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This question was asked in an examination a while back.

I was able to solve this question but the computation required was too much. The solution said that the trick to solving this lies in the fact that the product of $P$ with its Transpose is an Identity matrix.(I've come to learn that this is known as an orthogonal matrix).

My question is, how can I know something like this while solving the question. Given any Matrix $A$ , is it possible to identify whether it is an orthogonal matrix?

PS:I am still in high school, so please use simple words to explain. Thank you. enter image description here

Karan Singh
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2 Answers2

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Identifying an orthogonal matrix is fairly easy: a matrix is orthogonal if and only if its columns (or equivalently, rows) form an orthonormal basis. A set of vectors $\{v_1, \ldots, v_n\}$ is said to be an orthonormal basis if $v_i \cdot v_i = 1$ for all $i$ and $v_i \cdot v_j = 0$ for all $i \neq j$.

(If you want to see why this is true, think about what it means to say that $P^TP = I$ in terms of the columns of $P$.)

We can check this for the matrix $P$ you gave: $$P = \begin{pmatrix} \sqrt{3}/2 & 1/2 \\ -1/2 & \sqrt{3}/2 \end{pmatrix}$$ We have two columns: $$e_1 = {\sqrt{3}/2 \choose -1/2} \quad e_2 = {1/2 \choose \sqrt{3}/2}$$ with $e_1 \cdot e_1 = 3/4 + 1/4 = 1$, $e_2 \cdot e_2 = 1$ and $e_1 \cdot e_2 = \sqrt{3}/4 - \sqrt{3}/4 = 0$, so $P$ is indeed orthogonal.

Of course, another way of checking is just to multiply $P^T$ by $P$ and see if you get $I$!

In the context of the question, how would you go about solving it? If you write out in full what it's asking you to do, you see it's asking you to compute $P^T(PAP^T)^{2005}P$, which is a bit terrifying. But it's hoping that you'll notice that if you start expanding this out you get $P^TPAP^TPAP^TPA\ldots P^TP$.

It's normally reasonable to assume that the people who set exams aren't completely evil, and that they expect you to solve the question, and that multiplying out all these matrices is an unreasonable expectation in the time you're given. So we notice that $P^TPA$ occurs again and again, and we might hope that it's something nice.

When you multiply $P^TPA$ out, you get $A$ itself back! (Because $P$ is orthogonal, as we just showed.) So the question reduces to calculating $A^{2005}$. This still seems a little unreasonable to do directly, so we are missing another trick. If we start working out $A, A^2, A^3, \ldots$ we notice that $$A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$$ This is straightforward to prove using induction, but from the looks of the question it's just a multiple choice so don't need to.

Josh Hunt
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  • I am not sure what you mean by orthonormal basis but I assume I just have to take the dot product of each column with itself , and if it equals 1 then they are orthonormal? – Karan Singh May 02 '16 at 05:27
  • Sorry, the sentence following that (starting "That is") was intended to explain what an orthonormal basis is.

    You have to take the dot product of each column with itself and it has to equal 1, and also take the dot product of each column with all the other columns and it has to equal 0. If all of that holds then they are orthonormal. I'll revise my answer to make that a little clearer.

     – Josh Hunt May 02 '16 at 07:17
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Notice that $P$ is in the form of a rotation matrix. That is, $P$ has entries of the form

$$ P = \begin{bmatrix} \sqrt{3}/2 & 1/2 \\-1/2 & \sqrt3/2 \end{bmatrix} = \begin{bmatrix} \sin(60^\circ) & \cos(60^\circ) \\ -\cos(60^\circ) & \sin(60^\circ) \end{bmatrix} $$

This is known to be an orthogonal matrix. That's because orthogonal matrix are exactly those matrices that preserve distances in that space (I can show you the proof if you want), and rotation preserve distances in $\mathbb{R}^2$.

There's no surefire way to find out if a matrix is orthogonal. However, another way to characterize orthogonal matrices is to see that if you think of the matrix as vectors along columns,

$$P = \begin{bmatrix} \sin(60^\circ) \\ -\cos(60^\circ) \end{bmatrix} : \begin{bmatrix} \cos(60^\circ) \\ \sin(60^\circ) \end{bmatrix} $$

Consider the columns as vectors:

$$ v_1 = (\sin 60^\circ, -\cos 60^\circ) \\ v_2 = (\cos 60^\circ, \sin 60^\circ) $$

Now, notice that the matrix decomposed as vectors has orthonormal vectors - that is, the vectors have length 1 (you can check this by taking the dot product with itself). However, the dot product with any other vector is $0$.

I've done some partial computation here. You can check it for the other vectors as well $$v1 \cdot v1 = \sin^2 60^\circ + \cos^2 60^\circ = 1 \\ v1 \cdot v2 = \sin 60^\circ \cdot \cos 60^\circ - \cos 60^\circ \cdot \sin 60^\circ = 0$$

This is another way to characterize an orthogonal matrix - the matrix whose columns are pairwise orthogonal. This acts as a quick test to check if the matrix is orthogonal or not.

V.G
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Siddharth Bhat
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  • What is the relation between vectors and matrices? I am still in class 12 so I would be glad if you could explain me that in simple words.Thank you :) – Karan Singh May 02 '16 at 05:23
  • Well, before I launch into that, are you preparing for the IIT's by any chance? If so, I'd seriously recommend going through some book / a few video lectures on linear algebra, since in my experience, it helps with the exams.

    That aside, matrices are things that take vectors and transform them. If you have a matrix $A$, and a column vector $v$, the operation $w = Av$ will give you a new vector $w$ which is $v$ transformed by $A$. An orthogonal matrix is just a matrix that ensures that $|v| = |w|$ (that is, it preserves the lengths of vectors it acts upon)

     – Siddharth Bhat May 02 '16 at 05:26
  • Yes, JEE .This is a past year question and I didn't get the logic of the solution given hence the question. – Karan Singh May 02 '16 at 05:31
  • Right, so, matrices are things that act on vectors. – Siddharth Bhat May 02 '16 at 05:37
  • Ok, so if I just take the dot product of any column with itself and if it comes out to be 1 (or 0 with other columns), then I have an orthogonal vector right? – Karan Singh May 02 '16 at 05:43
  • No, you have an orthogonal matrix. (Orthogonal means "perpendicular"). The vectors themselves are pairwise orthogonal unit vectors. This means that each vector $v_i$ has length 1. So, $v_i \cdot v_i = 1$. And the vectors are orthogonal to each other which is why $v_i \cdot v_j = 0$. The orthogonal matrix is made up of $n$ orthogonal vectors all put one next to each other in the columns. – Siddharth Bhat May 02 '16 at 05:45
  • If I wasn't clear enough - each column of the matrix is a vector. The vectors put together form a matrix. In an orthogonal matrix, the vectors are orthonormal to each other – Siddharth Bhat May 02 '16 at 05:46
  • Oh yes I get it. I meant matrix not vector(in the comment).Thank you. – Karan Singh May 02 '16 at 05:47
  • You're welcome :) – Siddharth Bhat May 02 '16 at 05:48
  • What about higher dimensional matrices? – Jdeep Mar 09 '21 at 13:09
  • @NoahJ.Standerson it's the same: one can check if $AA^T = I$. – Siddharth Bhat Mar 09 '21 at 14:13
  • @SiddharthBhat Yes thats the definition of orthogonal matrix. But is there any trick or pattern through which we can identify it? – Jdeep Mar 09 '21 at 17:27
  • Let ${v=(a_1,a_2,...,a_n)}$ and ${w= (b_1,b_2,...,b_n)}$, are vectors in ${K^n}$ then we define the scalar or dot product of them ${v \cdot w}$ as ${v \dot w = a_1 b_1 + a_2 b_2}+...+ a_n b_n$. if ${v \cdot w =0}$ then both are orthogonal. for ex. ${v=11001 , w=01101}$ then ${v \cdot w = 1 \cdot 0 + 1 \cdot 1 + 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 1 = 1+1 =0 }$ – SSA Jul 29 '22 at 09:25