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Geometrical meaning of $\tau: \mathbb{R^2} \rightarrow \mathbb{R^{2,2}}, \begin{pmatrix} a\\ b \end{pmatrix} \mapsto \begin{pmatrix} a&-b\\ b&a \end{pmatrix}$

I have troubles of reading matrices and interpreting them. How can you interpret them? Here my best guess is, that a vector is mapped into a higherdimensional space. In the image, do I have to look at the columns to see the vectors, does that even make sense or any of the images below?

EDIT: enter image description here

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    See also [this question](http://math.stackexchange.com/questions/1128528/form-of-matrix-for-reflection-about-a-line). – 2017-01-20
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    You've drawn the green vector as $(a,-b)$, it should be $(-b,a)$. – 2017-01-20

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First, note that your pictures don't actually convey the correct column vectors - the second vector should have $-b$ as its $x$-component, not its $y$.

But this isn't a very useful way of visualizing the situation anyway; a matrix isn't just an assemblage of vectors. The "right" way to think about matrices is as a transformation. A $2\times 2$ matrix is a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. In this case, the particular matrix you describe has a very nice interpretation as a transformation: it's a rotation and a scaling together. To be specific, the matrix $\left(\begin{array}{cc}\cos(\theta) & -\sin(\theta)\\\sin(\theta) & \cos(\theta)\end{array}\right)$ is a counterclockwise rotation by the angle $\theta$. Conveniently, the vector $(\cos(\theta), \sin(\theta))$ is a unit vector inclined at an angle $\theta$ from the $x$-axis. So if $\mathbf{u}$ is a unit vector, $\tau(\mathbf{u})$ is the transformation of rotating counterclockwise by the angle $\mathbf{u}$ makes with the $x$-axis.

If $\mathbf{a}$ is not a unit vector, then it's $||\mathbf{a}||\mathbf{u}$ for some unit vector $\mathbf{u}$. $\tau(\mathbf{a})$ turns out to be $||\mathbf{a}||\tau(\mathbf{u})$ (I'll let you check that). So we get a scaling by a factor of $||\mathbf{a}||$ as well.

For example, if $\mathbf{a} = (1,1)$ (so $a = b = 1$), then $\tau(\mathbf{a})$ is a counterclockwise rotation by $45^{\circ}$, followed by a stretching by a factor of $\sqrt{2}$.

Note: above, I've written vectors in the format $(a,b)$ rather than as column vectors; that's just for my convenience.

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    So $A := \begin{pmatrix} a&-b\\ b&a \end{pmatrix}$ is a transformation of any vector you put in, so you should actually get another Vector out of it. I find it confusing that we have a function with$\mathbb{R^2} \rightarrow \mathbb{R^{2,2}}$, instead of $\mathbb{R^2} \rightarrow \mathbb{R^2}$. Also, shouldn't there be a Matrix with $B\cdot v$ so that we get the same image? – 2017-01-20
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    @WhatAMesh Yes, $A$ is a transformation that *would* yield another vector, but we're not *giving* it a vector. $A$ is itself the output of $\tau$. Think of it like a cookbook - the pages are recipes, and therefore should yield meals, but $\tau$ is just giving you the page numbers. And I don't know what you mean by "a matrix with $B\cdot v$ so that we get the same image". – 2017-01-20
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You drew your vector wrong. The green vector should have x-coordinate -b and y-coordinate a. If you remember from algebra class that if a line has slope m, then a perpendicular line has slope -1/m. Therefore it is a mapping of a vector to itself and a perpendicular vector.

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The space $\mathbb{R}^{2,2}$ is isomorphic (as a vector space over $\mathbb{R}$) to $\mathbb{R}^4$, and there are no simple way to graphically represent this four dimensional space.

You can interpret geometrically the matrix as the linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$, that transforms the basis vector $[1,0]^T$ to the vector $[a,b]^T$ and $[0,1]^T$ to $[-b,a]$, and note that these two vectors are orthogonal.