Nullity of the trace of a matrix

[JKCB]

Homework Statement

What is the nullity of the trace (A), A is an element of all nxn square matrices.

The Attempt at a Solution

the null space would be when the sum of the diagonal is equal to 0. So the Σaii for i=1 to n must equal 0 which would be when aii = -aii. Therefore the nullity of the trace would equal 1? Does this look right?

 

[arkajad]

Take n=4. Contemplate:

1+2+3+(-6)=0

 

[HallsofIvy]

Saying that $\sum a_{ii}= 0$, that is, the sum of numbers, implies that each number must be 0 is absolutely false! The space of all n by n matrices has dimension $n^2$. Adding that the trace is 0 adds one equation that must be satisfied and so reduces the "degrees of freedom" in choosing the entries by 1.

 

[JKCB]

Thank you all, I'm feeling quite not so smart!
HallsofIvy that makes sense, so if I had just said the sum of the diagonal must equal zero and not used the notation that I used that would have been correct? Which means the nullity of the trace is equal to 1?

 

[HallsofIvy]

NO! You have this exactly backwards! The space of 2 by 2 matrices, of the form
$$\begin{bmatrix}a & b \\ c & d\end{bmatrix}$$
has dimension 4 because you are free to choose each of the four numbers a, b, c, and d, any way you want. Saying that the trace is 0 means that a+ d= 0. You are still free to choose b and c to be anything you want. And you could choose either a or b to be anything you want- then the other is given by the equation: either a= -d or d= -a.

That was what I meant when I said that the one equation "reduces the "degrees of freedom" in choosing the entries by 1." You can now choose 3 of the entries to be anything so the "number of degrees of freedom" in choosing entries, which is the same as the dimension of the vector space, is 4- 1= 3.

The space of all 2 by 2 matrices, whose trace is 0, is 3.

More generally, since "trace equals 0" gives one equation and reduces the "degrees of freedom" in choosing entries for the matrix by 1, the dimension of the space of all n by n matrices, whose trace is 0, is $n^2- 1$, not 1.

But I may be confused by your phrase "the nullity of the trace". To me "nullity" is the dimension of the null space of a linear transformation and "trace" is a number. You cannot have a "nullity of the trace". I thought perhaps you were talking about a linear transformation that took a general matrix to a matrix that has trace equal to 0 but that is not "well-defined". In any case, what I am saying is that the dimension of the space of all n by n matrices, with trace 0, is $n^2- 1$, which is the best interpretation of this problem I can come up with.

 

[JKCB]

Yes I am talking about the linear transformation, sorry. For some reason, its hard for me to let the nullity be that large. Most of the problems I have worked had a nullity of 0 or 1. This one threw me a little(lot). Thank you. Your detailed explanation is wonderful. Sorry to be so dense.