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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?

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    $\begingroup$ Yes, this is a general fact. If a matrix is similar to a diagonal matrix, the diagonal matrix must contain eigenvalues. $\endgroup$ Jan 7, 2021 at 9:12

3 Answers 3

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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.

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  • $\begingroup$ So my explanation should be alright as it is now without having to show that this is the case? $\endgroup$
    – user831952
    Jan 7, 2021 at 9:20
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    $\begingroup$ Yes, your proof is correct. @mathstudent23 $\endgroup$ Jan 7, 2021 at 9:21
  • $\begingroup$ and by definitione the diagonal can not include 0 right? This is why it is true. $\endgroup$
    – user831952
    Jan 7, 2021 at 9:30
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    $\begingroup$ The diagonal can include $0$ -- you may have an eigenvalue of $0$. Note that it is the eigenvectors which can't be zero. $\endgroup$ Jan 7, 2021 at 9:31
  • $\begingroup$ Oh... But then I am not sure why this is not true? The eigenvalues of $\alpha$ over real numbers is only $3$. Then why could the diagonal not just be $diag(0,3,0)$ or any other coordinate of 3. Is it just since it has to be eigenvalues and thus there is bascically no numbers for the 2 coordinates? $\endgroup$
    – user831952
    Jan 7, 2021 at 9:33
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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.

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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 $A$.

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  • $\begingroup$ But I do not see how the similarities of 2 matrices relates to the problem I am having? Why does this matter? $\endgroup$
    – user831952
    Jan 7, 2021 at 9:01
  • $\begingroup$ Because $\delta=\beta^{-1}\alpha\beta$ and therefore $\delta$ and $\alpha$ are similar. $\endgroup$ Jan 7, 2021 at 9:04
  • $\begingroup$ So since $\alpha$ has the eigenvalues $3,-i,i$ these has to be the diagonal values of the matrix $\delta$. This seems a bit circular to me. Why do they have the same characteristic polynomials - that is probably where I am stuck. $\endgroup$
    – user831952
    Jan 7, 2021 at 9:08
  • $\begingroup$ I provided the answer to that question in first two sentences of my answer. What is it that you don't understand? $\endgroup$ Jan 7, 2021 at 9:09
  • $\begingroup$ But in why are they similar? and by similar do you mean same coordinates? $\endgroup$
    – user831952
    Jan 7, 2021 at 9:10

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