Is $M^TM=I$ not the same as $\mathbf{v}^TM^TM\mathbf{v}=\mathbf{v}^T\mathbf{v}$???

Question

I define a function $f[\mathbf{v}]$ as follows:

$$ f[\mathbf{v}]=\mathbf{v}^T\mathbf{v} $$

I am now interested in a group of linear transformation $M$ ($n\times n$ matrices) which leaves the function invariant:

$$ f[M\mathbf{v}]=f[\mathbf{v}] $$

To 'solve' for the properties of $M$, I consider two equality:

$$ M^TM=I\tag{1} $$

$$ \mathbf{v}^TM^TM\mathbf{v}=\mathbf{v}^T\mathbf{v}\tag{2} $$

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Working out the details, I get different results for $M$ depending I use (1) or (2).

We assume $\operatorname{Dim} {\mathbf{v}}=2$ and $\operatorname{Dim} {M}=2 \times 2$.

$$ \mathbf{v}=\pmatrix{x\\y}\\ M=\pmatrix{a&b\\c&d} $$

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For (1), I want to solve:

$$ \pmatrix{a&c\\b&d}\pmatrix{a&b\\c&d}=\pmatrix{1&0\\0&1}\tag{1a} $$

and for (2), I want to solve:

$$ \pmatrix{x&y}\pmatrix{a&c\\b&d}\pmatrix{a&b\\c&d}\pmatrix{x\\y} =x^2+y^2\tag{2a} $$

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Now, for (1) I get

$$ \left( \begin{array}{cc} a^2+c^2 & a b+c d \\ a b+c d & b^2+d^2 \\ \end{array} \right)=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)\tag{1b} $$

and for (2) I get

$$ (a x+b y)^2+(c x+d y)^2=x^2+y^2 \tag{2b} $$

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For (1) I clearly get the orthogonal group $O^TO=I$.

But for (2), using Mathematica:

Solve[(a x+b y)^2+(c x+d y)^2==x^2+y^2]

I get:

$$ \left\{\left\{d\to \frac{-\sqrt{-a^2 x^2 y^2-2 a b x y^3-b^2 y^4+x^2 y^2+y^4}-c x y}{y^2}\right\},\left\{d\to \frac{\sqrt{-a^2 x^2 y^2-2 a b x y^3-b^2 y^4+x^2 y^2+y^4}-c x y}{y^2}\right\},\left\{a\to -\sqrt{1-c^2},y\to 0\right\},\left\{a\to \sqrt{1-c^2},y\to 0\right\},\{x\to 0,y\to 0\}\right\} $$

which is anything but. So why (1) and (2) give me different results?

Answer 1

I don't see that anyone has written out $$ (a x+b y)^2+(c x+d y)^2=x^2+y^2 \tag{2b} $$

$$ a^2 x^2 + 2abxy + b^2 y^2 + c^2 x^2 + 2cdxy + d^2 y^2 = x^2 + y^2 $$ for all $x,y,$ which means the coefficients agree. $$ (a^2 +c^2) x^2 + (2ab+2cd)xy + (b^2 + d^2) y^2 = x^2 + y^2 = 1 x^2 + 0 xy + 1 y^2 $$

$$ a^2 + c^2 = 1 \; , \; \; 2(ab+cd) = 0\; , \; \; b^2 + d^2 = 1 $$

So your $$ \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) $$ has two unit vectors as columns, and these columns are orthogonal to each other. It is an orthogonal matrix.

Answer 2

The 2 equations $$M^TM=I$$ and $$v^TM^TMv=v^Tv$$ are equivalent if you put $\forall v$ in front of the second equation. This actually gives you infinitely many equations which together pin down the matrix to exactly $M^TM=I$. Solving the second equation for a single unknown vector is just using one of those infinitely many equations, which leaves more freedom on the matrix $M^TM$.

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To see why the two equations are equivalent, notice that $$(\forall c: c^Ta = c^Tb) \implies a = b \label{a}\tag{1}$$ Indeed, $$ \begin{aligned} (\forall c: c^Ta = c^Tb) &\iff (\forall c: c^Ta - c^Tb = 0)\\ &\iff (\forall c: c^T(a-b) = 0)\\ &\stackrel{*}\iff a-b = 0\\ &\iff a = b \end{aligned} $$ The equivalence indicated with $*$ says that 0 is the only vector that is orthogonal to all vectors. We notice that $a-b$ is orthogonal to all vectors so it must be $0$.

Also notice that $$(\forall v: v^TM^TMv=v^Tv) \implies (\forall v,w: v^TM^TMw=v^Tw) \label{b}\tag{2}$$ by using the polarisation identity twice: $$ \begin{aligned} v^TM^TMw &= \frac{1}{4}((v+w)^TM^TM(v+w) - (v-w)^TM^TM(v-w))\\ &= \frac{1}{4}((v+w)^T(v+w) - (v-w)^T(v-w))\\ &= v^Tw \end{aligned} $$

Applying \ref{a} with $a:=M^TMw$ and $b:=w$ and using \ref{b} we get $$ \begin{aligned} (\forall v: v^TM^TMv=v^Tv) &\implies (\forall v,w: v^TM^TMw=v^Tw)\\ &\implies (\forall w:M^TMw = w)\\ &\implies M^TM = I \end{aligned}$$

The other direction is trivial.

Answer 3

$x^2, xy, y^2$ are a basis for quadratic forms in $x$ and $y$, meaning

$$ax^2 + bxy + cy^2 = a'x^2 + b'xy + c'y^2 \text{ for all }x,y \in \mathbf{R}$$

if and only if $a = a', b = b', c = c'$.

If you set $x = 1, y = 0$, you get $a = a'$. If you set $x = 0, y = 1$ you get $c = c'$. Therefore we must also have $bxy = b'xy$ (for all $x, y$) from which it follows that $b = b'$.