Let $A$ be a symmetric matrix, and $B=J(p, q,\theta) A J^T(p, q,\theta)$, where $J$ is the Givens rotation matrix that rotates in the $p, q$ plane by an angle $\theta$.

Consider the magnitude $\text{off}(B)$, as the square root of the sum of squares of the off-diagonal terms of $B$:

$$\text{off}(B)^2={\left\lVert B \right\rVert}_F - \sum_{i=1}^n b_{ii}^2$$

In relating it to the value of that magnitude for the original matrix, $\text{off}(A)$, and reach the known result: $\text{off}(B)^2 = \text{off}(A)^2 -2a_{pq}^2$ (see ref. [3]), it is made the following:

\begin{array}{ll} \text{off}(B)^2 &= {\left\lVert A \right\rVert}_F - \sum_{i=1}^n b_{ii}^2 = {\left\lVert A \right\rVert}_F - \sum_{i \neq p, q} b_{ii}^2 - (b_{pp}^2+b_{qq}^2) &= {\left\lVert A \right\rVert}_F - \sum_{i \neq p, q} a_{ii}^2 - (b_{pp}^2+b_{qq}^2)\\ &\uparrow &\uparrow \\ & \left\lVert A \right\rVert_F=\left\lVert B \right\rVert_F \ \textrm{, because} \ J \ \text{is orthogonal} &\textrm{only}\ p\textrm{-th} \ \textrm{and} \ q\textrm{-th} \ \textrm{rows and columns of A change} \\ \\ &={\left\lVert A \right\rVert}_F - \sum_{i \neq p, q} a_{ii}^2 - (a_{pp}^2+ 2a_{pq}^2 +a_{qq}^2) \\ &\uparrow \\ &?? \ \textrm{(my question)} \end{array}

I understand this implies that $$b_{pp}^2+b_{qq}^2 = a_{pp}^2+ 2a_{pq}^2 +a_{qq}^2\quad (1)$$

To verify this, I calculate the value of the transformed $p$-th and $q$-th diagonal terms:

\begin{align} b_{pp} &= c^2a_{pp}-2csa_{pq}+s^2a_{qq}\\ b_{qq} &= s^2a_{pp}+2csa_{pq}+c^2a_{qq} \end{align}

with $c$ and $s$ being $\cos(\theta)$ and $\sin(\theta)$ respectively.

I then square and add them, leading to a long expression that seems far from being the stated result $(1)$. For example, the coefficient of $a_{pp}$ turns out to be $c^4+ s^4$, instead of $1$, and so does the coefficient of $a_{qq}$.

Does this resulting long trigonometric expression just ends up leading to eq. $(1)$? Or just am I missing something in the procedure?

Possibly this not a general result and other conditions need to be imposed for that to happen. This result is used to prove that the Jacobi eigenvalue algorithm converges to a diagonal matrix upon the repeated application of this transformation. In it, the angle $\theta$ is chosen such that the entries in the $(p,q)$ and $(q,p)$ positions are zeroed after the transformation, i.e. $b_{pq}$, $b_{qp}=0$. As pointed in the comments, for size-2 matrices and $\theta=-\frac{\pi}{2}$, it does not hold.


[1] https://en.wikipedia.org/wiki/Givens_rotation
[2] https://en.wikipedia.org/wiki/Jacobi_eigenvalue_algorithm
[3] https://web.stanford.edu/class/cme335/lecture7.pdf


1 Answer 1


Equality $(1)$ is false in general. E.g. $$ \pmatrix{0&-1\\ 1&0}\pmatrix{0&1\\ 1&0}\pmatrix{0&1\\ -1&0}=\pmatrix{0&-1\\ -1&0}. $$ However, in the context of the linked lecture notes, $p,q$ and $\theta$ are chosen such that $b_{pq}=b_{qp}=0$. It follows that $$ a_{pp}^2+a_{pq}^2+a_{qp}^2+a_{qq}^2 =b_{pp}^2+\underbrace{b_{pq}^2+b_{qp}^2}_{0+0}+b_{qq}^2 =b_{pp}^2+b_{qq}^2. $$ For instance, in the example above, if we use a Givens rotation for an angle $\pi/4$, we have $$ \frac{1}{\sqrt{2}}\pmatrix{1&1\\ -1&1}\underbrace{\pmatrix{0&1\\ 1&0}}_A\frac{1}{\sqrt{2}}\pmatrix{1&-1\\ 1&1}=\underbrace{\pmatrix{1&0\\ 0&-1}}_B, $$ so that $a_{pp}^2+2a_{pq}^2+a_{qq}^2=b_{pp}^2+b_{qq}^2$ and $\operatorname{off}(B)^2=0<2=\operatorname{off}(A)^2$.

  • $\begingroup$ Giving another thought to this, I am still thinking how this generalizes to higher dimensions. In the $n=2$ case you are able to write the full norms explicitly (only have four terms). In the general case we know the norms are preserved as well, however not only the $(p, p)$, $(q, p)$, $(p, q)$ and $(p, p)$ entries are changed, but the rest of elements in their row and column as well. $\endgroup$
    – abcd
    Aug 14, 2021 at 18:41
  • $\begingroup$ I understand then the relation would be in general $\sum_{i \neq p, q} \left(a_{ip}^2 + a_{iq}^2 + a_{pi}^2 + a_{qi}^2\right) + a_{pp}^2 +a_{pq}^2+a_{qp}^2+a_{qq}^2 =\sum_{i \neq p, q} \left(b_{ip}^2 + b_{iq}^2 + b_{pi}^2 +b_{qi}^2\right) + b_{pp}^2+b_{pq}^2+b_{qp}^2+b_{qq}^2$ How can we prove the terms within the sum symbols cancel? $\endgroup$
    – abcd
    Aug 14, 2021 at 19:03
  • 1
    $\begingroup$ @abcd This has already been explained in the lecture notes. Suppose $(p,q)=(1,2)$. Let $R$ denotes the $2\times2$ rotation matrix for an angle $\theta$. Partition $A$ as $\pmatrix{X&Y\\ Z&W}$ where $X$ is $2\times2$. Then $$B=\pmatrix{R\\ &I}\pmatrix{X&Y\\ Z&W}\pmatrix{R^T\\ &I}=\pmatrix{RXR^T&RY\\ ZR^T&W}...$$ $\endgroup$
    – user1551
    Aug 14, 2021 at 19:14
  • 1
    $\begingroup$ ...In the $2\times2$ case, we have $\operatorname{off}(RXR^T)^2<\operatorname{off}(X)^2$. Therefore \begin{aligned}\operatorname{off}(B)^2&=\operatorname{off}(RXR^T)^2+\|RY\|_F^2+\|ZR^T\|_F+\operatorname{off}(W)^2\\ &=\operatorname{off}(RXR^T)^2+\|Y\|_F^2+\|Z\|_F+\operatorname{off}(W)^2\\ &<\operatorname{off}(X)^2+\|Y\|_F^2+\|Z\|_F+\operatorname{off}(W)^2\\ &=\operatorname{off}(A)^2.\end{aligned} $\endgroup$
    – user1551
    Aug 14, 2021 at 19:15

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