3
$\begingroup$

What is the dimension of the space of all $n \times n$ matrices with real entries which are such that the sum of the entries in the first row and the sum of the diagonal entries are both zero?

I tried by finding number of independent entries. The number of independent entries on the diagonal is $n-1$. The number of upper triangular independent entries is $\frac{n(n-1)}{2}-1(n-1)$, and the number of lowertriangular independent entries is $\frac{n(n-1)}{2}$. Now adding them will give dimension. Am I right?

  • 0
    Well yes, of course. Repeating exactly the same constraint (or a linearly dependent one) changes nothing. It was a suggestion of how to proceed, not a recipe to be followed blindly. You still need to think!2012-05-08

2 Answers 2

2

All entries except the lower and upper right can be chosen freely, and no matter what you chose them to be, there exists exactly one value for each of those two spots which gives you a matrix on the form you want. This gives $n^2 - 2$.

Roughly speaking, when you have a vector space over $\mathbb R$, $\mathbb C$, or other fields, each equation relating the coordinates will lower the total dimension by one. In this case you have two equations relating coordinates in a vector space of dimension $n^2$ (forgetting for a moment that it is a matrix, and just look at it as an $n^2$-dimensional vector).

2

Assuming $n\geq2$, you have $n^2$ variables (the entries) and 2 linear equations, which are independent. Hence you have $n^2-2$ free variables. Your space is the space of solutions to those equations, so the dimension of this space will be $n^2-2$