$A$ is not similar to a diagonal matrix over the reals

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.

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

• $A$ 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 $\mathbb{R}$.

• 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. Jun 23 '14 at 14:52
• Suppose there existed three real diagonal components then they would have to be different from $\{2,\pm i\}$ but then that would contradict uniqueness... Jun 23 '14 at 15:28

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 $\operatorname{Trace(D)}$ ?

• $Tr(A) = 2$ and $Tr(D) = 1 \pm i$ Jun 23 '14 at 15:00
• @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 ? Jun 23 '14 at 15:02
• sorry yes you are right $Tr(A) =2$ $Tr(D)= 6$ but in my matrix $D$ i have that $Tr(D) = 2 = 2 + i - i$ Jun 23 '14 at 15:04
• @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 ? Jun 23 '14 at 15:05
• i can see it from your matrix $D$ but why you get that matrix $D$ in your answer ? Jun 23 '14 at 15:07

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.