# Why is this diagonal matrix not possible over the reals

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?

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

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.

• So my explanation should be alright as it is now without having to show that this is the case?
– user831952
Jan 7, 2021 at 9:20
• Yes, your proof is correct. @mathstudent23 Jan 7, 2021 at 9:21
• and by definitione the diagonal can not include 0 right? This is why it is true.
– user831952
Jan 7, 2021 at 9:30
• The diagonal can include $0$ -- you may have an eigenvalue of $0$. Note that it is the eigenvectors which can't be zero. Jan 7, 2021 at 9:31
• 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?
– user831952
Jan 7, 2021 at 9:33

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.

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 $$A$$.

• But I do not see how the similarities of 2 matrices relates to the problem I am having? Why does this matter?
– user831952
Jan 7, 2021 at 9:01
• Because $\delta=\beta^{-1}\alpha\beta$ and therefore $\delta$ and $\alpha$ are similar. Jan 7, 2021 at 9:04
• 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.
– user831952
Jan 7, 2021 at 9:08
• I provided the answer to that question in first two sentences of my answer. What is it that you don't understand? Jan 7, 2021 at 9:09
• But in why are they similar? and by similar do you mean same coordinates?
– user831952
Jan 7, 2021 at 9:10