# Invertible matrix($T:M_{2*2}(R) \rightarrow M_{2*2}(R)$ defined by T( $\bigl( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)$

$$T:M_{2*2}(R) \rightarrow M_{2*2}(R)$$ 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). $$rank(T)=dim(V)$$

• Yes, you can. Feb 11 '20 at 16:48

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.

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),

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