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Thread: Find a basis and the dimension of the following subspaces:

  1. #1
    Newbie JCS007's Avatar
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    Find a basis and the dimension of the following subspaces:

    1). The space of solutions to the linear system Ax=0, where $\displaystyle A=\begin{pmatrix} 1&2&-1&1 \\ 3&0&2&-1\end{pmatrix}$

    2.) The set of all quadratic polynomials $\displaystyle p(x) = ax^2+bx+c$ that satisfy $\displaystyle p(1)=0$

    3.) The space of all solutions to the homogeneous ordinary differential equation $\displaystyle u^m-u^n+4u'-4u=0$
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by JCS007 View Post
    1). The space of solutions to the linear system Ax=0, where $\displaystyle A=\begin{pmatrix} 1&2&-1&1 \\ 3&0&2&-1\end{pmatrix}$

    2.) The set of all quadratic polynomials $\displaystyle p(x) = ax^2+bx+c$ that satisfy $\displaystyle p(1)=0$

    3.) The space of all solutions to the homogeneous ordinary differential equation $\displaystyle u^m-u^n+4u'-4u=0$
    1.) Suppose $\displaystyle
    x = \left( {\begin{array}{*{20}c}
    {x_1 } \\
    {x_2 } \\
    {x_3 } \\
    {x_4 } \\

    \end{array} } \right)
    $ thus: $\displaystyle
    A \cdot x = \left( {\begin{array}{*{20}c}
    1 \\
    3 \\

    \end{array} } \right) \cdot x_1 + \left( {\begin{array}{*{20}c}
    2 \\
    0 \\

    \end{array} } \right) \cdot x_2 + \left( {\begin{array}{*{20}c}
    { - 1} \\
    2 \\

    \end{array} } \right) \cdot x_3 + \left( {\begin{array}{*{20}c}
    1 \\
    { - 1} \\

    \end{array} } \right) \cdot x_4 = 0
    $

    We must have: $\displaystyle
    \left\{ \begin{gathered}
    x_1 + 2 \cdot x_2 - x_3 + x_4 = 0 \hfill \\
    3 \cdot x_1 + 2 \cdot x_3 - x_4 = 0 \hfill \\
    \end{gathered} \right.
    $ note there that we may take $\displaystyle x_1$ and $\displaystyle x_2$ as parameters and write $\displaystyle x_3$ and $\displaystyle x_4$ according to the values of them easily.

    Just summing both equations we get; $\displaystyle
    x_3 = - \left( {4 \cdot x_1 + 2 \cdot x_2 } \right)
    $ and: $\displaystyle
    x_4 = - \left( {5x_1 + 4 \cdot x_2 } \right)
    $

    Thus: $\displaystyle
    x = \left( {\begin{array}{*{20}c}
    {x_1 } \\
    {x_2 } \\
    { - \left( {4 \cdot x_1 + 2 \cdot x_2 } \right)} \\
    { - \left( {5x_1 + 4 \cdot x_2 } \right)} \\

    \end{array} } \right)
    $ , $\displaystyle
    x_1 ,x_2 \in \mathbb{R}
    $ are the parameters and note that this can be written as: $\displaystyle
    x = \left( {\begin{array}{*{20}c}
    {x_1 } \\
    {x_2 } \\
    { - \left( {4 \cdot x_1 + 2 \cdot x_2 } \right)} \\
    { - \left( {5x_1 + 4 \cdot x_2 } \right)} \\

    \end{array} } \right) = \left( {\begin{array}{*{20}c}
    1 \\
    0 \\
    { - 4} \\
    { - 5} \\

    \end{array} } \right) \cdot x_1 + \left( {\begin{array}{*{20}c}
    0 \\
    1 \\
    { - 2} \\
    { - 4} \\

    \end{array} } \right) \cdot x_2
    $ so any vector in that subspace is a linear combination of $\displaystyle
    \left( {\begin{array}{*{20}c}
    1 \\
    0 \\
    { - 4} \\
    { - 5} \\

    \end{array} } \right)
    $ and $\displaystyle
    \left( {\begin{array}{*{20}c}
    0 \\
    1 \\
    { - 2} \\
    { - 4} \\

    \end{array} } \right)
    $ (and they are LI) thus $\displaystyle
    B = \left\{ {\left( {\begin{array}{*{20}c}
    1 \\
    0 \\
    { - 4} \\
    { - 5} \\

    \end{array} } \right);\left( {\begin{array}{*{20}c}
    0 \\
    1 \\
    { - 2} \\
    { - 4} \\

    \end{array} } \right)} \right\}
    $ is a base and the dimension of the subspace is 2

    2.) Let $\displaystyle
    p\left( x \right) = a_2 \cdot x^2 + a_1 \cdot x + a_0
    $ that polynomial is in our set iff: $\displaystyle
    p\left( 1 \right) = a_2 + a_1 + a_0 = 0
    $ then $\displaystyle
    a_0 = - \left( {a_2 + a_1 } \right)
    $ consider $\displaystyle a_2$ and $\displaystyle a_1$ as parameters, then $\displaystyle
    p\left( x \right) = a_2 \cdot x^2 + a_1 \cdot x - \left( {a_2 + a_1 } \right)
    $ where $\displaystyle
    a_1 ,a_2 \in \mathbb{R}
    $ are our parameters

    $\displaystyle
    p\left( x \right) = a_2 \cdot \left( {x^2 - 1} \right) + a_1 \cdot \left( {x - 1} \right)
    $ thus any polynomial is in the subspace iff it is a combination of $\displaystyle
    {x^2 - 1}
    $ and $\displaystyle
    {x - 1}
    $ these 2 are LI, thus $\displaystyle
    B = \left\{ {x^2 - 1;x - 1} \right\}
    $ is a base of this subspace and its dimension is 2.
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