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Math Help - Subspaces

  1. #1
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    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?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    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.
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  3. #3
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    Re: Subspaces

    Quote Originally Posted by FernandoRevilla View Post
    (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".
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