On determining the angle between two matrices using the Euclidean Inner Product

linear algebramatrices

So I was bored and decided to teach myself about matrices for absolutely no reason, and found out about the Euclidean Inner Product. Assuming matrices $A_{2,2}$ and $B_{2,2}$, the angle between those two matrices would be calculated as$$\theta=\arccos\left(\dfrac{A\cdot B}{|A||B|}\right)$$where $|A|$ and $|B|$ are the magnitudes of matrices $A$ and $B$. Now, assuming that the two matrices $A_{2,2}$ and $B_{2,2}$ are defined as$$A=\begin{bmatrix}a_1&b_1\\ c_1&d_1\end{bmatrix}$$$$B=\begin{bmatrix}a_2&b_2\\ c_2&d_2\end{bmatrix}$$we then define the formula for finding the angle between the two matrices as$$\theta=\arccos\left(\dfrac{a_1a_2+b_1c_2+a_1b_2+b_1d_2+c_1a_2+d_1c_2+c_1b_2+d_1d_2}{\sqrt{a_1^2+b_1^2+c_1^2+d_1^2}\sqrt{a_2^2+b_2^2+c_2^2+d_2^2}}\right)$$However, something to note here is that this is in degrees. To do this formula in radians, the new formula would be$$\theta=\arccos\left(\dfrac{180(a_1a_2+b_1c_2+a_1b_2+b_1d_2+c_1a_2+d_1c_2+c_1b_2+d_1d_2)}{\pi\sqrt{a_1^2+b_1^2+c_1^2+d_1^2}\sqrt{a_2^2+b_2^2+c_2^2+d_2^2}}\right)$$Now say I have two $2$ by $2$ matrices,$$A=\begin{bmatrix}2&3\\ 1&2\end{bmatrix}$$$$B=\begin{bmatrix}3&2\\ 1&2\end{bmatrix}$$Then, the angle between them would be calculated as$$\theta=\arccos\left(\dfrac{6+3+4+6+3+2+2+4}{\sqrt{4+9+1+4}\sqrt{9+4+1+4}}\right)$$$$\theta=\arccos\left(\dfrac{30}{\sqrt{18}\sqrt{18}}\right)$$$$\theta=\arccos\left(\dfrac{30}{18}\right)$$$$\theta=\arccos\left(\dfrac{5}{3}\right)$$$$\theta=i\ln(3)\quad\text{in both degrees and radians}$$

My question

Would my understanding of the Euclidean Inner Product be correct, or what could I do to understand it more easily?

To clarify

  1. How I figured out that the angle of the two matrices would be calculated as $\theta=\arccos\left(\dfrac{A\cdot B}{|A||B|}\right)$ is (from what I understand) because two vectors being multiplied together would be calculated as $|\overrightarrow i||\overrightarrow j||\cos\theta$, where $\cos\theta$ is the angle between the $2$ vectors, and I have a feeling that it probably would be similar when it comes to determining the angle between two matrices.

Best Answer

A fix to my formula:

we then define the formula for finding the angle between the two matrices as$$\theta=\arccos\left(\dfrac{a_1a_2+b_1c_2+a_1b_2+b_1d_2+c_1a_2+d_1c_2+c_1b_2+d_1d_2}{\sqrt{a_1^2+b_1^2+c_1^2+d_1^2}\sqrt{a_2^2+b_2^2+c_2^2+d_2^2}}\right)$$However, something to note here is that this is in degrees. To do this formula in radians, the new formula would be$$\theta=\arccos\left(\dfrac{180(a_1a_2+b_1c_2+a_1b_2+b_1d_2+c_1a_2+d_1c_2+c_1b_2+d_1d_2)}{\pi\sqrt{a_1^2+b_1^2+c_1^2+d_1^2}\sqrt{a_2^2+b_2^2+c_2^2+d_2^2}}\right)$$

As pointed out by @$0$XLR:

If you want to change angles, you should be multiplying the $\dfrac{180}\pi$ outside the $\arccos$[,] otherwise you will be doing $\arccos\left(\text{"some bizzare value greater than }1\text{"}\right)$ and get complex values.

And as pointed out by @mildbison:

Note that for any inner product space, you would have $|A\cdot B|\leq|A||B|$ which is known as the Cauchy-Schwarz inequality. Importantly, you would see that in your particular example that this inequality is violated. I do not know much about the example you have given, but you should probably not be multiplying by $\dfrac{180}\pi$. Essentially, the formula for the inner product is treating the matrix as a regular $4$ vector, so you should probably stick to the formula for that inner product.

Note the fact that I should be treating the matrices as a regular $4$ vector as mildbison pointed out. Therefore, the formula should be$$\theta=\arccos\left(\dfrac{a_1a_2+b_1b_2+c_1c_2+d_1d_2}{\sqrt{a_1^2+b_1^2+c_1^2+d_1^2}\sqrt{a_2^2+b_2^2+c_2^2+d_2^2}}\right)$$for calculating this in degrees, and in radians,$$\theta=\dfrac{180}\pi\arccos\left(\dfrac{a_1a_2+b_1b_2+c_1c_2+d_1d_2}{\sqrt{a_1^2+b_1^2+c_1^2+d_1^2}\sqrt{a_2^2+b_2^2+c_2^2+d_2^2}}\right)$$And, when we plug in my example, we get$$\theta=\arccos\left(\dfrac{6+6+1+4}{18}\right)$$$$=\arccos\left(\dfrac{17}{18}\right)\approx19.19\unicode{xB0}$$And in radians,$$\theta\approx0.335\text{rad}$$

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