# Visualizing transpose of matrices

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!

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 $$\cos(\theta) = \frac{\langle v, w \rangle}{|v||w|}.$$

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 $$|Av|^{2} = |v|^{2}, \quad \Longleftrightarrow \quad \langle Av, Av \rangle = \langle v, v \rangle.$$ 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 $$\frac{\langle v, w \rangle}{|v||w|} = \cos(\theta) = \frac{\langle Av, Aw\rangle}{|Av||Aw|} = \frac{\langle Av, Aw \rangle}{|v||w|}.$$ 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 $$\langle v, w \rangle = \langle Av, Aw \rangle = \langle v, A^{T} A w \rangle.$$ Since this is supposed to be true for all $v,w \in V$ we may conclude that $A^{T}A = \text{Id}$.

You know that the determinant of a rotation matrix is one by definition and you know that the product of a matrix with its inverse gives the identity. Making explicit the definition of the inverse of a matrix (the transpose of the adjoint matrix divided by its determinant) you can easily see what you don't see: that the inverse of the rotation matrix coincides with its transpose.

Maybe this helps:

• If a transformation is given by a matrix $B$, then the inverse transformation is given by the inverse matrix $B^{-1}$. Performing a transformation and its inverse leaves all vectors unchanged, you have $BB^{-1} = B^{-1}B = I$ with $I$ the identity matrix.
• Some special transformations have the property that their inverse matrix is the same as the transpose, i.e. $B^{-1}=B^T$. These transformation (and corresponding matrices) are called orthogonal and thus have the property $BB^T = B^TB = I$. Their determinant is $\pm 1$.
• Rotations are examples of such orthogonal transformations, you could for example verify that the determinant is indeed $1$. This means that for rotations, the matrix transpose is the same as the inverse, which also means that the composition of the matrix with its transpose simply gives you the identity.

So in summary: $BB^T = B^TB = I$ is not true in general, but it is for rotations (and all other so called orthogonal transformations), because $B^T = B^{-1}$ in that case.