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

• Jun 23rd 2008, 04:11 PM
JCS007
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 $u^m-u^n+4u'-4u=0$
• Jun 23rd 2008, 06:35 PM
PaulRS
Quote:

Originally Posted by JCS007
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 $
x = \left( {\begin{array}{*{20}c}
{x_1 } \\
{x_2 } \\
{x_3 } \\
{x_4 } \\

\end{array} } \right)
$
thus: $
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: $
\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 $x_1$ 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; $
x_3 = - \left( {4 \cdot x_1 + 2 \cdot x_2 } \right)
$
and: $
x_4 = - \left( {5x_1 + 4 \cdot x_2 } \right)
$

Thus: $
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)
$
, $
x_1 ,x_2 \in \mathbb{R}
$
are the parameters and note that this can be written as: $
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 $
\left( {\begin{array}{*{20}c}
1 \\
0 \\
{ - 4} \\
{ - 5} \\

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

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

$
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 $
{x^2 - 1}
$
and $
{x - 1}
$
these 2 are LI, thus $
B = \left\{ {x^2 - 1;x - 1} \right\}
$
is a base of this subspace and its dimension is 2.