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

1. ## 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 $\displaystyle u^m-u^n+4u'-4u=0$

2. Originally Posted by JCS007
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.