Verifying that a matrix T represents a projection operation

[Seydlitz]

Hello guys,

I want to verify or rather show that a given matrix $T$ does represent a projection from $\mathbb{R^{3}}$ to a particular plane, also lying in $\mathbb{R^{3}}$. Would it be enough to pre-multiply that matrix to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane?

Or do I need to row reduce the matrix $T$ until I can see the basis vectors used in the original $T$, and verify that they all lie on the plane? Or rather since I can also get the basis of the kernel, will showing that the basis of the kernel is parallel with the normal of the plane enough? Geometrically I imagine that the kernel space is all of the vectors that are orthogonal to the plane and their projection to that plane will be 0.

Thanks

 

[jbunniii]

Hello guys,

I want to verify or rather show that a given matrix $T$ does represent a projection from $\mathbb{R^{3}}$ to a particular plane, also lying in $\mathbb{R^{3}}$. Would it be enough to pre-multiply that matrix to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane?

This is necessary but not sufficient. If $n$ is normal to the plane and $\langle n, Tx\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix
$$T = \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
This matrix has the property that $\langle n, Tx\rangle = 0$, where $n = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1.

For another example, consider the matrix
$$T = \begin{pmatrix}0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}$$
Once again, we have $\langle n, Tx\rangle = 0$, where $n = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this is still not a projection because it stretches vectors lying in the plane, e.g. it maps $\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$ to $\begin{pmatrix}0 \\ 2 \\ 0\end{pmatrix}$. So you also need a constraint ensuring that no such stretching (and also no rotating) occurs. This is neatly captured by the condition $T^2 = T$. Indeed, a matrix represents a projection if and only if it satisfies $T^2 = T$.

So to summarize, if you want to show that a 3x3 matrix is a projection onto a particular plane, you need to verify all of the following:

1. $T^2 = T$, so $T$ is a projection
2. $\dim(\ker(T)) = 1$ or equivalently, $\dim(\text{im}(T)) = 2$, so $T$ projects onto a plane
3. $\langle n, Tx\rangle = 0$ where $n$ is normal to the plane and $x$ is arbitrary, so $T$ projects onto the specified plane

You can find equivalent conditions which will allow you to do less work. [strike]For example, you can replace condition 3 with $Tn = 0$.[/strike] [correction: If condition 3 is replaced with $Tn=0$ then not only is $T$ a projection, it is in fact an orthogonal projection.] But the basic idea remains the same.

 

[Seydlitz]

Thanks for the comprehensive information jbunniii, I really appreciate it.

By the way I just realized the fact that the kernel space of a projection matrix is orthogonal to the image of the the projection. Is this true in general? For example if I have a subspace $W$ and a linear transformation from a vector space $V$ to $W$. Can we consider the kernel of that transformation as the orthogonal complement of $W$?

 

[jbunniii]

Thanks for the comprehensive information jbunniii, I really appreciate it.

By the way I just realized the fact that the kernel space of a projection matrix is orthogonal to the image of the the projection.

Actually, that's only true of orthogonal projections. For general projections, it need not be true. Consider for example
$$T = \begin{pmatrix}0 & 0 \\ c & 1 \\ \end{pmatrix}$$
This is a projection matrix, because $T^2 = T$. The image is the subspace consisting of all scalar multiples of $\begin{pmatrix}0 \\ 1 \end{pmatrix}$. The kernel is the subspace consisting of all $\begin{pmatrix}x \\ y \end{pmatrix}$ satisfying $cx + y = 0$, or in other words, all scalar multiples of $\begin{pmatrix}1 \\ -c \end{pmatrix}$.

You can think of $T$ as a source of light aimed in the direction of $\begin{pmatrix}1 \\ -c \end{pmatrix}$, which projects a given vector onto its "shadow" on the image subspace. If the source of light is directly overhead ($c = 0$) then it's an orthogonal projection, otherwise it's called an oblique projection.