Suppose $\Vert\vec v\Vert=1$. Does $\vec v\cdot A\vec v=1$ imply $A=I$?

Question

I'm trying to show that $\vec v\cdot A\vec v=1 \iff A=I$. The reverse implication is straightforward. Somehow I'm unable to proceed for the forward part. I'm not even sure if its true. Any help would be much appreciated.

Answer 1

Let $A:\vec{i}\mapsto\vec{i}+\vec{j}, \vec{j}\mapsto -\vec{i}+\vec{j}$ is given on an orthonormal basis $(\vec{i}, \vec{j})$ of a $2$-dimensional dot product space. Let $|\vec{v}|=1$ for a vector $\vec{v}$, and express $\vec{v}=x\vec{i}+y\vec{j}$ in co-ordinates in this basis. Now:

$$\vec{v}\cdot A(\vec{v})=(x\vec{i}+y\vec{j})\cdot(x(\vec{i}+\vec{j})+y(-\vec{i}+\vec{j}))=x(x-y)+y(x+y)=x^2+y^2=1$$

but $A\ne I$.

Update: In an orthonormal base $e$, the value of $\vec{v}\cdot A\vec{v}$ is given as $\vec{v}_e^TA_e\vec{v}_e$ where $\vec{v}_e$ is the column of co-ordinates of $\vec{v}$ and $A_e$ is the matrix of $A$ in $e$. However, note that, for any column $x$ and square matrix $A$ of appropriate size we have $x^TAx=x^TA^Tx=x^T(\frac{1}{2}(A+A^T))x$ so we can always replace matrix $A$ with its symmetric component $\frac{1}{2}(A+A^T)$.

Now, let's go back: we know that $\vec{v}\cdot I\vec{v}=1$ for $|\vec{v}|=1$, however, the previous consideration makes it clear that we can just start with another operator $A\ne I$ if we can somehow make $\frac{1}{2}(A_e+A_e^T)=I_e=I$ (identity matrix). This is the motivation for me to build a counterexample which will, as a matrix, have something like $\begin{bmatrix}1&-1\\1&1\end{bmatrix}$.

Update 2: One can see that the chosen map $A$ above is the rotation by $45^\circ$ followed by the multiplication by the factor $\sqrt{2}$. In other words, $A\vec{v}$ is rotated from $\vec{v}$ by $45^\circ$, which reduces the dot product by the factor $\sqrt{2}=\sec(45^\circ)$ but then is multiplied by $\sqrt{2}$ to get to the initial value of the dot product. You may take any angle instead of $45^\circ$.