$$T\binom{x}{y} = \binom{2x-y}{2x-y}$$ is a projection of $\mathbb{R}^2$ onto the line $y = x$, but not with respect to the usual inner product. Is it an orthogonal projection with respect to some inner product?
If so, I'd like to find that inner product; if not, prove that no such inner product exists.

I see that the matrix of the transformation (with respect to the standard basis) is a projection matrix, i.e., $$ P = \begin{bmatrix} 2 & -1 \\ 2 & -1 \end{bmatrix} $$ And as usual for a projection, $P^2 = P$. However, $P \neq P^T$, unlike what we'd expect for an $\textit{orthogonal}$ projection. Thus $P$ is not symmetric, and the kernel of its transformation is not orthogonal to its range.

So I suspect no such inner product exists. But I'm unsure how to relate all this back to machinery of the inner product happening here, in order to address the prompt. Any insights on this would be much appreciated!


1 Answer 1


There actually do exist such inner products. Note that $v_1=(1,2)$ spans the kernel and that $v_2=(1,1)$ spans the range of $P$. Let $S=\begin{bmatrix}a & b \\ b & c \end{bmatrix}$ be a symmetric matrix which defines an inner product $\langle v,w \rangle = v^T S w $. We want that the kernel and range of $P$ are orthogonal. So the inner product must satisfy $v_1^T S v_2 = 0$ or $$ \begin{align*} \begin{bmatrix} 1 & 2 \end{bmatrix} \begin{bmatrix} a & b \\ b & c \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = 0\\ a+3b+2c=0. \end{align*} $$ The matrix also has to be positive definite. By Sylvester's criterion, this is the case if $a > 0$ and $ac-b^2 > 0$.

There are infinitely many solutions $a,b,c$ satisfying these conditions, so there are many inner products such that $P$ is an orthogonal projection w.r.t. to them. For instance, take $a=5$, $b=-3$ and $c=2$.

I also want to remark that w.r.t. to the basis $\{(1,2),(1,1)\}$ the matrix of the projection becomes $$ \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, $$ which is symmetric.

  • 1
    $\begingroup$ Just a minor detail, $S$ also has to be positive definite. (Which your example is.) $\endgroup$
    – J_P
    Jun 3, 2019 at 17:15
  • 1
    $\begingroup$ @J_P Good point. I added an extra argument in the answer. $\endgroup$
    – Ernie060
    Jun 3, 2019 at 20:26

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