I need help verifying a few linear algebra proofs for trace of a matrix transposed and scalar distribution

Question

First, I would like to verify two entry proofs for trace matrices and one very trivial entry proof. I greatly appreciate any help or hints. Please don't go easy on me, serious criticism wanted to help my journey to formal proof writing. Alternative proofs more than welcomed.

1. given: definition of a trace(A)=a11+a22+...+ann. prove tr(cA)=ctr(A), where c is a scalar. my proof: We know that by the definition of scalar multiplication,the (i, j)-entry of cA is caij. The trace of cA, by definition, must be ca11+ca22+...+cann.If we factor out the scalar multiple, we get c(a11+a22+...+ann). by definition, this is ctr(A). This next part I am unsure of: For tr(cA)...we again know the (i, j)-entry of cA is caij. Mult. is associative, so we can write it as c * aij, if we take the tr(cA) we can write it as tr(A) * c because of associativity (?), that would be tr(A)=a11+a22+...+ann * c or c * tr(A) Questions: 1. is this right? 2. which side of the equation is best to start with in proofs?

2. prove trace(A^t)=trace(A) my proof: we known A^T=A for a symmetric matrix. trace(A^t)-trace(A)=0. factor trace--> (A^T - A)trace=0. by definition, the trace of matrix A is a11+a22+...+ann.

3. Lastly, if anyone has the the time to verify, prove (s+t)A=sA+tA my proof: we know by def. of scalar multiplication that the (i j) entry of sA is saij, and for tA taij, so sA+tA is saij +taij (correct wording for this?). Question: I am not sure where to go with (s+t)A. besides showing that if it distributes, that it would be saij +taij and the entries would be the same as above.

Answer 1

As noted by Robert in the comments, you are using the distributive property, not the associative property. The proofs of 1) and 3) are correct, but for 2) you have only shown that $\operatorname{tr}(A) = \operatorname{tr}(A^T)$ when $A$ is a symmetric matrix. You can't really factor out the trace as you did in 2), but you can use the fact that taking the trace is a linear operation to 'factor out' trace as $$ \operatorname{tr}(A) + \operatorname{tr}(B) = \operatorname{tr}(A + B). $$ However, you should prove this before you use it (you don't need it to show that $\operatorname{tr}(A) = \operatorname{tr}(A^T)$ but it is a nice exercise to practise your proof-writing. As a bonus, together with 1) this shows that trace is a linear map).

When writing proofs, you should be clear about what assumptions you are making. For instance, it is clear from context that $A$ is a square $n\times n$ matrix and that $c$ is a scalar value, however it is not clear whether $c$ and the entries of $A$ are real numbers, complex numbers, or even elements of some ring $R$. In general, it is better to be more explicit than less.

This also applies to the justifications of the steps you take. In general, the way you write your proof depends on your target audience. You have to use your own judgment here, and this is something that will come with experience.

For completeness sake, here is how I would write your proofs (I will assume you are working with real numbers):

Let $A$ be a real $n \times n$ matrix, and let $c \in \mathbb R$. Prove that $\operatorname{tr}(cA) = c\operatorname{tr}(A)$.

By definition of the trace, we have that $\operatorname{tr}(cA) = (cA)_{11} + (cA)_{22} + \ldots + (cA)_{nn}$. By definition of scalar multiplication, we have that $(cA)_{ij} = cA_{ij}$ for $1 \leq i, j \leq n$, so that $\operatorname{tr}(cA) = cA_{11} + \ldots + cA_{nn}$. By the distributive property, we can factor out $c$ and find $\operatorname{tr}(cA) = c(A_{11} + \ldots + A_{nn}) = c\operatorname{tr}(A)$, where in the last equality we used that $\operatorname{tr}(A) = A_{11} + \ldots + A_{nn}$.

Let $A$ be a real $n \times n$ matrix. Prove that $\operatorname{tr}(A) = \operatorname{tr}(A^T)$.

Note that by definition of the transpose, we have that $(A^T)_{ij} = A_{ji}$ for $1 \leq i, j \leq n$. In particular, this means that $(A^T)_{ii} = A_{ii}$. Thus we see that $$ \operatorname{tr}(A^T) = (A^T)_{11} + \ldots + (A^T)_{nn} = A_{11} + \ldots + A_{nn} = \operatorname{tr}(A) $$

Let $A$ be a real $n \times n$ matrix, $s, t \in \mathbb R$. Prove that $(s + t)A = sA + tA$.

We know by definition of scalar multiplication that $(sA)_{ij} = sA_{ij}$, and similarly for $tA$. Thus $(sA + tA)_{ij} = (sA)_{ij} + (tA)_{ij} = sA_{ij} + tA_{ij}$ by definition of matrix addition. By the distributive property, this is equal to $(s + t)A_{ij}$. On the other hand, we have that $((s + t)A)_{ij} = (s + t)A_{ij}$ by definition of scalar multplication, so $((s + t)A)_{ij} = (sA + tA)_{ij}$. Thus the entries of $(s + t)A$ and $sA + tA$ are equal, hence $(s + t)A = sA + tA$.