# Thread: [SOLVED] Linear Algebra, trace similar matrices

1. ## [SOLVED] Linear Algebra, trace similar matrices

a) Show that if $\displaystyle A$ and $\displaystyle B$ are $\displaystyle n$-by-$\displaystyle n$ matrices with coefficients in an arbitrary field $\displaystyle F$, then $\displaystyle Tr(AB) = Tr(BA)$.

b) use (a) to show that if $\displaystyle A$ and $\displaystyle B$ are similar matrices, then $\displaystyle Tr(A) = Tr(B)$.

c) Show that the matrices of trace zero form a subspace of $\displaystyle M{_n}(F)$ of dimension $\displaystyle n{^2} - 1$. [Hint: The mapping $\displaystyle Tr: M{_n}(F) \rightarrow F$ is a linear transformation.]

d) Prove that the subspace of $\displaystyle M{_n}(F)$ defined in (c) is generated by the matrices $\displaystyle AB - BA,\ A,\ B \in M{_n}(F)$.

2. Originally Posted by JJMC89
a) Show that if $\displaystyle A$ and $\displaystyle B$ are $\displaystyle n$-by-$\displaystyle n$ matrices with coefficients in an arbitrary field $\displaystyle F$, then $\displaystyle Tr(AB) = Tr(BA)$.

b) use (a) to show that if $\displaystyle A$ and $\displaystyle B$ are similar matrices, then $\displaystyle Tr(A) = Tr(B)$.

c) Show that the matrices of trace zero form a subspace of $\displaystyle M{_n}(F)$ of dimension $\displaystyle n{^2} - 1$. [Hint: The mapping $\displaystyle Tr: M{_n}(F) \rightarrow F$ is a linear transformation.]

d) Prove that the subspace of $\displaystyle M{_n}(F)$ defined in (c) is generated by the matrices $\displaystyle AB - BA,\ A,\ B \in M{_n}(F)$.
a) If $\displaystyle A = (a_{ij})$ and $\displaystyle B = (b_{ij})$ then $\displaystyle \text{Tr}(AB) = \sum_{i,j=1}^na_{ij}b_{ji}$, and $\displaystyle \text{Tr}(BA)$ is given by the same formula.

b) By a), $\displaystyle \text{Tr}(P^{-1}(AP)) = \text{Tr}((AP)P^{-1})$.

c) Use the hint. The subspace consisting of matrices in $\displaystyle M_n(F)$ with trace 0 is the kernel of the map in the hint.

d) The identity $\displaystyle A = \bigl(A-\tfrac1n\text{Tr}I_n\bigr) + \tfrac1n\text{Tr}I_n$ shows that every n×n matrix can be written as the sum of a matrix with trace 0 (namely the expression in parentheses) and a multiple of the identity.

Write $\displaystyle E_{ij}$ for the matrix having a 1 in the (i,j)-position and zeros everywhere else. If $\displaystyle i\ne j$ then $\displaystyle E_{ij} = E_{ii}E_{ij} - E_{ij}E_{ii}$. By a linear combination of such matrices you can construct any matrix with zeros along the diagonal as a linear combination of commutators. Next, $\displaystyle E_{11}-E_{ii} = E_{1i}E_{i1} - E_{i1}E_{1i}$ for all $\displaystyle i>1$. Check that by forming linear combinations of such matrices you can form any diagonal matrix that is not a scalar multiple of the identity.

That shows that any matrix with trace 0 can be expressed as sum of commutators.