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I have a question about why the square of the matrix Q, below, is equal to the identity matrix.

Q =

   cos X  -sin X
   sin X   cos X

My knowledge of trigonometry seems to have rotted away, and I can't figure out what rules are used to justify the statement made by professor Strang at 11:55 of this video > Lec 20 | MIT 18.06 Linear Algebra, Spring 2005 (he says the square of Q is equal to I)

The thing that has me stuck is the following: In order to calculate the first cell of Q * Q we take the dot product of the first row and the first column. That would be

  cos^2 Q  -  sin^2 Q

But how is that equal to the "1" that we want in the upper left most cell of the identity matrix ?

If this were cos^2 Q + sin^2 Q

then it would be equal to 1.. but we have a difference here, not a sum.

Thanks in advance ! chris

epilogue: thanks for the answers ! I don't know how I missed the fact that he was not making this claim for Q^2, but Q-transpose * Q, instead. Appreciate your pointing it out.

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    $\begingroup$ Is mayhaps $X = \pi$? For $$\begin{pmatrix}\cos\varphi & -\sin\varphi\\\sin\varphi&\cos\varphi\end{pmatrix}^2 = \begin{pmatrix}\cos(2\varphi) & -\sin(2\varphi)\\\sin(2\varphi)&\cos(2\varphi)\end{pmatrix}$$ $\endgroup$ – Daniel Fischer Sep 13 '13 at 22:41
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    $\begingroup$ I took the square and I did not end up with the identity matrix either $\endgroup$ – imranfat Sep 13 '13 at 22:41
  • $\begingroup$ Are you sure he didn't say $Q^TQ=I$ instead? I can't find the statement in the video; at 11:55 he's talking about a different $4\times4$ matrix. $\endgroup$ – user856 Sep 13 '13 at 22:44
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Look at the video you referenced around 21:30, he is not saying that the square of this matrix is the identity matrix. He is saying that if the columns of $Q$ are orthonormal, then $QQ^T=I$ provided the matrix is square. That is to say the matrix is an $n \! \times \! n$ matrix.

Concisely: If $Q$ is a square ($n \! \times \! n$) matrix with orthonormal column vectors, then the product of $Q$ and its transpose equals the identity.

Also note that the column vectors of your matrix are orthonormal as their dot product is zero.

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The square of this matrix is not equal to the identity matrix! We compute:

$Q^2 = \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{bmatrix} = \begin{bmatrix} \cos^2 x - \sin^2 x & -2\sin x \cos x \\ 2 \sin x \cos x & \cos^2 x - \sin^2 x \end{bmatrix}, \tag{1}$

in fact this is

$Q^2 = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}; \tag{2}$

but if we try $Q^TQ$, we get

$Q^TQ = \begin{bmatrix} \cos^2 x + \sin^2 x & -\cos x \sin x + \cos x \sin x \\ -\sin x \cos x + \sin x \cos x & \cos^2 x+ \sin^2 x \end {bmatrix} = I, \tag{3}$

$I$ being the $2 \times 2$ identity matrix.

So someone, somewhere must have gotten $Q^2$ and $Q^TQ$ mixed up!

Hope this helps clear things up!

Cheers!

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$$ \color{magenta}{ \left(% \begin{array}{cr} \cos\left(X\right) & \sin\left(X\right) \\ \sin\left(X\right) & -\cos\left(X\right) \end{array}\right)} $$

really does square to give the identity.

For an $n$ by $n$ matrix with real entries, call it $A,$ if $A$ is symmetric and $A^2 = I$ and $\operatorname{trace} A = n-2,$ then $A$ is a reflection. This applies to the example above with $n=2.$ The eigenvalues are, indeed, $-1,1.$

If we have real numbers $\alpha, \beta, \gamma$ such that $$ \color{blue}{ \alpha^2 + \beta^2 + \gamma^2 = 1 }, $$ then $$ \color{green}{ \left( \begin{array}{ccc} 1 - 2 \alpha^2 & - 2 \alpha \beta & -2 \alpha \gamma \\ - 2 \alpha \beta & 1 - 2 \beta^2 & - 2 \beta \gamma \\ -2 \alpha \gamma & - 2 \beta \gamma & 1 - 2 \gamma^2 \end{array}\right)} $$ is a reflection. It negates the vector with entries $(\alpha, \beta, \gamma)$ and fixes vectors orthogonal to that, such as $(\beta, -\alpha, 0)$ and $(\alpha \gamma, \beta \gamma, \gamma^2 - 1).$ So the eigenvalues are $-1,1,1$ and the trace is $1 = 3-2.$

Oh, if you simply negate all the entries in the 3 by 3 matrix, what you get fixes $(\alpha, \beta, \gamma)$ and rotates in the orthogonal plane by $180^\circ$

Given a column vector $v$ with transpose $v^T$ a row vector, we know that $v^T v$ is the 1 by 1 matrix whose only entry is $v \cdot v = |v|^2.$ In contrast, $v v^T$ is a symmetric $n$ by $n$ matrix of rank $1.$ If, in addition, $|v| = 1,$ then the matrix of the reflection in $v$ is $$ \color{magenta}{ I - 2 v v^T}. $$ So, given any $w$ orthogonal to $v,$ we get $v^T w = ( v \cdot w) = (0);$ so $$(I - 2 v v^T) w = Iw - 2 v (v^T w) = w - 2 v (0) = w.$$ But $$(I - 2 v v^T) v = Iv - 2 v (v^T v) = v - 2 v (1) = -v.$$ Meanwhile, $$(I - 2 v v^T)^2 = I - 4 v v^T + 4 v v^T v v^T = I - 4 v v^T + 4 v (v^T v) v^T = I.$$

If $ a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 + h^2 = 1, $ the reflection that negates $(a,b,c,d,e,f,g,h)$ is $$ \color{magenta}{\left( \begin{array}{cccccccc} 1 - 2 a^2 & -2ab & -2ac & -2ad & -2ae & -2af & -2ag & -2ah \\ -2ab & 1-2b^2 & -2bc & -2bd & -2be & -2bf & -2bg & -2bh \\ -2ac & -2bc & 1-2c^2 & -2cd & -2ce & -2cf & -2cg & -2ch \\ -2ad & -2bd & -2cd & 1-2d^2 & -2de & -2df & -2dg & -2dh \\ -2ae & -2be & -2ce & -2de & 1-2e^2 & -2ef & -2eg & -2eh \\ -2af & -2bf & -2cf & -2df & -2ef & 1-2f^2 & -2fg & -2fh \\ -2ag & -2bg & -2cg & -2dg & -2eg & -2fg & 1-2g^2 & -2gh \\ -2ah & -2bh & -2ch & -2dh & -2eh & -2fh & -2gh & 1-2h^2 \end{array}\right)} $$ with trace 6.

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Your matrix is the rotation matrix of any vector of length 1 about the origin. Why would that square end up being the identity matrix? It is, however for a selective number of values, as Daniel pointed out.

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  • $\begingroup$ Just saw your answer and other comments! :-) $\endgroup$ – Arash Sep 13 '13 at 22:49
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This matrix $Q$ represents a rotation around origin by $X$. Then $Q^2$ represents two consecutive rotation which means a rotation by $2x$: $$ Q^2=\left[ \begin{array} {ll} \cos(2x)&-\sin(2x)\\ \sin(2x)&\cos(2x) \end{array} \right] $$ So $Q^2$ is not identity. However if you rotate back by $-x$ then you expect to go back to your original point. So the true inverse of $Q$ is: $$ Q^*=\left[ \begin{array} {ll} \cos(x)&\sin(x)\\ -\sin(x)&\cos(x) \end{array} \right] $$ which is indeed its transpose.

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$$ \left(% \begin{array}{cr} \cos\left(X\right) & -\sin\left(X\right) \\ \sin\left(X\right) & \cos\left(X\right) \end{array}\right) = \cos\left(X\right) - {\rm i}\sin\left(X\right)\,\sigma_{y}\,, \quad \sigma_{y} \equiv \left(% \begin{array}{cr} 0 & -{\rm i} \\ {\rm i} & 0 \end{array}\right)\,, \quad \left\vert% \begin{array}{l} \mbox{Notice that} \\ \sigma_{x}^{2} = 2\times 2\ \mbox{identity}. \end{array}\right. $$

Then $$ \left(% \begin{array}{cr} \cos\left(X\right) & -\sin\left(X\right) \\ \sin\left(X\right) & \cos\left(X\right) \end{array}\right) = \cos\left(X\right) - {\rm i}\sin\left(X\right)\,\sigma_{y} = {\rm e}^{-{\rm i}X\sigma_{y}} $$ So, $$\color{#ff0000}{\large% \left(% \begin{array}{cr} \cos\left(X\right) & -\sin\left(X\right) \\ \sin\left(X\right) & \cos\left(X\right) \end{array}\right)^{n} = {\rm e}^{-{\rm i}nX\sigma_{y}} = \left(% \begin{array}{cr} \cos\left(nX\right) & -\sin\left(nX\right) \\ \sin\left(nX\right) & \cos\left(nX\right) \end{array}\right)} $$

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