91影视

Consider two linear spaces Vand W. A function Tfrom Vto Wis called linear transformation if

1. $\mathit{T}\left(f+g\right)\mathbf{=}\mathit{T}\left(f\right)\mathbf{+}\mathit{T}\left(g\right)$
   
2. $\mathit{T}\left(kg\right)\mathbf{=}\mathit{k}\mathit{T}\left(f\right)$
   

for all elements f and g ofand for all scalars k.

If $\left(v\right)$is finite dimensional, then

$\mathit{d}\mathit{i}\mathit{m}\left(v\right)\mathbf{=}\mathit{r}\mathit{a}\mathit{n}\mathit{k}\left(T\right)\mathbf{+}\mathit{n}\mathit{u}\mathit{l}\mathit{l}\mathit{i}\mathit{t}\mathit{y}\left(T\right)\mathbf{=}\mathit{d}\mathit{i}\mathit{m}\left(imT\right)\mathbf{+}\mathit{d}\mathit{i}\mathit{m}\left(kerT\right)$

A subset Wof a linear space Vis called a subspace of Vif

1. W contains the neutral element 0of V.
2. W is closed under addition (if fand gare in Wthen so is $\mathit{f}\mathbf{+}\mathit{g}$)
3. W is closed under scalar multiplication (if fis in Wand kis scalar, then kfis in W).

Consider the linear transformation$T:P_n鈫�\mathrm{鈩�}$given by$T\left(f\left(t\right)\right)=f\left(0\right)$.

Let$z_n=\left\{f鈭�P_n:f\left(0\right)=\left(0\right)\right\}$.

Since$f鈭�ker\left(T\right)$if and only if $T\left(f\left(t\right)\right)=f\left(0\right)=0$.

We know that$ker\left(T\right)=z_n$ , thus$ker\left(T\right)$ is a subspace of $p_n$.

Therefore,$z_n$ is a subspace of $p_n$.

Given, that for any real numberwe have,

$T\left(t+a\right)=0+a=a$.

This implies $a鈭�im\left(T\right)$.

$thus,\mathrm{鈩�}鈯�im\left(T\right)$Thus, (i.e.) .

Hence,$rank\left(T\right)=dim\left(im\left(T\right)\right)=1$ .

Since,$dim\left(p_n\right)=n+1$ .

Then,

$$\begin{gathered} nullity\left(T\right)=dim\left(p_n\right)-rank\left(T\right) \\ =n+1-1 \\ =n \end{gathered}$$

Thus,$nullity\left(T\right)=dim\left(ker\left(T\right)\right)-DIM\left(Z_n\right)=n$