Assume that $A$ is an $n\times n$ symmetric positive-definite matrix.
Prove that:
the element of $A$ with maximum magnitude must lie on the diagonal.
Asked
Active
Viewed 272 times
5
Samrat Mukhopadhyay
- 16,277
sami
- 133
-
Ok, what do you need, where are you stuck, which hint do you want? What did you do yourself? – Karl Dec 26 '14 at 13:39
3 Answers
4
Let $a_{max}$ be the element of $\mathbf{A}$ with maximum magnitude. Assume that it is not lie on the diagonal. Thus, there is exist a $2\times 2$ principal minor equal to $ab- a_{max}^2 \leq 0$ for some $a$ and $b$ in the main diagonal. Hence, as the matrix is positive definite, $a_{max}$ should be actually in the main diagonal.
Alex Silva
- 3,557
-
1you can choose a vector $x$ with 1 and -1 at the right spots and everywhere else 0 to make $x^TAx < 0$ – abel Dec 26 '14 at 13:58
-
Yes! Choose this vector is related to choose the right principal minor. – Alex Silva Dec 26 '14 at 14:02
1
For an $n\times n$ Hermitian positive semidefinite matrix $A=(a_{ij})$ and $1\leq i,j\leq n$, $$|a_{ij}|^2\leq a_{ii}a_{jj}\leq\max\limits_{k}a_{kk}^2 \quad \Rightarrow \quad|a_{ij}|\leq\max_k a_{kk}.$$
Algebraic Pavel
- 22,928
0
Hint: Assume that this is not the case, i.e., the entry of maximum magnitude is an off-diagonal entry $A_{ij}=A_{ji}$, with $j \neq i$. To show that $A$ cannot be PSD, it suffices to construct an $x$ (a certificate) such that $$ x^{T}Ax < 0. $$ How can we find one such $x$? In order to answer that, you have to make clear what $x^{T}Ax$ really is. How can you write it as a sum? Which entries of $x$ multiply $A_{ij}$?
megas
- 2,276