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Actually i am not asking about specific projection operator, this is question about about projection in general.

So, if we are talking about projections in three dimensions (for example $\mathbb{R}^3$), we actually can project a vector onto:

1) a line spanned by one vector from $\mathbb{R}^3$

2) a plane spanned by two vectors from $\mathbb{R}^3$

In case 1) image of matrix of projection would be a line (geometrically), since it maps every vector onto a line, now, since vector that lays on line maps onto itself that would mean that $Px=x$ which means that $1$ is one of the eigenvalues, now, since we are talking about three dimensional space, we can find a vector that is orthogonal to the vector laying on a line and since it's orthogonal on that vector it is orthogonal to the whole line so that vector maps onto a zero, so zero is one eigenvalue, we can repeat this with another vector so zero is eigenvalue with algebraic multiplicity two.

Similar thing is in case 2) only difference is that we would have algebraic multiplicity of one to be $2$, and of zero is one.

Now, to the actual question, if we have a matrix $3 \times 3$ whose eigenvalues are $1, 0, 0$ is it always a matrix of projection onto a line, furthermore, if we have this matrix, how we can find it's mapping rule? I mean, let's say we have a projection onto line spanned by $a=(a_1, a_2, a_3) \in \mathbb{R}^3$, and let's say $x$ is an argument, is mapping rule for projection operator always $P(x)=(x,a)a=x^Ta\vec{a}$ regardless of line we map onto or which argument we choose? I'd like an explanation on this topic in terms of geometry behind this.

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  • $\begingroup$ @M.Winter exactly, it was a typo, thanks for the heads up! $\endgroup$ Commented Oct 13, 2017 at 11:56

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A matrix $P$ is called a projection if $P^2=P$. From this, it is easy to see that all eigenvalues must be $0$ or $1$. On the other hand, a matrix with only eigenvalues $0$ and $1$ may not necessarily a projection.

Example. Take

$$A=\begin{pmatrix} \color{red}1&0&0\\ 0&\color{red}0&\color{blue}1\\ 0&0&\color{red}0 \end{pmatrix}.$$

This matrix is not diagonalizable. It is already in its Jordan normal form. Its eigenvalues are $\{\color{red}1,\color{red}0,\color{red}0\}$ but

$$A^2=\begin{pmatrix} \color{red}1&0&0\\ 0&\color{red}0&\color{blue}0\\ 0&0&\color{red}0 \end{pmatrix}\not=A.$$

Any projection onto a line (through the origin) spanned by a normalized vector $a\in\Bbb R^3$ can be given as

$$x\quad\mapsto\quad a\cdot\langle x,\;a\rangle_M$$

for some inner product $\langle x,\;a\rangle_M:=x^\top M a$ (where $M$ is a symmetric positive definite matrix). So there are more such projections like the one you have given, but the only difference is that the inner product can be a general one and must not be the standard product of $\Bbb R^3$.

Such projections with $M\neq I$ (where $I$ is the identity matrix) might not be "perpendicular" as you know them from a geometric point of view. But they still obey the definition of a projection.

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  • $\begingroup$ Isn't it enough for matrix to be non-diagonisable in order to prove it's not projection matrix, i mean, projection matrix has three linearly independent eigenvectors, is that right? $\endgroup$ Commented Oct 13, 2017 at 12:11
  • $\begingroup$ @cdummie Probably. But any proof of this has to show that $A^2\not=A$ anyway, so why not giving this result explcitiely? ;) $\endgroup$ Commented Oct 13, 2017 at 12:14
  • $\begingroup$ I know, i am just trying to get some deeper understanding of this so i could more easily handle any problems regarding projections, one more thing, in your definition of inner product you added index $M$, what does it actually mean? Is it with respect to another basis, while i was using standard one in my definition (and therefore there was no need for matrix between x and a since it was identity matrix, is that right)? $\endgroup$ Commented Oct 13, 2017 at 12:21
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    $\begingroup$ @cdummie Yes, you used $M=\Bbb I$, i.e. $x^\top \Bbb I a=x^\top a$. And yes, for $M\not=\Bbb I$ you can think about it as the "perpendicular" inner product in some other basis. In the end, what is considered "perpendicular" is only defined by an inner product. Usually we assume our notion of being perpendicular is given by $x^\top a$. $\endgroup$ Commented Oct 13, 2017 at 12:27
  • $\begingroup$ Great, now i understand it much better, thanks! $\endgroup$ Commented Oct 13, 2017 at 12:31

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