# Diagonalize the matrix A (complex numbers)

Diagonalize the matrix A

$A=\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}$

So, i began the problem by finding the characteristic polynomial which was

$λ^3-7λ^2-15λ-27$

using long division i got $(λ-9)(λ^2+2λ+3)$

so i used the quadratic formula and got

$λ_1=-1+i\sqrt{2}$ and $λ_2=-1-i\sqrt{2}$ and $λ_3=9$

I decided to start with $λ_1$

$A-\left(-1+i\sqrt{2}\right)I=\begin{pmatrix}2-i\sqrt{2} & 2 & 4 \\3 & 6-i\sqrt{2} & 2 \\2 & 6 & 2-i\sqrt{2}\end{pmatrix}$

Now i understand how to diagonalize when i have all numbers but once i get these $i$'s in the equation it's like my brain doesn't comprehend the steps i need to take the get the diagonal form..

I would like to think i'm supposed to start by getting the inverse of this new matrix but fail to see how to do that with $i$'s involved.

• Do you wish to find the eigenvectors? Or are you trying to find the matrices $P$ and $Q$ such that $A = PDQ$ where $D$ is your diagonalized matrix? Because once you've found the eigenvalues, that's usually enough. – user123 Aug 10 '15 at 19:45
• I mean for me to diagonalize the matrix A i guess i don't know if i need to find the eigenvectors or not.. but i think i do need to find $D=P^{-1} AP$ and i don't know how to do that without getting the eigenvectors. – Charlene Aug 10 '15 at 19:53
• Ineed, you have to find a basis of eigenvectors. Note that, as there are two conjugate roots, you'll get two conjugate eidenvectors. So you have only two eigenvectors to determine. – Bernard Aug 10 '15 at 20:08
• So am i correct to try to find the inverse of $A+1-i√2$? – Charlene Aug 10 '15 at 20:20
• You just need to find a generator for the kernel of that matrix. The kernel of that matrix will be the eigenspace relative to that eigenvalue; once you find the three generators for the kernels you're done. You simply take a $3 \times 3$ matrix with these three vectors as columns: that's your transition matrix. – Lonidard Aug 10 '15 at 20:29

To diagonalize that matrix you should look for the kernel of $A-\lambda_1\mathbb{I}_3$, which is a fancy way to indicate the family of vectors for which $A(v)=\lambda_1v$. The kernel is defined as

$$\text{ker}(A+(1-i\sqrt 2)\mathbb{I}_3)=\{v \in \mathbb C^n | (A+(1-i\sqrt 2)\mathbb{I}_3)(v)=0_{\mathbb{C}^n}\}$$

That is the set of vectors that are mapped to zero by the matrix $(A+(1-i\sqrt 2)\mathbb{I}_3)$. With real numbers, it can be intuitive; with complex ones it seldom is without quite a bit of practice. You can systematically (no pun intended) understand for which vectors this happens using any kind of technique for the resolution of a linear system of equations.

In your case the eigenspace relative to $\lambda_1$ is $\langle (-1-i\sqrt 2, \frac{i}{\sqrt 2},1)^T \rangle$: this will be the first column vector of you transition matrix. You must repeat this process for the other two eigenvalues to find the other two columns.

Can you take it from here?

• $\begin{pmatrix}9 & 0 & 0 \\0 & -i\left(-i+\sqrt{2}\right) & 0 \\0 & 0 & i\left(i+\sqrt{2}\right)\end{pmatrix}$ is the diagonal matrix i got, does it look right even though it doesnt match the eigenvalues or did i mess up on a step ? – Charlene Aug 10 '15 at 21:02
• Note: with real matrices, once you have a complex eigenvector, the complex conjugate will be another eigenvector. In this case, we can take as our second eigenvector $$(-1 + i \, \sqrt{2}, -i/\sqrt{2}, 1)^T$$ – Ben Grossmann Aug 10 '15 at 21:02
• @Charlene $$-i(-i + \sqrt{2}) = (-i)^2 - i\sqrt{2} = -1 - \sqrt{2}\,i$$ your diagonal matrix is correct. – Ben Grossmann Aug 10 '15 at 21:03
• That's right. Good job @Charlene! – Lonidard Aug 10 '15 at 21:03

A strange way to find the eigenvectors is to use the Cayley-Hamilton Theorem, $$(A-9I)(A^{2}+2A+3I) = 0.$$ Because of this, the columns of $A^{2}+2A+3I$ are eigenvectors of $A$ with eigenvalue $9$. For example, $$\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix} \begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}+ 2\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}+ 3\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{pmatrix} \\ =\begin{pmatrix}15 & 36 & 12 \\ 22 & 43 & 24 \\22 & 40 & 21 \end{pmatrix} + \begin{pmatrix}2 & 4 & 8 \\ 6 & 10 & 4 \\ 4 & 12 & 2\end{pmatrix} + \begin{pmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{pmatrix} \\ = \begin{pmatrix}20 & 40 & 20 \\ 28 & 56 & 28 \\ 26 & 52 & 26\end{pmatrix}$$ Similarly, the columns of the following are eigenvectors with eigenvalue $-1-\sqrt{2}i$ $$(A-9I)(A-(-1+\sqrt{2}i)I)=A^{2}+(-8-\sqrt{2}i)A+(-9+9\sqrt{2}i)I$$ You only need the first column: $$\begin{pmatrix} 15 \\ 22 \\ 22\end{pmatrix}+(-8-2\sqrt{2}i)\begin{pmatrix} 1 \\ 3 \\ 2\end{pmatrix} + (-9+9\sqrt{2}i)\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} \\ =\begin{pmatrix} -2+7\sqrt{2}i \\ -2-6\sqrt{2}i \\ 6-4\sqrt{2}i\end{pmatrix}$$ And the conjugate of this vector is going to have the conjugate of its eigenvalue. So a basis of eigenvectors is $$\begin{pmatrix}10 \\ 14 \\ 13\end{pmatrix}, \begin{pmatrix} -2+7\sqrt{2}i \\ -2-6\sqrt{2}i \\ +6-4\sqrt{2}i\end{pmatrix}, \begin{pmatrix} -2-7\sqrt{2}i \\ -2+6\sqrt{2}i \\ +6+4\sqrt{2}i\end{pmatrix}$$