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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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up vote 1 down vote accepted

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? – CodyBugstein 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 $2 \times 2$? – CodyBugstein 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

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.

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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? – CodyBugstein 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

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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