Prove $tr(\mathbf{A} + \mathbf{B} ) = tr(\mathbf{A}) + tr(\mathbf{B})$

Question

Supposedly, this is an easy proof. But I'm really inexperienced and have little mathematical sophistication (trying to improve).

Prove $tr(\mathbf{A} + \mathbf{B} ) = tr(\mathbf{A}) + tr(\mathbf{B})$

Since the size of either A or B is not defined, I guess the summation definition of trace of a matrix must be used in the proof.

So,

$tr\mathbf(A) = \sum_{k=1}^n \mathbf{A}_{k,k}$ where $\mathbf{A}\in\mathbb{R}^{n,n}$

Any intuitive hints would be nice

Answer 1

Realized how basic this proof is thanks to Winther's comment.

So the definition is that summing 2 matrices means adding each $ij$th element to get the corresponding element in a new matrix. Or,

if $\mathbf{C} = \mathbf{A} + \mathbf{B}$, where $\mathbf{A},\mathbf{B} \in \mathbb{R}^{m,n}$, then $\mathbf{C}_{ij} = \mathbf{A}_{ij} + \mathbf{B}_{ij}$

So,

Statement:

$tr(\mathbf{A} + \mathbf{B} ) = tr(\mathbf{A}) + tr(\mathbf{B})$

Proof:

if $tr(\mathbf{A}) = \sum_{k=1}^n \mathbf{A}_{k,k}$, then

$tr(\mathbf{A+B}) = \sum_{k=1}^n \mathbf{(A+B)}_{k,k}$

By definition $\mathbf{(A+B)}_{ij} = \mathbf{A}_{ij} + \mathbf{B}_{ij}$ thus, $\sum_{k=1}^n \mathbf{(A+B)}_{k,k} = \sum_{k=1}^n (\mathbf{A}_{k,k}+\mathbf{B}_{k,k})$

Distribute the summation notation on the right hand side to get $\sum_{k=1}^n \mathbf{(A+B)}_{k,k}=\sum_{k=1}^n \mathbf{A}_{k,k} + \sum_{k=1}^n \mathbf{B}_{k,k}$

Finally, replace the summation notation with trace to get

$tr(\mathbf{A} + \mathbf{B} ) = tr(\mathbf{A}) + tr(\mathbf{B})$