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Matrix representation of orthogonal projection onto the line defined by $L=span(x_1,x_2,x_3)$ where $x_1^2+x_2^2+x_3^2=1$

My attempt:

The projection is defined by $$p_L(v)=\frac{v\cdot(x_1,x_2,x_3)}{\mid\mid x_1,x_2,x_3\mid\mid}*(x_1,x_2,x_3)\implies p_L(1,0,0)=x_1*(x_1,x_2,x_3)=(x_1^2,x_1x_2,x_1x_3)$$

repeating this process for $(0,1,0)$ and $(0,0,1)$ we get the matrix representation of the projection:

$$ M= \begin{bmatrix} x_1^2 & x_1x_2 & x_1x_3 \\ x_1x_2 & x_2^2 & x_2x_3 \\ x_1x_3 & x_2x_3 & x_3^2 \end{bmatrix} $$

The follow up question is whether or not this is diagonalizable, going through the steps I don't see any cancellations that give nice eigenvalues, so I am suspecting something is wrong with my matrix. Any help is appreciated, thanks!

Mike
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  • I think it's worth to note that the projection to a normal vector $n$ is just $n \circ n$, with $(n \circ n)_{ij}=n_i n_j$. – Botond Aug 08 '18 at 05:39
  • Hint: what are the ranges and kernels of the projection? – amd Aug 08 '18 at 07:36

2 Answers2

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Hint: consider any orthonormal basis (of $\mathbb{R}^3$) that contains your special vector $(x_1, x_2, x_3)$ as one of the basis elements. Are these basis elements eigenvectors?

angryavian
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$M$ is real symmetric, so not only is it diagonalizable, it is orthogonally diagonalizable.

If you didn’t happen to remember this fact about symmetric matrices, you could proceed directly from what you know about $M$. By construction, its column space is $L$ and its null space is $L^\perp$, with dimensions $1$ and $2$, respectively. The null space of $M$ is obviously the eigenspace of $0$. Moreover, for $v\in L$, $p_L(v)=v$, i.e., it is the identity map on $L$, so $1$ is an eigenvalue of $M$ with eigenspace $L$. Finally, $\mathbb R^3 = L\oplus L^\perp$, the direct sum of eigenspaces of $M$, which means that $M$ is diagonalizable.

As an aside, every projection matrix $M$, orthogonal or not, satisfies $M^2=M$, so the only possible eigenvalues of a projection are $0$ and $1$.

amd
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