0
$\begingroup$

Question is to Prove/Disprove :

If a triangular matrix $A$ is similar to diagonal matrix, then $A$ is already diagonal matrix.

I guess the answer is Yes.

I am not able to write the solution precisely.

What i have thought is Suppose $A$ is similar to diagonal matrix, then its eigen values are all distinct and as $A$ is already triangular, A has to be diagonal

I clearly see there is a gap in my justification, But do not have an idea to proceed..

Another Question is to prove that :

Suppose $A$ is a nilpotent $n \times n$ matix then trace of $A$ is zero.

My justification for this is :

As $A$ is nilpotent, we have $A^k=0$

As we know eigenvalues determines the minimal polynomial for $A$ (as roots of minimal polynomial) and sum of roots is being coefficient of $A$ in its minimal polynomial gives that sum of eigen values is zero and thus trace is zero (as trace equals to sum of eigen values )

$\endgroup$
  • $\begingroup$ What do you mean by "already diagonal" ? $\endgroup$ – Dutta Sep 20 '13 at 6:33
  • $\begingroup$ Hmm, for a lot of matrices $A$ (for instance diagonalizable matrices) it is possible to do a similarity-transformation with an orthogonal factor $B$ such that we arrive at a triangular matrix $C$ like $ B A B^T = C$ But since $A$ is also diagonalizable by some matrix $M$ such that $ D = M A M^{-1}$ there is a similarity transformation from triangular $C$ to diagonal $D$ by $D = M B^T C B M^{-1}$ $\endgroup$ – Gottfried Helms Sep 20 '13 at 6:34
  • 1
    $\begingroup$ It means to show, that there in fact are non-diagonal, triangular matrices ($C$), which are still be diagonalizable - I thought that was your question. $\endgroup$ – Gottfried Helms Sep 20 '13 at 6:45
  • 1
    $\begingroup$ What the excellent examples below have in common, they are $2 \times 2$ matrices in triangular form, with two different values on the diagonal. Now these values are precisely the eigenvalues, and the fact that they are distinct implies that the matrix is diagonalizable. So what (non-zero) element you put off-diagonal is immaterial. $\endgroup$ – Andreas Caranti Sep 20 '13 at 6:48
  • 1
    $\begingroup$ Just a note: A nilpotent matrix can have only zero eigenvalues. This is of course stronger than just having their sum equal to zero. $\endgroup$ – Algebraic Pavel Sep 20 '13 at 10:46
3
$\begingroup$

Since $A$ is similar to a diagonal matrix you know that, $$A = P^{-1}BP, $$ for some invertible matrix P and diagonal matrix B. A is called diagonalizable. Such a matrix is diagonalizable if it has all distinct eigenvalues. Such a matrix does not have to be diagonal to begin with. $$A = \begin{pmatrix} -1 & 2 \\ 0 & 3 \end{pmatrix}, $$ is diagonaliable and hence similar to a diagonal matrix but is not diagonal.

$\endgroup$
  • $\begingroup$ I got it :) Thank You $\endgroup$ – user87543 Sep 20 '13 at 6:49
1
$\begingroup$

Your first part is wrong. Let $$ A = \begin{pmatrix} 1 & 1 \\ 0 & 0\end{pmatrix} $$ A is triangular, and diagonizable, as its eigenvalues are $1$ and $0$.

The second part looks fine.

$\endgroup$
1
$\begingroup$

If a triangular matrix A is similar to diagonal matrix, then A is already diagonal matrix

This holds true only when we add a constraint that all the diagonal elements of A are equal

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy