Give 2x2 matrices such that for $x \in \Bbb R$

Question

$Ax$ is the vector obtained projecting $x$ onto the line $x_1=x_2$

I can not understand what the exercise asks, or how to start to solve it.

The funny thing is that this exercise is within the portion of linear transformations, and I can not find the relationship.

I hope someone can help me and guide me to bring it out.

Answer 1

This makes no sense to me. $x$ must be an element of $\mathbb R^2$ if you want to take it as an argument of a linear transformation $A:\mathbb R^2\to\mathbb R^2$. In this case, I guess the exercise to be: Find the orthogonal projection mapping $\mathbb R^2$ onto the subspace spanned by $e_1+e_2$.

Answer 2

The line $x_1 = x_2$ is given by the direction vector $[1;1]$. $A$ is the "projection onto the direction given by $[1;1]$" matrix. So, $A x = \frac{<[1;1], x>}{||[1;1]||} [1;1]$ (the inner product with normalization gives you the projection's magnitude, and the $[1;1]$ gives direction). Now, note that $<[1;1],x> = [1;1]^T x = x^T [1;1]$, and re-arrange it so that you get $(stuff)x$. Then, $A=stuff$.

You can read the vector projection article on wikipedia for more details (in particular, the section on "Matrix Representation".

Answer 3

Well, I presume you are looking for an orthogonal projection (that is $Ax$ is the nearest point to the line $x_1=x_2$).

It takes a little bit of work to demonstrate that the projection operator is a linear operator. The nice thing about a linear operator is that you only need to figure out how it affects a set of basis vectors and then you know how it behaves everywhere.

I am assuming that you are asking how to compute the $A$ rather than to demonstrate that the orthogonal projection is linear.

One way is to see how the matrix $A$ operates on the basis vectors $e_1=(1,0)^T$, $e_2 = (0,1)^T$.

If you draw a picture, you will see that $e_1$ is mapped to $({1 \over 2}, {1 \over 2})^T$, and $e_2$ is also mapped to $({1 \over 2}, {1 \over 2})^T$.

Hence the matrix is $A = \begin{bmatrix} {1 \over 2} & {1 \over 2} \\ {1 \over 2} & {1 \over 2}\end{bmatrix}$.

Note that we could have created a projection that was not orthogonal, but we need to ensure that the operator is a projection, that is $A^2 = A$. For example, we could map $e_1 \to (1,1)^T$, which would require that $A^2 e_1 = A (1,1)^T = A(e_1+e_2) = e_1+e_2$, which in turn gives $A e_2 = 0$, so the matrix $A = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$ would also work.