$ is a subspace of $\mathbb{R}^2$.
This is an equation of a parabola (you might recognize that if I expressed it as $y=x^2$).
The general rule of thumb is, that if there is multiplication, division or power of variables, the space is not a subspace.

Counter Example
I am going to choose two vectors that are part of the space,

$$\vec{x} = \begin{bmatrix}1\\1\end{bmatrix}$$
$$\vec{y} = \begin{bmatrix}2\\4\end{bmatrix}$$

When these two vectors are added, the resulting vector is

$$\vec{x}+\vec{y} = \begin{bmatrix}1+2\\1+4\end{bmatrix}$$
$$\vec{x}+\vec{y} = \begin{bmatrix}3\\5\end{bmatrix}$$

Since $5\neq3^2$, it is not closed under addition.
The space does not satisfy one of the conditions, therefore the space is not a subspace.

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