Why is this diagonal matrix not possible over the reals

Question

Let $\alpha = \begin{pmatrix} 7 &3 &-4 \\ -2&-1 &2 \\ 6&2 &-3 \end{pmatrix}$ over the reals.

Show that there does not exist a invertible real matrix $\beta$, so that $\delta = \beta ^{-1} \alpha \beta$ is a diagonal matrix

My "proof"

the characterisitc polynomial is $-\lambda ^3 + 3\lambda ^2 -\lambda + 3=-(\lambda - 3)(\lambda ^2 +1)$. The solutions would then be $3,i,-i$. Now, we have that $\delta = \beta ^{-1} \alpha \beta$ where the diagonal values of $\delta$ would then be eigenvalues of $\alpha$. if both $\alpha , \beta$ are real $3 \times 3$-matrices then $\delta$ must also be a real $3 \times 3$-matrix. But that is not possible since the eigenvalues of $\alpha$ are $3,i,-i$ and thus $\beta$ must be complex.....

Now I am stuck since I would think I would have to show that the diagonal of $\delta$ must be the eigenvalues of $\alpha$. Is this a generel fact or should this be proven? If so - how?

Answer 1

What you said is a general fact and your proof is correct.

If any matix $\alpha$ is similar to a diagonal matrix $\delta=\text{diag}[\delta_{11},\delta_{22},\ldots,\delta_{nn}]$ i.e. $\exists$ an invertible matrix $B$ such that $\delta=B^{-1}\alpha B$, the diagonal entries of $\delta$ must be the eigenvalues of $\alpha$. You can prove this fact easily.

We have $\alpha B=B\delta$. Let the columns of $B$ be $B_i$. Then$$\alpha B=\alpha[B_1~B_2~\ldots B_n]=[\alpha B_1~\alpha B_2\ldots\alpha B_n]$$and$$B\delta=[\delta_{11}B_1~\delta_{22}B_2\ldots\delta_{nn}B_n]$$Equating the columns we get $\alpha B_i=\delta_{ii}B_i$, i.e. $B_i$ are eigenvectors of $\alpha$ and $\delta_{ii}$ their corresponding eigenvalues.

Answer 2

If two matrices $A$ and $B$ are similar, $A-\lambda\operatorname{Id}$ and $B-\lambda\operatorname{Id}$ are similar too, and therefore they have the same determinants. In other words, $A$ and $B$ have the same characteristic polynomials. But, if $B$ is a diagonal matrix and the entries of the main diagonal are $b_1,b_2,\ldots,b_n$, then the characteristic polynomial of $B$ is $(b_1-\lambda)(b_2-\lambda)\cdots(b_n-\lambda)$, whose roots are $b_1,b_2,\ldots,b_n$. But $b_1,b_2,\ldots,b_n$ are the eigenvalues of $B$ and therefore of $A$.

Answer 3

The characteristic polynomial is satisfied by $\alpha$. $$(\alpha-3I)(\alpha^2+I)=0$$ This is still true if $\alpha$ is replaced by the diagonal matrix $\delta$ $$(\delta-3I)(\delta^2+I)=\beta^{-1}(\alpha-3I)(\alpha^2+I)\beta=0$$ If $\delta$ consists of the diagonal terms $x$, then $$(x-3)(x^2+1)=0$$ Over the reals, this only gives $x=3$ as a solution, so $\delta=3I$ and $\alpha=\beta(3I)\beta^{-1}=3I$, which is not the case.