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Math Help - How to find a basis for this

  1. #1
    lo2
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    How to find a basis for this

    We have a vectorspace (is it a vectorspace) R^{2x2} we have a subspace U \subseteq R^{2x2} where U is the amount of symmetric matrices, this means that the matrix A {2x2}
    belongs to U if and only if A=A^T

    Then I have to find a basis. I would say that:

    E_1=\begin{bmatrix}1&0 \\ 0 & 0 \end{bmatrix} , E_2=\begin{bmatrix}0&1 \\ 0 & 0 \end{bmatrix} , E_3=\begin{bmatrix}0&0 \\ 1 & 0 \end{bmatrix} , E_4=\begin{bmatrix}0&0 \\ 0 & 1 \end{bmatrix}

    Is a basis.

    Is that true and how can I show that?

    Furthermore is it true that the dimension of U is 4?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by lo2 View Post
    We have a vectorspace (is it a vectorspace) R^{2x2} we have a subspace U \subseteq R^{2x2} where U is the amount of symmetric matrices, this means that the matrix A {2x2}
    belongs to U if and only if A=A^T

    Then I have to find a basis.
    A basis of which vector space ? \mathbb{R}^{2\times 2} or U ? (what you've done is correct if the answer to the former question is \mathbb{R}^{2\times 2}, otherwise it's wrong since E_2\not \in U)

    Furthermore is it true that the dimension of U is 4?
    No. 4 is the dimension of \mathbb{R}^{2\times 2}. As U\subset\mathbb{R}^{2\times 2}, saying that \dim U = 4 = \dim \mathbb{R}^{2\times 2} amounts to saying that U=\mathbb{R}^{2\times 2}... It can't be true since not every 2\times 2 matrix is symmetric.
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  3. #3
    lo2
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    Quote Originally Posted by flyingsquirrel View Post
    Hi,

    A basis of which vector space ? \mathbb{R}^{2\times 2} or U ? (what you've done is correct if the answer to the former question is \mathbb{R}^{2\times 2}, otherwise it's wrong since E_2\not \in U)
    A basis for U. So I guess I have not found the proper basis.


    Quote Originally Posted by flyingsquirrel View Post
    No. 4 is the dimension of \mathbb{R}^{2\times 2}. As U\subset\mathbb{R}^{2\times 2}, saying that \dim U = 4 = \dim \mathbb{R}^{2\times 2} amounts to saying that U=\mathbb{R}^{2\times 2}... It can't be true since not every 2\times 2 matrix is symmetric.
    Yeah that makes sense. I can see that. But what is the dimension then?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by lo2 View Post
    A basis for U. So I guess I have not found the proper basis.
    Let A=\begin{pmatrix}a & b \\ c & d \end{pmatrix} be a (real) symmetric matrix. One has A=A^T \Longleftrightarrow b=c hence every symmetric matrix looks like A=\begin{pmatrix}a & b \\ b & d \end{pmatrix} where a,b and d are real numbers. That should help you find a basis.

    But what is the dimension then?
    How many "vectors" are there in the basis of U you've found ?
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  5. #5
    lo2
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    Quote Originally Posted by flyingsquirrel View Post
    Let A=\begin{pmatrix}a & b \\ c & d \end{pmatrix} be a (real) symmetric matrix. One has A=A^T \Longleftrightarrow b=c hence every symmetric matrix looks like A=\begin{pmatrix}a & b \\ b & d \end{pmatrix} where a,b and d are real numbers. That should help you find a basis.
    I guess:

    \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}


    Quote Originally Posted by flyingsquirrel View Post
    How many "vectors" are there in the basis of U you've found ?
    3.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by lo2 View Post
    I guess:

    \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}
    That's it. Now to show that it is a basis you have to say that the three matrices are linearly independent and you have to show that


    \left\{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}\right\}

    is a spanning set of U (that's more or less what I did in my previous post).

    3.
    I agree.
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  7. #7
    lo2
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    Thank you for your help!

    But how can you show that span\{E_1, E_2, E_3\}=V?
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by lo2 View Post
    But how can you show that span\{E_1, E_2, E_3\}=V?
    By showing that U\subseteq S and S\subseteq U where S = \mathrm{span}\left(\left\{<br />
E_1=\left(\begin{smallmatrix}<br />
1 & 0\\<br />
0 & 0<br />
\end{smallmatrix}\right),<br />
E_2=\left(\begin{smallmatrix}<br />
0 & 1\\<br />
1 & 0<br />
\end{smallmatrix}\right),<br />
E_3=\left(\begin{smallmatrix}<br />
0 & 0\\<br />
0 & 1<br />
\end{smallmatrix}\right)\right\}<br />
\right).

    Remember that every real, symmetric matrix can be written as \left(\begin{smallmatrix}a & b \\ b & d \end{smallmatrix}\right) where a,b and d are real numbers.
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