A problem about the matrix equation $a^{\sf T}A_1a=0$.

Question

Let $A_1$ be an $m\times m$ real constant matrix in $\mathbb{R}^{m\times m}$. The matrix $A_1$ satisfies $a^{\sf T}A_1a=0$ for all $a\in\mathbb{R}^m$, where $a^{\sf T}$ is the transpose of $a$. Show that $$A_1=-A_1^{\sf T}.$$

Answer 1

For $a, b \in \Bbb R^m$, consider$ (a + b)^TA_1(a + b)$: we have

$0 = (a + b)^TA_1(a + b) = (a^T + b^T)A_1(a + b)$ $= a^TA_1a + a^TA_1b + b^TA_1a + b^TA_1b = a^TA_1b + b^TA_1a. \tag{1}$

We observe that

$a^TA_1b = (a^TA_1b)^T = b^TA_1^Ta, \tag{2}$

since these are all scalar quantities and hence are symmetric with respect to transposition. Thus (1) may be written

$b^TA_1^Ta + b^TA_1a = 0, \tag{3}$

or

$b^T(A_1^Ta + A_1a) = 0, \tag{4}$

which holds for all $b$, whence

$0 = A_1^Ta + A_1a = (A_1^T + A_1)a = 0 \tag{5}$

holds for all $a \in \Bbb R^m$; but (5) implies

$A_1^T + A_1 = 0, \tag{6}$

or

$A_1^T = -A_1, \tag{7}$

as was required. QED.

Cheerio,

and as always,

Fiat Lux!!!

Answer 2

If $(b_1,\cdots,b_m)$ is the canonical basis of $\mathbb R^m$, can you compute $(b_i+b_j)^{\sf T}A (b_i+b_j)$ in two different ways ?

Answer 3

Let's consider, ${\cal A} \equiv A_{1} + A_{1}^{\sf T}$ which is a hermitian matrix: ${\cal A}^{\dagger} = {\cal A}$. ${\cal A}$ has real eigenvalues: Let's $\left\{n\right\}$ and $\left\{\lambda_{n}\right\}$ the orthonormalized eigenvectors and real eigenvalues of ${\cal A}$, respectively:

\begin{equation} {\cal A}n = \lambda_{n}n\,, \qquad n^{\sf T}n' = \delta_{nn'} \tag{1} \end{equation}

Obviously, $a^{\sf T}{\cal A}\,a = 0,\ \forall\ a \in {\mathbb R}^{m}$. In particular, $n^{\sf T}{\cal A}n = 0,\ \forall$ eigenvectors $n$ of ${\cal A}$. From $\left(1\right)$, we get

\begin{equation} n^{\sf T}{\cal A}n = \lambda_{n}n^{\sf T}n = \lambda_{n} \quad\Longrightarrow\quad \lambda_{n} = 0. \quad\Longrightarrow\quad {\cal A}n = 0. \tag{2} \end{equation}

From $\left(2\right)$, it's clear that \begin{equation} {\cal A} = \sum_{n}{\cal A}\,nn^{\sf T} = 0 \end{equation}

Then, $$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% A_{1} = -A_{1}^{\sf T} \quad} \\ \\ \hline \end{array} $$

Answer 4

We have $(a^TA_1a)^T=(a^TA_1^Ta)=0$ for each $a\in \mathbb{R}^m$, where I employed the fact that the transpose of the scalar is itself. Now since both quantities are zero we can add them together and get that $a^T(A_1+A_1^T)a=0$ for each $a\in \mathbb{R}^m$. Now matrix $B=A+A^T$ is symmetric and $a^TBa$ is then quadratic form which is zero for all $a\in\mathbb{R}^m$. Hence $B=0$ and you have that $A=-A^T$.