Image and Columns Procedure 1

Theorem

1. Fix bases $\alpha =\{v_1,\dots ,v_n\}$, $\beta =\{w_1,\dots ,w_k\}$ of $V,W$.
2. Then image $Image(T)$ is spanned by columns of $[T{]}_{\alpha }^{\beta }$ which contains basic variables of homogeneous system $[T{]}_{\alpha }^{\beta }[v{]}_{\alpha }=0$

Down to Earth Version

Consider any $p\in P_3(\mathbb{R})$ If we take its derivative, we get: $\frac{d}{dx}[p]$ $=\frac{d}{dx}[a_0+a_1\mathbf{x}+a_2{x}^2+a_3{x}^3]$ $=0+a_1+a_2\ast 2x+a_3\ast 3x^2$ $\in span\{D(x),D(x^2),D(x^3)\}$ $=span\{1,2x,3x^2\}$

Example

Where $D$ is the differentiation function, find a basis for the image $D:P_3(\mathbb{R})\to P_2(\mathbb{R})$

Soln

1. Set basis $P_3(\mathbb{R})=span\{1,x,x^2,x^3\}=span(\alpha )$
2. Set basis $P_2(\mathbb{R})=span\{1,x,x^2\}=span(\beta )$
3. Then, the matrix $[D{]}_{\alpha }^{\beta }$
   1. $D(1)=\frac{d}{dx}[1]=0\ast 1+0\ast x+0\ast x^2$
   2. $D(x)=\frac{d}{dx}[x]=1\ast 1+0\ast x+0\ast x^2$
   3. $D(x^2)=\frac{d}{dx}[x^2]=0\ast 1+2\ast x+0\ast x^2$
   4. $D(x^3)=\frac{d}{dx}[x^3]=0\ast 1+0\ast x+3\ast x^2$
4. This gives: $$[D]_{\alpha}^{\beta} = \left[\begin{array}{cccc|c} 0 & 1 & 0 & 0 & 0\ 0 & 0 & 2 & 0 & 0\ 0 & 0 & 0 & 3 & 0\ \end{array}\right]

5. Find the [[Basic and Free Variables|Basic]] solns of $[D]_{\alpha}^{\beta} [\overrightarrow{v}]_{\alpha} = \overrightarrow{0}$. We get: 1. $\{ D(x), D(x^{2}), D(x^{3})\} = \{ 1,2x,3x^{2} \}$