# How can this equation hold $$\sum_{m=0}^{N-1} \sum_{n=0}^{N-1} |u(m,n)|^2 = \sum_{k=0}^{N-1}\sum_{l=0}^{N-1} |v(k,l)|^2$$?

I am trying to learn digital Image processing by myself and now stuck at a problem in the two dimensional unitary transformations. It states that let $U$ be the input image and $V$ be the transformed image and $A$ is the unitary transformation matrix. To prove above equation I am rather trying to prove $$U^2=V^2$$ which I think is equivalent to proving equation in question.We have $$V=A U A^T$$ now to prove the above relation: $$V^2=V^{*T} V$$ where $$V=A U A^T \mbox{ and } V^{*T}=A^*[A^*U^*]^T$$ which on solving gives me $$A^* U^{*T} U A^T$$ which gives $$A^* U^2 A^T$$but it should be equal to $$U^2$$, as I understand. Though $$A^T A^*=I$$ as $A$ is Unitary matrix, but as much as I know we can't change order.

You can prove this using trace of the matrix and the similarity invariance property:

\begin{align} \sum\limits_{k=0}^{N-1} \sum\limits_{l=0}^{N-1} \left| v(k,l) \right|^2 &= \text{tr} \left(V^{*} V\right) \\ &= \text{tr} \left(A U^{*} A^{*} A U A^{*} \right) \\ &= \text{tr} \left(A U^{*} U A^{*} \right) \\ &= \text{tr} \left(U^{*} U \right)~~\text{See attached image} \\ &= \sum\limits_{m=0}^{N-1} \sum\limits_{n=0}^{N-1} \left| u(m,n) \right|^2 \end{align}

The following snapshot has been taken from wikipedia:- Trace of a Matrix • Thanks a lot! just one thing, I hope you mean to say first equation is equal to$$tr(V^{*T}V)$$ – Userhanu May 5 '16 at 2:41
• Actually $*$ includes transpose. Look at complex conjugate of a matrix. – Amal May 5 '16 at 2:43
• Nice! I hadn't interpreted the question that way. Good work @user3288586 ! – Peter K. May 5 '16 at 14:02

OK, so let's have a look at whether

$$\sum_{m=0}^{N-1} \sum_{n=0}^{N-1} |u(m,n)|^2 = \sum_{k=0}^{N-1}\sum_{l=0}^{N-1} |v(k,l)|^2\tag{1}$$

is equivalent to

$$U^2=V^2\tag{2}$$

Let's have a look at a simple $2 \times 2$ example: $$U = \left [ \begin{array}{cc} u_a & u_b\\ u_c & u_d \end{array} \right ] \\ V = \left [ \begin{array}{cc} v_a & v_b\\ v_c & v_d \end{array} \right ]$$ so that $$U^2 = \left [ \begin{array}{cc} u_a & u_b\\ u_c & u_d \end{array} \right ] \left [ \begin{array}{cc} u_a & u_b\\ u_c & u_d \end{array} \right ] = \left [ \begin{array}{cc} u_a^2 + u_b u_c & u_au_b + u_bu_d\\ u_cu_a + u_d u_c & u_c u_b + u_d^2 \end{array} \right ]$$ and $$V^2 = \left [ \begin{array}{cc} v_a^2 + v_b v_c & v_av_b + v_bv_d\\ v_cv_a + v_d v_c & v_c v_b + v_d^2 \end{array} \right ]$$

Then (1) becomes $$|u_a|^2 + |u_b|^2 + |u_c|^2 + |u_d|^2 = |v_a|^2 + |v_b|^2 + |v_c|^2 + |v_d|^2$$

The two are completely different. Why do you think they are the same?

• i wonder if Ankit means that $V$ this the result of applying a unity-gain linear transform, e.g. the DFT (with the $\frac{1}{\sqrt{N}}$ scaler in both directions), to the vector $U$. would that mean that $|V|^2 \ = |U|^2$ ? – robert bristow-johnson May 4 '16 at 21:17
• @robertbristow-johnson : Could be! But $U$ and $V$ are explicitly called images... so they can't really be vectors, unless they're doing the Matlab U(:) operation... – Peter K. May 4 '16 at 21:54
• yeah, so the unitary 2-dim DFT has a $\frac1N$ in both directions. if it were that linear transform, i think then the energy of the output is the same as the energy of the input. – robert bristow-johnson May 4 '16 at 22:01
• Yup. That'd be Parseval, which would make sense.... Hmmm. Let me see if I can rejig the question to ask that. – Peter K. May 4 '16 at 22:05
• Nope. My brain is fried for the day. Will have a look later, if I can. – Peter K. May 4 '16 at 22:07