# Todai Entrance Exam: Math 2018 – Problem 1

**(1) **Use Gaussian Elimination to address (i). All the rest can be easily inferred.

**(2)** Proof by contradiction:

Assume and the solution of the simultaneous linear equation does not exist.

If then . However, and , so is guaranteed to have linearly dependent columns. Therefore, we can always represent as a linear combination of other column vectors in . As the solution of the simultaneous linear equation is guaranteed to exist, this contradicts the assumption.

So the statement follows.

**(3) **There are various solutions to this problem. But I am not sure if it is appropriate to apply the normal equation directly.

Let . We set the partial derivative of to be zero to find critical points:

Given , if then (proof). So is invertible. Therefore, we can write:

Now we need to prove that critical point is the local minimum (and also global minimum as we only have one critical point):

is a positive-semidefinite matrix (as ), so the found critical point with is the local minimum and also the global minimum.

**(4) **Let’s formulate the problem:

We have:

Let:

We obtain:

Let’s find :

So: . Given , if then . Therefore, is invertible and we can write .

Prove that the found critical point is also the constrained local minimum:

So, the critical point with is the constrained local minimum and also the constrained global minimum.

**(5)** How can we compute Pseudoinverse for any Matrix? is the pseudoinverse of .

**(6)** for (3) and for (4). I really don’t know what this question is for.

## 1 comment on “Todai Entrance Exam: Math 2018 – Problem 1”