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

1. $\mathbf{T}\mathbf{(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{)}\mathbf{=}\mathbf{T}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{+}\mathbf{T}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$
2. $\mathbf{T}\mathbf{\left(}\mathbf{kf}\mathbf{\right)}\mathbf{=}\mathbf{kT}\mathbf{\left(}\mathbf{f}\mathbf{\right)}$

for all elements f and g of Vand for all scalars.

Ifis finite dimensional, then

$\mathbf{dim}\mathbf{\left(}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{rank}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{+}\mathbf{nullity}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{\left(}\mathbf{imT}\mathbf{\right)}\mathbf{+}\mathbf{dim}\mathbf{\left(}\mathbf{kerT}\mathbf{\right)}$

An invertible linear transformation Tis called an isomorphism.

Let V and W be two vector spaces.

Let$0_v$ and$0_w$ be the neutral elements V of W and respectively.

Consider the linear transformation $T:V鈫�W$.

Then, $T\left(0_v\right)+T\left(0_v\right)=T\left(0_v+0_v\right)=T\left(0_v\right)$.

Add$-T\left(0_v\right)$ on both sides of the above equation,

Then, $T\left(0_v\right)+T\left(0_v\right)+\left(-T\left(0_v\right)\right)=T\left(0_v+0_v\right)+\left(-T\left(0_v\right)\right)$.

Since,$\mathrm{T}\left(0_{\mathrm{v}}\right)鈭�\mathrm{W}$

$T\left(0_v\right)+\left(-T\left(0_v\right)\right)=0_w$

This implies, $T\left(0_v\right)+T\left(0_v\right)+\left(-T\left(0_v\right)\right)=T\left(0_v\right)+0_w=T\left(0_v\right)$.

Thus,$T\left(0_v\right)=0_W$

Given that T is an isomorphism, so multiplerole="math" localid="1659426688760" $T^{-1}$ on both sides

$T^{-1}\left(T\left(0_v\right)\right)=T^{-1}\left(0_W\right)$

Therefore, T is an isomorphism and $T^{-1}\left(0_w\right)=0_v$.