Suppose B represents the matrix of orthogonal (perpendicular) projection of $\mathbb{R}^{3}$ onto the plane $x_{2} = x_{1}$. Compute the eigenvalues and eigenvectors of B and explain their geometric meaning.
What I have I come up so far as an attempt to deduce this question down is, for instance. If we pick an arbitrary point in space ($\mathbb{R}^{3}$), then we must project this point onto a plane (in particular $x_{2} = x_{1}$) which I imagine on a three dimensional axis of ($x,y,z$), if we choose $x_{1}$ to represent the $x$-axis, $x_{2}$ to represent the $y$-axis, and $x_{3}$ to represent the $z$-axis then we would have a plane of the equation that looks like $y=z$ or conversely ($x_{2}=x_{1}$) which is its equivalent. Once you project this point onto the plane, I see that it is true to be perpendicular and its vector is coming out of the plane. My troubles are finding the new coordinates of the new point that is projected onto the plane.
Here is a skeleton sketch of what I had in math-ese.
$\left[\begin{array}{c}
?\\
?\\
? \end{array} \right] =
\left[\begin{array}{ccc}
\Box & \Box & \Box \\
\Box & \Box & \Box \\
\Box & \Box & \Box
\end{array} \right]
\left[\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3}
\end{array} \right] $, $~~$ where B is the matrix with empty boxes for elements.
These question marks inside of the first matrix represent the coordinates in which I am trying to find. Once these our found, making some appropriate choices for the entries in the coefficient matrix labeled B in the question can be found, so that when B is multiplied by the last matrix $\left(\left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array} \right]~\right) $, we will get back the matrix with the ? marks in the entries. I believe this is known as doing a linear transformation. I didn't know how to include graphics, but I hope the words was enough detail to be able to duplicate what I am saying on paper in a graphical meaning. If not, please let me know how I can clarify anything up.
Some help would be very appreciated.
Thanks
Best Answer
Kristi, first of all if you are projecting $\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$ onto the $x_{1}=x_{2}$ plane, then you are projecting onto the plane $x=y$, not $y=z$ (since you defined $x=x_{1}$, $y=x_{2}$, and $z=x_{3}$).
Now, as you are trying to find the coordinates of the projection vector, imagine the geometric meaning -- $z$, the 'height' of the vector will not ever change, as it is not relevant to the equation, but $x$ and $y$ will, depending on where the vector lies. When we are trying to find a projection on an n-dimensional subspace $W$, we can use a formula of ${proj_{W}}{\vec{x}}$=$(\vec{u_1}\cdot \vec{x})$$\vec{u_1}$+$(\vec{u_2}\cdot \vec{x})$$\vec{u_2}$$+\cdots +$$(\vec{u_n}\cdot \vec{x})$$\vec{u_n}$, where $\vec{u_1}, \vec{u_2}\dots \vec{u_n}$ form an orthonormal basis of the subspace $W$. Here, $W$ is defined as $x=y$, meaning it can be spanned by vectors $\vec{v_1}$ = $ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ and $\vec{v_2}$ = $ \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$, for example. To find an orthonormal basis of our space (meaning that all vectors in it will be mutually orthogonal/perpendicular, as well as of a length one), let's use the Gram-Schmidt process. An orthonormalized version of the vector $ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ would be $\vec{u_1}$ = $\frac{1}{\sqrt{2}}$$ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$, as that will make it of length one. Now, by Gram-Schmidt, $\vec{u_2}=\vec{v_2}-\frac{\vec{v_2}\cdot\vec{u_1}}{\vec{u_1}\cdot\vec{u_1}} \vec{u_1}$, since we are basically subtracting the $\vec{u_1}$ component from our second vector, in order to get a vector perpendicular to $\vec{u_1}$ as a result. Calculations result into the following: $\vec{u_2}$= $\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$ $-$$\frac{\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}}{\frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}} \frac{1}{\sqrt{2}} \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ = $\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$ - $\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$=$\begin{bmatrix} 0\\ 0\\ 2 \end{bmatrix}$. Normalizing the resulting vector, we get $\vec{u_2}$ = $\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$.
Now that we have an orthogonal basis $\vec{u_1}$ = $\frac{1}{\sqrt{2}}$$\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ and $\vec{u_2}$ = $\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$, we can calculate the projection.
So, to find the projection of your vector $\begin{bmatrix} x_1\\\ x_2\\\ x_3 \end{bmatrix}$ we use our orthonormal basis and the projection formula: ${proj_{W}}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$=($\frac{1}{\sqrt{2}}$$\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} \cdot \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix})(\frac{1}{\sqrt{2}}$$\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix})$$+$$(\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} \cdot \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix})(\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix})$. After arithmetic, this results into ${proj_{W}}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$=$(\frac{x_1+x_2}{\sqrt{2}})$$(\frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix})$+$\begin{bmatrix} 0\\ 0\\ x_3 \end{bmatrix}$=$\begin{bmatrix} \frac{x_1+x_2}{2}\\ \frac{x_1+x_2}{2}\\ x_3 \end{bmatrix}$.
So, now you have your coordinates.
To find out the eigenvalues, think of the nature of the transformation -- the projection will not do anything to a vector if it is within the plane onto which you are projecting, and it will crash it if the vector is perpendicular to the plane. So, your eigenvalues are 1 and 0. A basis of eigenspace of 1 $\xi_{1}$ will have two vectors, as the plane is spanned by two of them. You could choose them to be your original $v_1$ and $v_2$, which were $ \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$. To find a basis of eigenspace of 0 $\xi_{0}$, you need to find a vector perpendicular to this plane. You could use a property of cross-product, which states that $\vec{v_1} \times \vec{v_2}$ produces a vector $\vec{v_3}$ perpendicular to both. Crossing the aforementioned vectors, you get $\vec{v_3}=\begin{bmatrix} 2\\ -2\\ 0 \end{bmatrix}$.
Now that you know all of this, finding the matrix B is very easy by inspection; consider $\begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}$.