# Norm of off-diagonal matrix terms decreases under similarity transformation with Givens matrix. Detail of proof.

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. ), 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.

References:

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 $$\operatorname{off}(B)^2=0<2=\operatorname{off}(A)^2$$.
• 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.
• 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?
• @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 $$B=\pmatrix{R\\ &I}\pmatrix{X&Y\\ Z&W}\pmatrix{R^T\\ &I}=\pmatrix{RXR^T&RY\\ ZR^T&W}...$$ Aug 14, 2021 at 19:14
• ...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} Aug 14, 2021 at 19:15