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Say we have an $n \times n$ matrix $M$ and a vector $v \in \Bbb{R}^n$.

If we must have $v^T M v = v^T.v$ for all vectors $v$, this must necessarily mean $M=I$. Its obvious that $M=I$ will work, but how to prove its the only matrix that will work?

My attempt:

$M$ must be full rank, because otherwise, there is a non-trivial $v$ that will satisfy $Mv = 0$. Since $v^Tv > 0$ but $v^T M v = 0$, we get a contradiction.

Next, let's think about the eigen values of $M$. If $M$ has a single eigen value that is not $1$, we'll similarly get a contradiction. So, all its eigen values must be $1$. And the identity is the only matrix that satisfies this.

The issue with this is that not all matrices will be diagonalizable and also does having all eigen values equal $1$ necessarily imply the identity?

Thomas Andrews
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Rohit Pandey
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    By taking specific choices for $v$ (e.g. standard basis vectors) you can isolate the components of $M$. – Joseph Expo Dec 22 '24 at 22:57
  • Standard basis vectors will only isolate diagonal entries of $M$, no? So we'll get all diagonals should be $1$ but what about off-diagonals? – Rohit Pandey Dec 22 '24 at 23:02
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    Don't say "The title says it all" and then give a longer form with more specifics. Clearly, the title didn't say it all. – Thomas Andrews Dec 22 '24 at 23:06
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    The claim is false. $v^\top M v=v^\top v$ for all $v$ if and only if $M-I$ is antisymmetric (more generally, if $\operatorname{char}F\ne2$ and $g_v(X)=v^\top Xv$, then $\bigcap_{v\in F^n}\ker g_v$ is the subspace of the antisymmetric matrices). – Sassatelli Giulio Dec 22 '24 at 23:08
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    Note that if $A$ is skew-symmetric then $v^T A v = 0$ for all $v$. So if $M$ satisfies $v^T M v = v^T v$ for all $v$, then so does $M + A$ for all skew-symmetric matrices $A$. – littleO Dec 22 '24 at 23:27
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    Be careful, complex eigenvalues involve complex vectors. The claim that the eigenvalues must be $1$ is false. Using the characterization given in comment and in the separated answer, the eigenvalues actually belong to $1+i\mathbb{R}$. – reded Dec 22 '24 at 23:51

2 Answers2

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First, the fact that all eigenvalues are equal to $1$ does not imply that your matrix is $I_d$. Take for example any triangular matrix with $1$'s on the diagonal.

Second, the identity $v^\top M v$ only involves the symmetric part of $M$, see my post here.

With $S$ the symmetric part of $M$, for all $v\in\mathbb{R}^n$: \begin{align*} v^\top (S-I_d)v = 0. \end{align*} Within an orthonormal change of basis the symmetric matrix $S-I_d$ is diagonal by the spectral theorem, a diagonal that must be equal to $0$ by considering the vectors of the canonical basis.

The matrix $M$ is thus equal to $I_d+A$ with $A$ skew-symmetric.

Conversely, for any $M := I_d + A$ with $A$ skew-symmetric, it is that $v^\top M v = v^\top v$ for all vectors $v\in\mathbb{R}^n$.

reded
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First, we can take for choices of $v$, the basis vectors. Example, $v=[1,0,0, \dots , 0]$. Plugging this into $v^TM v = v^T v$ will lead to $M_{1,1} = 0$. And similarly, all the diagonals will be $1$.

Next, let's plug in $v = [1,1,0, \dots 0]$. This will lead to:

$$M_{11}+M_{22}+M_{21}+M_{12} = 2$$ Since $M_{11}=M_{22} = 1$, $$M_{21}+M_{12} = 0$$.

As long as $M_{12}$ and $M_{21}$ have opposite signs, this works.

In the special case where $M = A^TA$, the matrix $M$ must be symmetric. And so, $M_{12} = M_{21} = 0$ and similarly all off-diagonal entries must be $0$.

Rohit Pandey
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