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

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

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

    Is a basis.

    Is that true and how can I show that?

    Furthermore is it true that the dimension of $\displaystyle U$ is $\displaystyle 4$?
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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)$\displaystyle R^{2x2}$ we have a subspace $\displaystyle U \subseteq R^{2x2} $ where U is the amount of symmetric matrices, this means that the matrix $\displaystyle A {2x2}$
    belongs to $\displaystyle U$ if and only if $\displaystyle A=A^T$

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

    Furthermore is it true that the dimension of $\displaystyle U$ is $\displaystyle 4$?
    No. 4 is the dimension of $\displaystyle \mathbb{R}^{2\times 2}$. As $\displaystyle U\subset\mathbb{R}^{2\times 2}$, saying that $\displaystyle \dim U = 4 = \dim \mathbb{R}^{2\times 2}$ amounts to saying that $\displaystyle U=\mathbb{R}^{2\times 2}$... It can't be true since not every $\displaystyle 2\times 2$ 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 ? $\displaystyle \mathbb{R}^{2\times 2}$ or $\displaystyle U$ ? (what you've done is correct if the answer to the former question is $\displaystyle \mathbb{R}^{2\times 2}$, otherwise it's wrong since $\displaystyle E_2\not \in U$)
    A basis for $\displaystyle U$. So I guess I have not found the proper basis.


    Quote Originally Posted by flyingsquirrel View Post
    No. 4 is the dimension of $\displaystyle \mathbb{R}^{2\times 2}$. As $\displaystyle U\subset\mathbb{R}^{2\times 2}$, saying that $\displaystyle \dim U = 4 = \dim \mathbb{R}^{2\times 2}$ amounts to saying that $\displaystyle U=\mathbb{R}^{2\times 2}$... It can't be true since not every $\displaystyle 2\times 2$ 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 $\displaystyle U$. So I guess I have not found the proper basis.
    Let $\displaystyle A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ be a (real) symmetric matrix. One has $\displaystyle A=A^T \Longleftrightarrow b=c$ hence every symmetric matrix looks like $\displaystyle A=\begin{pmatrix}a & b \\ b & d \end{pmatrix}$ where $\displaystyle a,b$ and $\displaystyle d$ 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 $\displaystyle U$ you've found ?
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  5. #5
    lo2
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    Quote Originally Posted by flyingsquirrel View Post
    Let $\displaystyle A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ be a (real) symmetric matrix. One has $\displaystyle A=A^T \Longleftrightarrow b=c$ hence every symmetric matrix looks like $\displaystyle A=\begin{pmatrix}a & b \\ b & d \end{pmatrix}$ where $\displaystyle a,b$ and $\displaystyle d$ are real numbers. That should help you find a basis.
    I guess:

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


    Quote Originally Posted by flyingsquirrel View Post
    How many "vectors" are there in the basis of $\displaystyle U$ 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:

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


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

    is a spanning set of $\displaystyle U$ (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 $\displaystyle span\{E_1, E_2, E_3\}=V$?
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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 $\displaystyle span\{E_1, E_2, E_3\}=V$?
    By showing that $\displaystyle U\subseteq S$ and $\displaystyle S\subseteq U$ where $\displaystyle S = \mathrm{span}\left(\left\{
    E_1=\left(\begin{smallmatrix}
    1 & 0\\
    0 & 0
    \end{smallmatrix}\right),
    E_2=\left(\begin{smallmatrix}
    0 & 1\\
    1 & 0
    \end{smallmatrix}\right),
    E_3=\left(\begin{smallmatrix}
    0 & 0\\
    0 & 1
    \end{smallmatrix}\right)\right\}
    \right)$.

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