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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.

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  • $\begingroup$ Are you sure the statement is true? quora.com/… $\endgroup$ – avs Mar 18 at 17:55
  • $\begingroup$ @avs what statement you mean? the main one? $\endgroup$ – Stanisław Maksicki Mar 18 at 18:11
  • $\begingroup$ Yes. The statement that $A$ is similar to the defined $B$. Did you see the link in my previous comment? $\endgroup$ – avs Mar 18 at 18:37
  • $\begingroup$ 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 $A$). $\endgroup$ – Marc van Leeuwen Mar 25 at 10:06
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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 $$\pmatrix{-A'_{n + 1, 1} / \lambda&\cdot&1\\\cdot&I_{n - 1}&\cdot\\\cdot&\cdot&1} .$$

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