• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QA 9 • QA 10 • QA 11 • QA 12 • QA 13 • QA 14 • QA 15 • QA 16 • QA 17 • QA 18 • QA 19 • QA 20 • QA 21 • QA 22 • QA 23 • QA 24 • QA 25 • QA 26 • QA 27 • QA 28 • QA 29 • QA 30 • QB 1(a) • QB 1(b) • QB 1(c) • QB 2(a) • QB 2(b) • QB 3(a) • QB 3(b) • QB 3(c) • QB 4(a) • QB 4(b) • QB 4(c) • QB 4(d) • QB 5(a) • QB 5(b) • QB 5(c) • QB 6(a) • QB 6(b) • QB 6(c) • QB 6(d) • QB 6(e) •
Question A 28
Circle all the matrix products below that are the identity matrix for all ,,.
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
How can you conceptualize the transformations. If that fails, is there a way to define transformation matrices and actually compute the resulting matrix of the desired transformations?
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
By the position of this question inside the exam, and given that it is a multiple choice question, it is very likely that a conceptual answer is sufficient.
With a good understanding of vector geometry, we can quickly see which operations are equivalent to the identity matrix. Recall that the identity matrix means that the vector you finish with is the same as the vector you begin with. We will go through each scenario and argue whether the net transformation recovers the original vector.
Recall that a reflection is angle preserving. For example, if we have a vector u that we want to reflect over some vector v and there is a counter-clockwise angle from u to v then the reflection will produce a new vector x that is a counter-clockwise angle from v, i.e. counter-clockwise from u. Therefore, in this case we have a vector that we are instructed to reflect over a line at an angle from the -axis. We are then instructed to take the new vector and reflect it once again over the same line. In other words we are asking for a "mirror-image of a mirror-image" which is the original vector. Therefore, we do get that this operation is equivalent to an identity matrix.
In this example we take a vector and for the first operation, rotate it an angle . We then take the new vector and rotate it an angle . This is equivalent to spinning something clockwise and then counter-clockwise by the same amount. Doing so would result in the original vector and therefore we do get that this operation is equivalent to an identity matrix.
For this question we need to understand what a projection operation does. If we want to project a vector u onto a vector v then the net result will be to determine all the components of u so that it lies parallel to v or points in the same direction. Therefore, in this case if our vector v makes an angle with the x-axis and we want to project some vector u onto it then after the first projection operation we will have a vector, w that lies on top of v. Any future attempts to project this new vector onto v will result in an identity because there is nothing left to make parallel to v. Therefore applying a second projection operator will just map w to itself. Therefore we have that we started with a vector u and ended with a vector w which for arbitrary u is not equivalent to an identity matrix operation. In fact the only way it could be an identity matrix is if the original vector were v itself so that the projection does nothing to the vector.
Recall from part (c) that projection is an operation that traps vectors to the original vector being projected on. This means that after one operation, further projections onto the same vector will result in no changes. Therefore in this example after the first operation we have a resultant vector in the direction of a vector at an angle with the x-axis. After the second projection, our resultant vector is in the direction of a vector making an angle with the x-axis. Therefore, unless the original vector made an angle with the x-axis then this resultant vector will not be the same as the original. Therefore, for any arbitrary starting vector, this will not be an identity matrix operation.
In this example we consider taking an initial vector, reflecting it by an angle of and then rotating it counter-clockwise by an angle . Consider if the original angle is clockwise of the reflector at . In this case if we reflect it so that it is now counter-clockwise to the reflector and rotate it even further counter-clockwise then there is no way of returning to the original vector. In order for it to be an identity operator, the operation has has to return the original vector for any arbitrary initial vector. We have shown with this class of examples that the operation will not produce the original vector ever and so it can't hold for any arbitrary initial vector. Therefore this is not an identity matrix operation.
Therefore, we conclude that only (a) and (b) are equivalent to identity matrix operations.
This question can of course also be solved computationally in terms of transformation matrices if one were so inclined. We will present this solution now as an exercise to those wishing to improve their understanding of linear transformations and also to appease the worries of those who struggle with understanding the conceptual framework in the previous solutions.
For the purposes of short form we will denote the following matrices
Recall that to obtain transformation matrices we take as the columns, the transformations on the basis vectors. Therefore, the first column would represent the transformation on the basis vector in the x-direction.
Consider rotating the vector by a counter-clockwise angle . Since [1,0] has magnitude 1 then the new vector u can be written in component form as
and therefore this is the first column of . We choose counter-clockwise angles to be consistent with measuring angles from the positive x-axis towards the positive y-axis. If we rotate the y-basis [0,1] by the same angle we'd get a vector v (which also has magnitude 1),
Note the negative sign in the x-component comes from the counter-clockwise rotation of [0,1] into the negative-x, positive-y quadrant. Therefore we see that the rotation matrix can be written as,
Reflection Matrix: To formulate the reflection matrix we will think of it in terms of rotation. We want to reflect the basis vectors around a vector called the reflector, that makes an angle with the x-axis. Recall that reflection is angle preserving so whatever angle the reflector makes with the original vector it must maintain that same angle with the new vector but in the opposite direction. For example if an initial vector is clockwise from the reflector than the new vector will be counter-clockwise from the reflector. The basis vector [1,0] is, by definition of the reflector, at a clockwise angle of from the reflector. Therefore the new vector must be an angle of counter-clockwise from the reflector or from the positive x-axis. Therefore we can consider reflecting [1,0] the same as rotating it by . Therefore the new vector u is
where once again the magnitude is 1 because the vector [1,0] has magnitude 1. Now the vector [0,1] makes an angle of clockwise from the y-axis with the reflector. Therefore the new vector v must be an angle of clockwise from the y-axis. However, we require that we measure angles from the positive x-axis and thus v makes an angle of
with the positive x-axis. Therefore we can think of v as a rotation from the x-axis of . Now recall that
Therefore we can write,
so that the reflection matrix can be written as
Projection: For the projection operator, we want to have an initial vector get projected onto a vector that makes an angle with the x-axis. Therefore, if we know immediately that the resultant vector u can be written as,
We also need to recall that the formula for projecting some vector v onto another vector w is
Now, if we call the basis vector [1,0], i, then we have,
Similarly, for the other basis vector, [0,1], which we call j, we get,
Therefore, the projection matrix is,
Now that we have the transformation matrices, we can answer the questions.
This is equivalent to
where we have required the identity . Notice, this is indeed an identity matrix.
This is equivalent to
where we have taken advantage of the odd and even properties of and . Notice this is and identity matrix.
This is equivalent to
which is not an identity matrix.
This is equivalent to
where we have used trig identities to simplify the argument . This is not an identity matrix.
This is equivalent to
which is not an identity matrix. Therefore, we conclude, like before, that only (a) and (b) are equivalent to an identity matrix.