# Is this an invertible linear transformation?

Suppose you have a linear transformation $T: M_{2\times 2}\to M_{2\times 2}$ given by

$$\begin{pmatrix} a & b \\ c & d\end{pmatrix}\mapsto \begin{pmatrix} a+b & a \\ c & c+d\end{pmatrix}$$

How can I tell quickly, if it is invertible?

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Hint: 1) You need to check the linear transformation part.

2) If someone gives you an arbitrary matrix, can you uniquely identify the matrix that it came from?

For example, let $$B=\begin{pmatrix} 12 & -3 \\ 4 & \pi\end{pmatrix}.$$ What would be the only $A$ such that $T(A)=B$? Note that "$a"$ must be $-3$, and $c$ must be $4$. Thus $b$ must be $12-(-3)$ and $d$ must be $\pi -4$.

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Yes I suppose working backwards you could. For example, if I gave the random matrix $\begin{pmatrix}1 & 2 \\ 4 & 1\end{pmatrix}$ I could figure out $a=2$ and $c=1$ etc.. and determine the original input matrix. Is that right? If yes, is there a more general way to put it, for more complicated transformations? –  Imray Dec 9 '12 at 18:29
For a more complicated transformation, I would go to the $4\times 4$ matrix of the transformation, and use the usual tools (row reduction, or maybe determinant). The vector space of $2\times 2$ matrices under usual addition of matrices is nothing mysterious, just write the matrix "flat" as $(a, b, c, d)$. Same is true of the space of $m\times n$ matrices. –  André Nicolas Dec 9 '12 at 18:35
What do you mean by $4 \times 4$ matrix of the transformation? Isn't it $2 \times 2$? –  Imray Dec 9 '12 at 18:37
Because writing matrices is a nuisance, I will write them "flat" which may luckily help with the understanding. Your transformation $T$ takes $(a,b,c,d)$ to $(a+b,a,c,c+d)$. Now can you find a $\times 4$ matrix that does that? Probably you know how to do this. First row of the $4\times 4$: $1,1,0,0$. Second row: $1,0,0,0$. Third row: $0,0,1,0$. Last row: $0,0,1,1$. –  André Nicolas Dec 9 '12 at 18:45
Yes, I think you've got it. I hope it is conceptually, not "here are the mechanical steps that work." –  André Nicolas Dec 9 '12 at 19:20

Since the domain and codomain are of same dimension, it is enough if u can check that it has a trivial kernel. That will do.

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See if it's possible to find two elements of $M_{2\times2}$ that map to the same thing. If you can, it's not invertible, if not then explain why not rigorously.
Alternatively, $M_{2 \times2}$is isomorphic to $\mathbb{R^4}$ so we could write the map as a $4\times4$ matrix and see if it's invertible or not.
What do you mean by "find two elements that map to the same thing"? Do you mean take two arbitrary matrices in place of $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$, apply $T$ and see if the result is the same? Also, can you please explain the "Alternatively" a little further? –  Imray Dec 9 '12 at 18:21
@Imray A matrix is not invertible iff it has non trivial kernel. In this case that would mean that there are non-zero matrices that map to $0$, and hence there are non-identical matrices where $T\begin{pmatrix} a & b \\ c & d \end{pmatrix} = T\begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}$. For the second part, any real vector space of finite dimension n is isomorphic to \mathbb{R^n}. If we think of how $T$ acts on $M_{2 \times 2}$ when we picture $M_{2 \times 2}$ as $\mathbb{R^4}$ we can write $T$ in matrix form, and see if this matrix is invertible. –  Tom Oldfield Dec 9 '12 at 18:29