Question

What is the dimension of the vector space consisting of five-by-one column matrices where the rows sum to zero and the first row is equal to the second row?

a. 5
b. 4
c. 3
d. 2

Answer 1

Option c.

Explanation:

If we have a vector of N components (or variables), and we have K linear independent restrictions for these N components (such that K < N, we can't have more restrictions than components.)

The dimension of the vector will be given by N - K.

Here we know that we have a vector of 5 components, that can be written as:

`v = \left[\begin{array}{ccc}v_1\\v_2\\v_3\\v_4\\v_5\end{array}\right]`

And we have two restrictions, so we can expect that the dimension of the vector is:

5 - 2 = 3

But let's see it, the restrictions are:

"the first row is equal to the second row"

Then we can rewrite our vector as:

`v = \left[\begin{array}{ccc}v_1\\v_1\\v_3\\v_4\\v_5\end{array}\right]`

Notice that now we have only 4 variables, v₁, v₃, v₄, and v₅

We also know that the sum of the rows is equal to zero, thus:

v₁ + v₂ + v₃ + v₄ + v₅ = 0

we know that v₂ = v₁, so we can replace that to get:

2*v₁ + v₃ + v₄ + v₅ = 0

Now we can isolate one of the variables, to write it in term of the others, for example, let's isolate v₅:

v₅ = -2*v₁ - v₃ - v₄

Now if we replace that in our vector, we have:

`v = \left[\begin{array}{ccc}v_1\\v_1\\v_3\\v_4\\-2*v_1 - v_3 - v_4\end{array}\right]`

Notice that our vector depends on only 3 variables, v₁, v₃, and v₄, so we can define our vector in a 3-dimensional space.

Then the correct option is c, the dimension of the vector space is 3.