1
$\begingroup$

Let $A = \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix} $ then $A$ is not similar to a diagonal matrix over the reals and it is not similar to a diagonal matrix over the complexes.

We know that $A$ is similar to a diagonal matrix over the reals(complexes) if there exist $D$ diagonal matriz and $P$ invertible matrix both $n \times n$ with real entries (complex entries) such that $A = PDP^{-1}$

I find that the inverse of $A$ is $A^{-1} = \frac{1}{2}\begin{bmatrix} -7 & 1 & 4 \\ -8 & 2 & 4 \\ -10 & 0 & 6 \end{bmatrix}$ and i diagonalize the matrix $A$ and got $A = PBP^{-1}$ with $ P = \begin{bmatrix} 1 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 0 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 2 & 1 & 1 \end{bmatrix} $ , $ P^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -1 + 3i & 1-\frac{i}{2} & \frac{1}{2}-\frac{3i}{2} \\ -1 - 3i & 1+\frac{i}{2} & \frac{1}{2}+\frac{3i}{2} \end{bmatrix} $ and finally i got $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & i \end{bmatrix}$ then A is diagonalizable but my question is how can i conclude that $A$ is not similar to a diagonalize matrix over the reals? , please some help for this.

$\endgroup$
  • $\begingroup$ Do you know what the characteristic polynomial of a matrix is ? $\endgroup$ – Gabriel Romon Jun 23 '14 at 14:48
  • $\begingroup$ yea the characteristic polynomial of $A$ is $p(x) = 2-x +2 x^2-x^3$ $\endgroup$ – Knight Jun 23 '14 at 14:50
1
$\begingroup$

Suppose for contradiction that $A$ were diagonalizable over the reals.

There would be some real $D = \begin{bmatrix} \beta_1 & 0 & 0 \\ 0 & \beta_2 & 0 \\ 0 & 0 & \beta_3\end{bmatrix}$ similar to $A$.

Furthermore, the $\beta_i$ are real roots of $2-x +2 x^2-x^3$.

Thus, $\beta_1=\beta_2=\beta_3=2$

Hence $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}$

As a result, what can you say about $\operatorname{Trace(A)}$ and $\operatorname{Trace(D)}$ ?

$\endgroup$
  • $\begingroup$ $Tr(A) = 2$ and $Tr(D) = 1 \pm i $ $\endgroup$ – Knight Jun 23 '14 at 15:00
  • $\begingroup$ @Knight $Tr(D)$ in your comment is not right. I've edited a bit. Can you tell me what is $Tr(D)$ for the matrix $D$ that is in my answer ? $\endgroup$ – Gabriel Romon Jun 23 '14 at 15:02
  • $\begingroup$ sorry yes you are right $Tr(A) =2$ $Tr(D)= 6 $ but in my matrix $D$ i have that $Tr(D) = 2 = 2 + i - i$ $\endgroup$ – Knight Jun 23 '14 at 15:04
  • $\begingroup$ @Knight ok. So you see that if you suppose for contradiction that $A$ were diagonalizable over the reals, you derive that $Tr(A)\neq Tr(D)$. Now, do you see why $Tr(A)\neq Tr(D)$ is a contradiction ? $\endgroup$ – Gabriel Romon Jun 23 '14 at 15:05
  • $\begingroup$ i can see it from your matrix $D$ but why you get that matrix $D$ in your answer ? $\endgroup$ – Knight Jun 23 '14 at 15:07
2
$\begingroup$
  • $A$ is diagonalisable to a matrix over $\mathbb{C}$.
  • The diagonal components contain complex numbers.
  • Matrix diagonals are unique (up to a permutation).

Therefore it is not diagonalisable over $\mathbb{R}$.

$\endgroup$
  • $\begingroup$ matrix A is diagonalizable over C and the diagonal components are $\{ 2, \pm i\}$ but i don't understand "Matrix diagonals are unique"? and why this implies that A is not similar to a diagonalizable over R?, thanx for your patience. $\endgroup$ – Knight Jun 23 '14 at 14:52
  • $\begingroup$ Suppose there existed three real diagonal components then they would have to be different from $\{2,\pm i\}$ but then that would contradict uniqueness... $\endgroup$ – lemon Jun 23 '14 at 15:28
1
$\begingroup$

The entries on the diagonal matrix are unique, take the characteristic polynomial of $A$ and $D$ which are the same since they are similar matrices (why? $Det(A- y I ) = Det( P D P^{-1} - y I ) = Det ( P ( D - y I ) P^{-1}) = Det (P) Det (D - yI) Det (P) ^{-1}$ ), so the zeros are the same.

Now you notice that the diagonal entries of $D$ are precisely the zeros with multiplicity of the characteristic polynomial, so $p_A$ determine $D$ up to permutation. If there are some $D$ that has complex entries, no $D$ is possible with real entries.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.