Inverse of linear tranformation Let $T$ be a linear transformation from $\mathbb R^n$ to $\mathbb R^n$.Let ${x_{1},x_{2},......,x_{n}}$ be a basis of $\mathbb R^n$ suppose that $Tx_{i}\neq{0}$. Then we say that that $\mathrm{null}(T)=0$ and $\mathrm{rank}(T)=n$. Thus $T$ is invertible. This is true.
Well, this is not true for $n&gt;1$.

Take e.g. $v\in\mathbb R^n\setminus\{0\}$ and define $T$ s.t. $$Tx_i=v\text{ for all }i\in\{1,\dots,n\}.$$.

Then $\mathrm{img}(T)=\mathrm{span}(v)\subsetneq\mathbb R^n$ and thus $\mathrm{rank}(T)=1&lt;n$ and followingly $\mathrm{null}(T)=n-\mathrm{rank}(T)=n-1&gt;0$.
consider T for ex. $ \begin{vmatrix}  A_1 &amp;  A_2 &amp;  A_3 \\ B_1 &amp; B_2 &amp;  B_3\\C_1 &amp; C_2 &amp;  C_3 \end{vmatrix}$, also consider the basis B $ \begin{vmatrix}  1 &amp;  0 &amp;  0 \\ 0 &amp; 1 &amp;  0\\0 &amp; 0 &amp;  1 \end{vmatrix}$, clearly $TB=T$, $Tx_i\ne0$ but column vectors of T can be completely in some 2d plane or 1d, for ex. they can lie completely in xy plane. So $TB$ is not a basis then. if $TB$ is a basis, then off course T is full rank. Let $TB=C$ so $T=CB^{-1}$ as $C,B^{-1}$ both are full rank then T is also full rank.