2
$\begingroup$

If $v^TAv = v^TBv$ for all constant vectors $v$ and $A,B$ are matrices of size $n$ by $n$, is it true that $A=B$?

I have thought about using basis vectors but cannot get system of equations uniquely. Is there a trick here?

$\endgroup$
1
$\begingroup$

It is not true. Take $A$ to be the zero matrix, and $B$ to be any nonzero skew-symmetric matrix. You can prove that both sides are always zero, but $A$ and $B$ are not equal.

If you assume that both $A$ and $B$ are symmetric real matrices, then they are equal. You can see that $v^TAv = v^TBv$ is equivalent to $v^T(A-B)v=0$. We know that $A-B$ is also a symmetric matrix. When the last equality holds for all vectors $v$, $A-B$ must be a zero matrix. This is due to the spectral theorem, as the existence of eigenvectors of $A-B$ is justified by the spectral theorem. See the details in another post. Now get any eigenvector $v$ of $A-B$ to see that $v^T(A-B)v=\lambda\left\lVert v \right\rVert^2$, where $\lambda$ is the eigenvalue of $A-B$ corresponding to the eigenvector $v$. Since eigenvectors are by definition nonzero, it must be the eigenvalue that is zero. So all eigenvalues of $A-B$ are zero. $A-B$ is similar to the zero matrix, so that $A-B$ itself is also the zero matrix.

$\endgroup$
6
$\begingroup$

The question is the same as:

Is it true $A=0$, if $v^TAv=0$, for all vectors $v$?

If $A$ is anti-symmetric, it is not.

$\endgroup$
  • $\begingroup$ In case it's not clear, I'll just add that they're the same because we may subtract $v^TBv$ from both sides of the OP's equation to get $v^T (A-B) v = 0$. Then $A$ and $B$ are the same if and only if $A-B = 0$. Additionally, every matrix $M$ may be expressed as $A-B$ (i.e. set $A = M$ and $B = 0$), and so the OP's question reduces to the one in this answer. $\endgroup$ – Myridium Oct 17 '16 at 7:24
2
$\begingroup$

If you assume that the matrices are symmetric then it is true. Otherwise not. In fact, $M=A-B$ may be any anti-symmetric matrix, i.e. $M^T=-M$.

If $M$ is symmetric on the other hand it is diagonaliable with real eigenvalues and in an orthonormal basis. Picking an eigenvector $v$ for eigenvalue $\lambda$ you see that $0=v^T M v = \lambda v^T v$ so $\lambda=0$. Thus, all eigenvalues are zero and for a diagonalizable matrix this implies that the matrix is the zero matrix.

$\endgroup$
  • $\begingroup$ I assume you don't need both $A$ and $B$ to be symmetric. Surely only $A-B$ needs to be symmetric? Like if $B=-A^T$? $\endgroup$ – snulty Oct 17 '16 at 8:04
  • 1
    $\begingroup$ No you are right. On the other hand if you want to specify a category of matrices for which the statement is true, then the most natural is the category of symmetric matrices. $\endgroup$ – H. H. Rugh Oct 17 '16 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.