Do all diagonalisable matrices have orthogonal eigenvectors?

So. Let us label right-eigenvectors of $M$ as $|\lambda\rangle$ and left-eigenvectors of $M$ as $\langle\lambda|$. That is, \begin{gather} M |\lambda\rangle = \lambda |\lambda\rangle \text{ and } \langle\lambda| M = \langle\lambda| \lambda . \end{gather} This makes sense (I hazard) because a left-eigenvector is adjoint, which doesn't have to mean ‘conjugate transpose’, to a right-eigenvector via a shared eigenvalue. Now, we may write \begin{gather} M|\lambda\rangle = \lambda |\lambda\rangle \text{ and } \langle\mu| M = \langle\mu| \mu , \end{gather} which implies that both \begin{gather} \langle \mu | M | \lambda \rangle = \lambda \langle \mu | \lambda \rangle \text{ and } \langle \mu | M | \lambda \rangle = \mu \langle \mu | \lambda \rangle \end{gather} which immediately implies, upon the assumption that \begin{gather} \lambda \neq \mu , \end{gather} that \begin{gather} \langle\mu|\lambda\rangle = 0 . \end{gather} What have I done wrong?

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Left and right eigenvectors are biorthogonal, that's what you showed. – Algebraic Pavel Oct 13 '13 at 1:17

The answer is yes, but to a different question: Do all diagonalizable matrices have biorthogonal right and left eigenvectors?

To see this, let the matrix $A\in\mathbb{C}^{n\times n}$ be diagonalizable so that there exists a nonsingular matrix $X=[x_1,\ldots,x_n]$ and diagonal $\Lambda=\mathrm{diag}(\lambda_1,\ldots,\lambda_n)$ such that $$A=X\Lambda X^{-1}.$$

A nonzero vector $x$ is called right eigenvector of $A$ (associated with an eigenvalue $\lambda$), if $Ax=\lambda x$ for some scalar $\lambda$. A nonzero vector $y$ is called left eigenvector (associated with an eigenvalue $\lambda$), if $y^*A=\lambda y^*$ for some scalar $\lambda$ (or equivalently, $A^*y=\bar{\lambda} y$).

The columns $x_1,\ldots,x_n$ of the matrix $X$ above are the right eigenvectors of $A$. To see this, post-multiply $A=X\Lambda X^{-1}$ by $X$ to get $AX=X\Lambda$ and look on each column of the matrix equation, which states that $Ax_i=\lambda_ix_i$, $i=1,\ldots,n$.

We have also that $X^{-1}A=\Lambda X^{-1}$ (simply pre-multiplying $A=X\Lambda X^{-1}$ by $X^{-1}$), that is, $Y^*A=\Lambda Y^*$ with $Y=[y_1,\ldots,y_n]=(X^{-1})^*$. The columns $y_1,\ldots,y_n$ of $Y$ are the left eigenvectors of $A$. To see that, look on the rows of $Y^*A=\Lambda Y^*$: $y^*_iA=\lambda_i y_i^*$, $i=1,\ldots,n$.

The right eigenvectors $\{x_i\}_{i=1}^n$ and left eigenvectors $\{y_i\}_{i=1}^n$ are biorthogonal, that is, $y_i^*x_j=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta symbol. This can be easily verified using the relation between $X$ and $Y$, because $y_i^*x_j=\delta_{ij}$ iff $Y^*X=I$ (where $I$ is the identity matrix) and $Y^*X=((X^{-1})^*)^*X=X^{-1}X=I$.

P.S.: Sorry for not using your notation :-)

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Thanks! This is a great help. I had a quick question: is there an obvious way to normalise left- and right-eigenvectors separately? PS: I chose bra-ket notation because it reduced the number of symbols in the question. But I think bra-ket notation is actually what confused me! – P. Plowman Oct 13 '13 at 21:32
Well, you can normalize them separately (say, to have unit norm). Then, however, you have only $y_i^*x_j=0$ if and only if $i\neq j$. Generally $y_i^*x_i$ won't be 1 but some other (nonzero) number dependent on $i$. This corresponds to replacing $X$ and $Y$ by $\tilde{X}=XD_X$ and $\tilde{Y}=YD_Y$, where $D_{\cdot}$ are nonsingular diagonal matrices. Then $\tilde{Y}^*\tilde{X}=D_YD_X$. – Algebraic Pavel Oct 13 '13 at 21:43
Anyway, there's nothing wrong on your statement that right and left eigenvectors corresponding to different eigenvalues are orthogonal. The matrix even does not need to be diagonalizable. – Algebraic Pavel Oct 13 '13 at 21:51
But the matrix does have to be diagonalisable if we want an eigenbasis to exist, no? – P. Plowman Oct 14 '13 at 8:58
Of course, if the matrix is diagonalisable then the eigenvectors (both left and right) are basis of $\mathbb{C}^n$. That does not mean that a matrix, which is not diagonalisable, cannot have (some) left and right eigenvectors, which are still biorthogonal. – Algebraic Pavel Oct 14 '13 at 18:15

The answer is no: define $T:R^2→R^2$ by $T(1,0)=(1,0)$, $T(1,1)=(2,2)$. Then $T$ is diagonalizable, and the eigenvalues are even distinct, but not in an orthonormal basis, ie. $T$ is not self adjoint. The problem with what you have done is that you have implicitly assumed $M$ is self adjoint, since for self adjoint operators, $⟨μ|M≡(M|μ⟩)^∗$. What does $⟨u|M$ mean if $M$ is not self adjoint?

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$\langle u | M$ means exactly that. And $\langle u | M = \langle u | u$ means $\langle u |$ is a left-eigenvector of $M$. Can I not take a single matrix $M$ and compose left- and right-eigenvectors? Somehow I have a misunderstanding of the relation of left-eigenvectors to right-eigenvectors, which is what I would like to clarify. – P. Plowman Oct 12 '13 at 23:25
$\langle u|M$ is an operator, and as usual is defined by how it operates on kets. If you already have an orthonormal basis of eigenvectors then since $\langle u|M|\lambda\rangle=\langle u|(M|\lambda\rangle)=\lambda\langle u|\lambda\rangle=0$ if $\lambda\ne u, =u$ if $\lambda=u$, you can just identify $\langle u|M$ with $u\langle u|$ but otherwise it may not make sense to make that identification. – JLA Oct 12 '13 at 23:32
Okay, so I take $M$ and in the usual way diagonalise, or find a set of (right-)eigenvectors, by solving $M | \lambda \rangle = \lambda | \lambda \rangle$. Then I define the left eigenvectors as the $\langle\lambda|$ such that $\langle\lambda| M |\lambda\rangle = \lambda$. Then what goes wrong with my original question? I must apologise for the somewhat trivial question; I post because I know there is something I am very confused about. – P. Plowman Oct 12 '13 at 23:38
That definition doesn't make sense though. What would $\langle\lambda|u\rangle$ be? – JLA Oct 12 '13 at 23:41
Hmmm. Okay, let me think for a bit; I'll get back in a bit. Thank you very much! – P. Plowman Oct 12 '13 at 23:43