Prove or disprove: If $x^T A x = 0 $ for all $x$, then $ A = 0 $.

Question

Let $A$ be a square matrix and $x$ be a vector. Now consider the statement:

If $x^T A x = 0 $ for any $x$, then $A = 0$.

Is the above statement true or false? How would you prove it?

Answer 1

If $x ^t Ax = 0$ for every $x \in \mathbb R^n $ then the diagonal entries of $A$ are all zero by taking $x$ to be the column vector with the $i$-th component as $1$ and $ 0$ elsewhere, $\forall i = 1,2,\ldots, n$.

then choose your $x$ as follows

the column vector with the $i$-th and the $j$-th components as $1$ and $0$ elsewhere. Then try to ahow $A^t=-A$

Answer 2

It is not true, $A=\begin{pmatrix}0&1\\-1&0 \end{pmatrix}$ is a counterexample.

Take $x=e_i+e_j$, you can get $A_{ij}+A_{ji}=0$, which means $A^T+A=0$

Answer 3

furthermore to the disproves of the fellers up, I would like to mention that if the question was in the complex field, the statement was true.

in other words:

$x^*Ax=0 \, \forall x \in C^n \Leftrightarrow A=0 $

proof:

$\Leftarrow$ this direction is immediate

$\Rightarrow$

since the left side is zero for all x, lets choose few x's to show how A must be zero as well

• $x_1=e_k \Rightarrow x^*Ax=e^T_kAe_k=A_{kk} \Rightarrow A_{kk}=0 \, \forall k$
• $x_2=e_k+e_m \Rightarrow x^*Ax= (e^T_k+e^T_m)A(e_k+e_m)=A_{kk}+A_{km}+A_{mk}+A_{mm} $

since $A_{kk}=A_{mm}=0$ we get $\Rightarrow A_{km}=-A_{mk} \, \forall k\ne m$

• $x_3=e_k+ie_m \Rightarrow x^*Ax= (e^T_k-ie^T_m)A(e_k+ie_m)=A_{kk}+iA_{km}-iA_{mk}+A_{mm} $

$\Rightarrow A_{km}=A_{mk} \, \forall k\ne m$

from $x_2,x_3$ we get that $A_{km}=-A_{km}$ where this can only valid if $A_{km}=0 \, \forall k\ne m$

therefore $A=0 \,\blacksquare$

Answer 4

Since $x^tAx=\langle x,Ax\rangle$ (the inner product), the question may be stated:

Is there a transformation that maps every vector to something orthogonal to itself?

What should come to mind is a rotation matrix that rotates $\mathbb{R}^2$ by 90 degrees, which Shu Xiao Li has presented.