# Math Help - Subspaces

1. ## Subspaces

Hi there. I started learning about subspaces in linear algebra and I came across a question which i'm unsure how to solve. I understand that there are 'rules' which need to be passed in order for something to be a subspace, but I have no idea how to start with this problem:

Consider the set M23 of all 2 × 3 matrices with real entries under the usual operations of matrix addition and scalar multiplication.
Let
T=([a b c] : a + c= 0 and b + d + f =0)
[d e f]
Prove that T is a subspace of M23
(T is a 2x3 matrix if i made it unclear)
I know that T must contain a zero vector and I know that there must be closer of scalar multiplication and addition.

Can anyone help?

2. ## Re: Subspaces

(i) $\begin{pmatrix}{0}&{0}&{0}\\{0}&{0}&{0} \end{pmatrix}\in T$ because in this case $a+c=0+0=0$ and $b+d+f=0+0+0=0$

(ii) If $A=\begin{pmatrix}{a}&{b}&{c}\\{d}&{e}&{f} \end{pmatrix}\in T$ and $A'=\begin{pmatrix}{a'}&{b'}&{c'}\\{d'}&{e'}&{f'} \end{pmatrix}\in T$ then $a+c=0,\;b+d+f=0$ and $a'+c'=0,\;b'+d'+f'=0$ . Now, find $A+A'$ and verify the conditions for belonging to $T$ .

(iii) ...

Second method: The elements of $T$ are of the form $\begin{pmatrix}{-c}&{-d-f}&{c}\\{d}&{e}&{f} \end{pmatrix}$ . This means that we can express $T$ in the form $T=L[S]$ with $S\subset T$ which is a subspace by a well known theorem.

3. ## Re: Subspaces

Originally Posted by FernandoRevilla
(i) $\begin{pmatrix}{0}&{0}&{0}\\{0}&{0}&{0} \end{pmatrix}\in T$ because in this case $a+c=0+0=0$ and $b+d+f=0+0+0=0$

(ii) If $A=\begin{pmatrix}{a}&{b}&{c}\\{d}&{e}&{f} \end{pmatrix}\in T$ and $A'=\begin{pmatrix}{a'}&{b'}&{c'}\\{d'}&{e'}&{f'} \end{pmatrix}\in T$ then $a+c=0,\;b+d+f=0$ and $a'+c'=0,\;b'+d'+f'=0$ . Now, find $A+A'$ and verify the conditions for belonging to $T$ .

(iii) ...

Second method: The elements of $T$ are of the form $\begin{pmatrix}{-c}&{-d-f}&{c}\\{d}&{e}&{f} \end{pmatrix}$ . This means that we can express $T$ in the form $T=L[S]$ with $S\subset T$ which is a subspace by a well known theorem.
Indeed, you can write each member of this subset as
$\begin{pmatrix}-c & -d- f & c \\ d & e & f \end{pmatrix}= c\begin{pmatrix}-1 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}+ d\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0\end{pmatrix}+ e\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}+ f\begin{pmatrix}0 & -1 & 0 \\ 0 & 0 & 1\end{pmatrix}$
so this is specifically the subspace spanned by those four "vectors".