# Show that matrix $A$ is similar to a matrix $B$ with elements on diagonal $(0, …, 0, \operatorname{Tr(}A))$ respectively.

Let $$A$$ be a matrix $$n \times n, n \geq 2$$. Let's assume that not all entries outside of the diagonal are zeros (we don't know what entries are on the diagonal). Show that matrix $$A$$ is similar to a matrix $$B$$ with elements on diagonal $$(0, ..., 0, \operatorname{Tr}(A))$$ respectively.

We know that $$\operatorname{Tr}(A) = \sum_{i=1}^n \lambda_i$$, where $$\lambda_i$$ is the i-th eigenvalue of A.

I found a similar problem with the solution using Rational Canonical Form. However, so far on the course we have only developed the Jordan Form and I believe we should use it to solve this problem.

Using Jordan Normal Form to show $M$ is similar to a matrix whose diagonal is $(0, 0, \operatorname{Tr}(M))$?

Any help would be greatly appreciated.

• Are you sure the statement is true? quora.com/… – avs Mar 18 at 17:55
• @avs what statement you mean? the main one? – Stanisław Maksicki Mar 18 at 18:11
• Yes. The statement that $A$ is similar to the defined $B$. Did you see the link in my previous comment? – avs Mar 18 at 18:37
• The problem statement is confusing, even if correct. Saying we don't know what entries are on the diagonal of $A$ is not really helpful if it is already said that $A$ is not diagonal (actually we don't know what are the off-diagonal entries of $A$ either); however it would be good to stress that nothing is required about the entries of $B$ off the main diagonal. Basically the only requirement of $B$ is that all diagonal entries except the last are zero (and of course that $B$ is similar to $A$). – Marc van Leeuwen Mar 25 at 10:06

Hint Prove the claim by induction on the size of the matrix, with inductive hypothesis that the claim holds for $$n \times n$$ matrices.
For an $$(n + 1) \times (n + 1)$$ matrix $$A$$, decompose $$A = \pmatrix{\lambda&\ast\\\ast&B} ,$$ where $$B$$ has dimension $$n \times n$$. By the inductive hypothesis there is a matrix $$Q$$ such that $$Q B Q^{-1}$$ has $$B_{11} = \cdots = B_{n - 1, n - 1} = 0$$. Then, conjugate $$A$$ by $$P := \pmatrix{1&\cdot\\\cdot&Q}$$.
Doing so gives $$A' := P A P^{-1} = \pmatrix{\lambda&\ast\\\ast&Q B Q^{-1}}$$ and in particular its diagonal entries except perhaps the $$(1, 1)$$ and $$(n + 1, n + 1)$$ entries are zero. If $$\lambda = 0$$, $$A'$$ has the desired form. If not, it remains to find a matrix by which conjugation clears the $$(1, 1)$$ entry, and we can take $$\pmatrix{-A'_{n + 1, 1} / \lambda&\cdot&1\\\cdot&I_{n - 1}&\cdot\\\cdot&\cdot&1} .$$