0
$\begingroup$

So I finally grasped the idea of linear subspace in my linear algebra class and I came across a practice question in the exercise book. Let's say we have

$$B_1:=\left\{\ \begin {pmatrix} y_1\\ y_2 \\ y_3 \end{pmatrix} \in \mathbb{C^3} \ | \ y_1=(y_2 - y_3) \right\}\subseteq\mathbb{C^3}$$

and it says to show that $B_1$ is a subspace of $\mathbb{C^3}$ using the 3 properties of a liner subspace. So I get a vector $\vec{a}=\begin {pmatrix} a_2-a_3\\ a_2 \\ a_3 \end{pmatrix}$ and $\vec{b}=\begin {pmatrix} b_2-b_3\\ b_2 \\ b_3 \end{pmatrix}$.

  1. If $b_2=b_3=0$ that shows that the set is not empty therefore fulfills the first requirement.
  2. $\vec{a}+\vec{b}=\begin{bmatrix}a_2 -a_3\\ a_2 \\ a_3 \end{bmatrix} + \begin{bmatrix}b_2 - b_3\\ b_2 \\ b_3 \end{bmatrix}=\begin{bmatrix}(a_2+b_2) -(a_3+b_3)\\ (a_2+b_2) \\ (a_3+b_3) \end{bmatrix}$ which shows that the vectors are closed under addition therefore fulfills the second requirement.
  3. $\vec{a}=\begin {pmatrix} a_2-a_3\\ a_2 \\ a_3 \end{pmatrix} \in \mathbb{C^3}$ and $α \in\mathbb{R}$ then $α\begin{bmatrix}a_2 -a_3\\ a_2 \\ a_3 \end{bmatrix}=\begin{bmatrix}α(a_2 -a_3)\\ α(a_2) \\ α(a_3) \end{bmatrix}$, which shows that it's closed under scalar multiplication.

I'm pretty sure this is correct but I'm not a $100\%$ sure.

$\endgroup$
1
  • $\begingroup$ This indeed seems to be correct. You showed that the set is closed under all linear combinations as required. $\endgroup$
    – NDewolf
    May 4, 2018 at 21:59

2 Answers 2

1
$\begingroup$

Your solution is correct, as it shows that all linear combinations of a pair of elements are also an element of the set.

A faster attempt would be showing it directly:

$(0,0,0) \in B_1$, so $B_1$ is not empty.

Let $a = (a_1 - a_2,a_1,a_2) $ and $b = (b_1 - b_2,b_1,b_2)$ be elements of $B_1$, and $\alpha$, $\beta$ being some scalars.

Then $\alpha a + \beta b = (\alpha a_1 + \beta b_1 - \alpha a_2 - \beta b_2, \alpha a_1 + \beta b_1, \alpha a_2 + \beta b_2)$ which obviously sastifies the condition for a vector to be an element of $B_1$.

Therefore $B_1$ is a sub-space.

$\endgroup$
1
$\begingroup$

It is correct, but perhaps you can do it in a simpler way.

The map $f\colon\mathbb{C}^3\to\mathbb{C}$ defined by $$ f\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}=y_1-y_2+y_3 $$ is linear, because it is $$ f\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}= \begin{pmatrix} 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} $$ Your set $B_1$ is the null space (or kernel) of $f$, so it is a subspace.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .