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Math Help - 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 A=\begin{pmatrix} 1&2&-1&1 \\ 3&0&2&-1\end{pmatrix}

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

    3.) The space of all solutions to the homogeneous ordinary differential equation 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 A=\begin{pmatrix} 1&2&-1&1 \\ 3&0&2&-1\end{pmatrix}

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

    3.) The space of all solutions to the homogeneous ordinary differential equation u^m-u^n+4u'-4u=0
    1.) Suppose <br />
x = \left( {\begin{array}{*{20}c}<br />
   {x_1 }  \\<br />
   {x_2 }  \\<br />
   {x_3 }  \\<br />
   {x_4 }  \\<br /> <br />
 \end{array} } \right)<br />
thus: <br />
A \cdot x = \left( {\begin{array}{*{20}c}<br />
   1  \\<br />
   3  \\<br /> <br />
 \end{array} } \right) \cdot x_1  + \left( {\begin{array}{*{20}c}<br />
   2  \\<br />
   0  \\<br /> <br />
 \end{array} } \right) \cdot x_2  + \left( {\begin{array}{*{20}c}<br />
   { - 1}  \\<br />
   2  \\<br /> <br />
 \end{array} } \right) \cdot x_3  + \left( {\begin{array}{*{20}c}<br />
   1  \\<br />
   { - 1}  \\<br /> <br />
 \end{array} } \right) \cdot x_4  = 0<br />

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

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

    Thus: <br />
x = \left( {\begin{array}{*{20}c}<br />
   {x_1 }  \\<br />
   {x_2 }  \\<br />
   { - \left( {4 \cdot x_1  + 2 \cdot x_2 } \right)}  \\<br />
   { - \left( {5x_1  + 4 \cdot x_2 } \right)}  \\<br /> <br />
 \end{array} } \right)<br />
, <br />
x_1 ,x_2  \in \mathbb{R}<br />
are the parameters and note that this can be written as: <br />
x = \left( {\begin{array}{*{20}c}<br />
   {x_1 }  \\<br />
   {x_2 }  \\<br />
   { - \left( {4 \cdot x_1  + 2 \cdot x_2 } \right)}  \\<br />
   { - \left( {5x_1  + 4 \cdot x_2 } \right)}  \\<br /> <br />
 \end{array} } \right) = \left( {\begin{array}{*{20}c}<br />
   1  \\<br />
   0  \\<br />
   { - 4}  \\<br />
   { - 5}  \\<br /> <br />
 \end{array} } \right) \cdot x_1  + \left( {\begin{array}{*{20}c}<br />
   0  \\<br />
   1  \\<br />
   { - 2}  \\<br />
   { - 4}  \\<br /> <br />
 \end{array} } \right) \cdot x_2 <br />
so any vector in that subspace is a linear combination of <br />
\left( {\begin{array}{*{20}c}<br />
   1  \\<br />
   0  \\<br />
   { - 4}  \\<br />
   { - 5}  \\<br /> <br />
 \end{array} } \right)<br />
and <br />
\left( {\begin{array}{*{20}c}<br />
   0  \\<br />
   1  \\<br />
   { - 2}  \\<br />
   { - 4}  \\<br /> <br />
 \end{array} } \right)<br />
(and they are LI) thus <br />
B = \left\{ {\left( {\begin{array}{*{20}c}<br />
   1  \\<br />
   0  \\<br />
   { - 4}  \\<br />
   { - 5}  \\<br /> <br />
 \end{array} } \right);\left( {\begin{array}{*{20}c}<br />
   0  \\<br />
   1  \\<br />
   { - 2}  \\<br />
   { - 4}  \\<br /> <br />
 \end{array} } \right)} \right\}<br />
is a base and the dimension of the subspace is 2

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

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