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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?

3 Answers 3

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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 think you've got it. I hope it is conceptually, not "here are the mechanical steps that work."2012-12-09
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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.

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    @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.2012-12-09
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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.