I am fairly familiar with how to multiply matrices and add/subtract them. This particular question, however, I am not sure about. How can I find the identity matrix whilst preserving congruence? Please explain.
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3Identity matrix is usually the matrix that makes no change to any vector when multiplying with it. Maybe you mean something else? – mathreadler Apr 17 '17 at 21:33
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1Reflection across $y=-2$ does not map zero vector to zero vector, so it is not linear map. – Ennar Apr 17 '17 at 21:38
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sometimes called affine... – Will Jagy Apr 17 '17 at 21:41
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It is a linear map, just not over $R^2$. – mathreadler Apr 17 '17 at 21:42
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I guess that you are looking for the reflection matrix (with $\text{det}=-1$) which transforms a point $(x,y)$ to $(x,y+2)$ if $y<2$ and takes $(x,y)$ to $(x,y-2)$ if $y>2$. – E Be Apr 17 '17 at 21:43
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Here is one way to construct it what you are asking for. We first need knowledge of what this block does on a vector:
$$\begin{bmatrix}1&a\\0&1\end{bmatrix}$$
So lets try it.
$$\begin{bmatrix}1&a\\0&1\end{bmatrix} \begin{bmatrix}v_1\\1\end{bmatrix} = \begin{bmatrix}v_1+a\\1\end{bmatrix}$$
So the block is an adder.
In the same way the following matrix will add 2*third element to second element: $$M_{A} = \left[\begin{array}{ccc}1&0&0\\0&1&2\\0&0&1\end{array}\right]$$
So we pad our vector with a 1 for third position:
\begin{bmatrix}v_1\\v_2\\1\end{bmatrix}
Now the matrix for reflection $y$ axis:
$$M_R = \begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix}$$
And a last to move back into place:
$${M_{A}}^{-1} = \left[\begin{array}{ccc}1&0&0\\0&1&-2\\0&0&1\end{array}\right]$$
All together:
$$ {M_A}^{-1} M_R M_A = \left[\begin{array}{ccc}1&0&0\\0&1&-2\\0&0&1\end{array}\right]\begin{bmatrix}1&0&0\\0&-1&0\\0&0&1\end{bmatrix}\left[\begin{array}{ccc}1&0&0\\0&1&2\\0&0&1\end{array}\right]$$
mathreadler
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