[Math] Visualizing transpose of matrices

linear algebramatricestranspose

I am a new linear-algebra student. I find linear algebra really difficult ,however, visualizing concepts really helps me out. So I stumbled upon rotation matrices and the book stated that the transpose of B times B must equal to the identity matrix. So are there any sources I could use to maybe view transposes for myself and kind of figure it out. And if not, I would greatly appreciate any help. Thanks!

Best Answer

I am not sure what your view is on what the transpose of a matrix is, so let me give my point of view first.

Suppose that $V$ is a real (finite dimensional) vector space with an inner product, $\langle \cdot, \cdot \rangle: V \times V \rightarrow \mathbb{R}$.

Now given two vectors $v,w \in V$, we can compute the inner product $\langle v, Aw \rangle$. One can now ask if there is a matrix $B$ (dependent on $A$), such that $\langle Bv , w \rangle = \langle v, Aw \rangle$ for all $v,w \in V$. This turns out to be the transpose of the matrix $A$, that is $B = A^{T}$. (Exercise: verify this...)

Given a vector $v \in V$, its length squared is given by $|v|^{2} =\langle v, v \rangle$. Given two vectors $v,w$, there is a relation between the cosine of the angle between them, $\theta$, and their inner product \begin{equation} \cos(\theta) = \frac{\langle v, w \rangle}{|v||w|}. \end{equation}

Now, a rotation is a linear map $A:V \rightarrow V$, that preserves angles and preserves lengths. That is, for any $v \in V$ we want \begin{equation} |Av|^{2} = |v|^{2}, \quad \Longleftrightarrow \quad \langle Av, Av \rangle = \langle v, v \rangle. \end{equation} Now if $v,w \in V$ let $\theta$ be the angle between $v$ and $w$, then we want the angle between $Av$ and $Aw$ to be $\theta$ as well, or in other words \begin{equation} \frac{\langle v, w \rangle}{|v||w|} = \cos(\theta) = \frac{\langle Av, Aw\rangle}{|Av||Aw|} = \frac{\langle Av, Aw \rangle}{|v||w|}. \end{equation} We conclude that we want $\langle v, w \rangle = \langle Av, Aw \rangle$ to hold for all $v$ and $w$. We can rewrite this using the transpose \begin{equation} \langle v, w \rangle = \langle Av, Aw \rangle = \langle v, A^{T} A w \rangle. \end{equation} Since this is supposed to be true for all $v,w \in V$ we may conclude that $A^{T}A = \text{Id}$.

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