# Left/Right Eigenvectors

Let $M$ be a nonsymmetric matrix; suppose the columns of matrix $A$ are the right eigenvectors of $M$ and the rows of matrix $B$ are the left eigenvectors of $M$.

In one of the answers to a question on left and right eigenvectors it was claimed that $AB=I$. Is that true, and how would you prove it?

• How are the eigenvectors normalized? Dec 25, 2016 at 22:09
• Call $v_k^i$ the $i$-th element of the $k$-th eigenvector of $M$. Then $\sum_i |v_k^i|^2 = 1$ for all $k$. Dec 25, 2016 at 22:10

Most counter-examples found here have to do with the inherent ambiguities in defining the eigenvectors. There are essentially two kinds:

• For repeating eigenvalues, the eigenvectors spanning their space are not uniquely defined. Only their subspace is. Any linearly indepedent basis for their space will do (extreme example $M=I$: any full rank matrix $A$ qualifies as matrix of right singular vectors. Likewise, any full rank matrix $B$ can be chosen independently as a valid basis of left singular vectors).
• If we exclude this by forcing distinct eigenvalues we still have each eigenvector only being unique up to a scalar multiple, since any $q$ satisfying $M q = \lambda q$ will also lead to $\alpha q$ satisfying the same equation for any $\alpha \neq 0$. Forcing $\left\|q\right\|=1$ eliminiates this only partially, in the real-valued case there is still a sign ambiguity (e.g., Roberts reply), in the complex-valued case a phase ambiguity.

These ambiguities aside, let us not forget that left singular vectors of $M$ are right singular vectors of $M^T$. So if $M$ has an EVD of the form $M=A \Lambda A^{-1}$, then $M^T = A^{-T} \Lambda A^T$. This shows that $B = A^{-1}$ does qualify as a pair of left/right singular vectors. It's only that if we compute $A$ and $B$ independently, we are not guaranteed to pick the pair of inverses among the set of ambiguities. If eigenvalues are distinct we might get a diagonal matrix (due to the scaling ambiguities), if they are not we might get something pretty arbitrary.

Try e.g. $$M = \pmatrix{3 & 2\cr -1 & 0\cr}$$ Eigenvalues are $1$ and $2$. Normalized right eigenvectors form the matrix $$A = \pmatrix{-1/\sqrt{2} & -2/\sqrt{5} \cr 1/\sqrt{2} & 1/\sqrt{5}\cr}$$ Normalized left eigenvectors form $$B = \pmatrix{1/\sqrt{5} & 2/\sqrt{5}\cr 1/\sqrt{2} & 1/\sqrt{2}\cr}$$ These are not inverses.

• In your example, $BA$ is a multiple of the identity. Dec 25, 2016 at 23:33
• $\pmatrix{1/\sqrt{10} & 0\cr 0 & -1/\sqrt{10}\cr}$ happens to be diagonal, but is not a multiple of the identity. Dec 26, 2016 at 1:04
• $BA$ is diagonal because it commutes with the diagonal matrix $\Lambda$ formed from the eigenvalues: $M A = A \Lambda$ and $B M = \Lambda B$ so $B A \Lambda = B M A = \Lambda B A$. Dec 26, 2016 at 1:12

Here's a tentative proof that $A$ is the inverse of $B$.

The left eigenvectors of $M$ are the right eigenvectors of $M^T$. Therefore if $M=B⋅Λ⋅B^{-1}$ then $M^T=B^{-T}⋅Λ⋅B^T$ which is an EVD. If the eigenvalues are distinct, the EVD is unique (up to scaling of eigenvectors), which proves that left eigenvectors are rows of $B^{−1}$, i.e. $A=B^{-1}$.

(I'm pasting this answer from Power of a Nonsymmetric Matrix , it's due to Florian)

Of course the counterexamples below cast a dark shadow on this proof, but I can see nothing wrong with it.

• Each column of $A$ is a multiple of the corresponding column of $B^{-1}$, i.e. $BA$ is a diagonal matrix, but that doesn't say $A$ and $B^{-1}$ are the same. Dec 26, 2016 at 1:33
• Okay, then it's clear. Dec 26, 2016 at 8:29
• That proof is bogus: all you said in the question was that the rows of $B$ are eigenvectors, not that the $i$th one corresponds to the eigenvalue for the $i$th column of $A$. Any shuffle of the rows of $B$ is still a set of left eigenvectors, but $BA$ is not a multiple of the identity. It's probably true (by the "proof" above) that if the eigenvalues are distinct, and the $i$th column of $A$ is the right unit eigenvector for $\lambda_i$, and the $i$th row of $B$ is the left unit eigenvector for the same eigenvalue, then their product is the identity. Dec 26, 2016 at 16:45
• To see something wrong with the proof, you might want to simply compute the eigensets of the matrices described in the other answers. Dec 26, 2016 at 16:45
• Yes that's right, thank you. Dec 26, 2016 at 17:02

No. Let $$M = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 4 & 2 \\ 0 & 0 & 0 & 2 \end{bmatrix}$$ Then two right eigenvectors for $1$ are $e_1$ and $e_2$, and two left eigenvectors for $1$ are $(\sqrt{2}/2) ( e_1 \pm e_2 )$. When you put these into $A$ and $B$, the upper left corner of your product will not be the $2 \times 2$ identity.

I have a bad feeling this question's about to be edited to add the hypothesis that the eigenvalues be distinct...

When that happens, the answer will still be no, for if we permute the rows of $A$ to form $A'$ they'll still all be eigenvectors for $M$, but the product $AB$ will undergo the same row permutation, so that even if $AB = I$, we'll have $A'B \ne I$.

• Mmm... Suppose I add that? Dec 25, 2016 at 22:12
• Doing so is the first step to a "chameleon question"; at that point, I tune out. Dec 25, 2016 at 22:13

if all eigenvalues are distinct, then AB is a diagonal (with the correct ordering and normalization)

Let $aM = c_1*a$, $Mb=c_2*b$ $0 = aMb - aMb = c_1ab-c_2ab=(c1-c2)ab$, so the right and left eqgenvectors with the distinct eigenvalues are orthogonal

• You wrote "when" but you meant "then", right? Dec 25, 2016 at 23:05
• You must reorder rows and columns so what eigenvalue of i'th row of B and eqigenvalue of i'th column of A will be equals Dec 25, 2016 at 23:07

Consider the matrix A made up rows (0,0,- 1,0),(0,0,1,0),(0,0,1,0) and (0,0,0,1). You will find that R made up of rows 1,0,- 1,0/0,1,1,0/0,0,1,0/0,0,0,1 and L with rows 1,1,0,0/1,0,1,0/0,0,1,0/0,0,0,1 are the right and left eigenvector matrices. That is if D is the doagonal (0,0,1,1) then AR =RD, LA=DL together RL,LR not equal and niether is equal to identity.