# Become a Linear Algebra Expert

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:

begin{enumerate}

item $vec{0}$ must exist in the space

item When vectors in the space are added, then that resulting vector must still be in the original space (Closed Under Addition)

item 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)

end{enumerate}

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 $S={begin{bmatrix}x_1x_2end{bmatrix} | x_2 = 2x_1}$ 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

$vec{0} = begin{bmatrix}0end{bmatrix}$

(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 $vec{y}$.

$$vec{x} = begin{bmatrix}x_1x_2end{bmatrix}, vec{y} = begin{bmatrix}y_1y_2end{bmatrix}$$

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

$$vec{x} = begin{bmatrix}x_12x_1end{bmatrix}, vec{y} = begin{bmatrix}y_12y_1end{bmatrix}$$

Now, we add the two vectors, to produce a new vector, $vec{x}+vec{y}=begin{bmatrix}x_1+y_12x_1+2y_1end{bmatrix}$.

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,

$$vec{x} = begin{bmatrix}x_1x_2end{bmatrix}$$

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

$$vec{x} = begin{bmatrix}x_12x_1end{bmatrix}$$

Additionally, we define a scalar value, $t, in mathbb{R}$.

The resulting vector formed after scalar multiplying the vector is

$$tvec{x} = begin{bmatrix}tx_1t2x_1end{bmatrix}$$

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 $S={begin{bmatrix}x_1x_2end{bmatrix} | x_2 = {x_1}^2}$ 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,

$$vec{x} = begin{bmatrix}11end{bmatrix}$$

$$vec{y} = begin{bmatrix}24end{bmatrix}$$

When these two vectors are added, the resulting vector is

$$vec{x}+vec{y} = begin{bmatrix}1+21+4end{bmatrix}$$

$$vec{x}+vec{y} = begin{bmatrix}35end{bmatrix}$$

Since $5neq3^2$, it is not closed under addition.

The space does not satisfy one of the conditions, therefore the space is not a subspace.