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$T:M_{2*2}(R) \rightarrow M_{2*2}(R)$ defined by T( $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$= $\bigl( \begin{smallmatrix} a+b & a \\ c & c+d \end{smallmatrix} \bigr)$. Justify whether T is invertible.

Once I get that it is one-to-one, can I apply the theorem in here to argue it is also onto?

Theorem: Let V and W be vector spaces of equal dimension, and let $T: V \rightarrow W$ be linear. Then the following are equivalent. a). T is one-to-one, b). T is onto, c). $rank(T)=dim(V)$

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    $\begingroup$ Yes, you can.$ $ $\endgroup$
    – user1551
    Feb 11 '20 at 16:48
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As @user1551 mentioned in the comments, you may apply the stated theorem. Alternatively, one may observe that $$ T\begin{pmatrix} b & a-b\\ c & d-c \end{pmatrix} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}, $$

to see that $T$ is onto.

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This can also be seen directly from the definition of $T$, writing

$T \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = \begin{bmatrix} a + b & a \\ c & c + d \end{bmatrix} = \begin{bmatrix} x & y \\ z & w \end{bmatrix}, \tag 1$

comparing entries yields

$a + b = x, \tag 2$

$a = y, \tag 3$

$c = z, \tag 4$

$c + d = w; \tag 5$

thus,

$b = x - a = x - y, \tag 7$

$d = w - c = w - z; \tag 8$

therefore we may take

$T^{-1}\left (\begin{bmatrix} x & y \\ z & w \end{bmatrix} \right ) = \begin{bmatrix} y & x - y \\ z & w - z \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \tag 9$

which explicitly presents the inverse of $T$; indeed, we have from (9),

$T \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = T \left (\begin{bmatrix} y & x - y \\ z & w - z \end{bmatrix} \right )$ $= \begin{bmatrix} y + (x - y) & y \\ z & z + (w - z) \end{bmatrix} = \begin{bmatrix} x & y \\ z & w \end{bmatrix}. \tag{10}$

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