If $A,B$ have the same trace, then $A=T^{-1}BT$, for some invertible matrix $T.$ Let $A, B$ be $n\times n$ invertible matrices. If $A,B$ have the same trace, could someone advise me how to show $A=T^{-1}BT$, for some invertible matrix $T$ ? Is it possible to do so using knowledge only from a first course in linear algebra?
 A: The statement is not true.
Let $$A=\begin{bmatrix}1&1\\2&2\end{bmatrix}$$
$$B=\begin{bmatrix}2&1\\5&1\end{bmatrix}$$
Note that $A$ is singular and $B$ is invertible so they are not similar.
A: Eigenvalues are invariant under conjugation: $B$ and $T^{-1}BT$ have the same eigenvalues. The trace is the sum of the eigenvalues.
So what you're asking would imply that if the sum of the eigenvalues of $A$ and $B$ are the same, then the set of eigenvalues are identical. But this clearly can't hold because $1 + 1 = 0 + 2$.
Also, while $A = T^{-1}BT$ implies that $A$ and $B$ have the same eigenvalues, it is not true that if $A$ and $B$ have the same eigenvalues then $A = T^{-1}BT$ for some $T$. For example:
$$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. $$
$A$ and $B$ have the same eigenvalues: $1$ (with algebraic multiplicity $2$) but you can see that $T^{-1}BT = T^{-1}T = I \ne A$.
What is correct is that if $A$ and $B$ have the same Jordan decomposition then $A = T^{-1}BT$ for some invertible matrix $T$ (we say $A$ and $B$ are similar if this happens). An easy example of this is when $A$ and $B$ are diagonalizable with the same eigenvalues (counted with multiplicity).
A: The statement is false.
Let $A$ be any nonzero matrix with trace zero and let $B = 0$. Then $A$ and $B$ have the same trace but
$$T^{-1}BT = 0 \ne A$$
for all invertible matrices $T$.
A: Here are two matrices $A,B,$ with the same characteristic polynomials and minimal polynomials. But not similar.
$$
A =
\left(
\begin{array}{cccc}
7&0&0&0 \\
0&7&0&0 \\
0&0&7&1 \\
0&0&0&7 \\
\end{array}
\right)
$$ 
$$
A =
\left(
\begin{array}{cccc}
7&1&0&0 \\
0&7&0&0 \\
0&0&7&1 \\
0&0&0&7 \\
\end{array}
\right)
$$ 
