Decomposition for non symmetric matrix using left and right eigenvectors

Question

I have a few questions on decomposition of a matrix based on left and right eigenvectors.

Given a non-symmetric matrix,

1. Are its left and right eigenvectors always bi-orthonormal? Can the left and right eigenvectors be obtained from the singular value decomposition?
2. Can the matrix be decomposed in terms of its left and right eigenvectors?
3. Do either answers change if we assume that each eigenvalue has algebraic multiplicity equal to geometric multiplicity?

Either a direct answer to the question, or citations to references on this topics, both are welcome and appreciated!

Answer 1

For a square matrix $A$, $x$ is a right-eigenvector if $Ax = \lambda x$ for some scalar $\lambda$ and a left-eigenvector if $A^Tx = \lambda x$ for some scalar $\lambda$. $A^T$ denotes the transpose of $A$ (notably, not the Hermitian transpose in the case that $A$ or its eigenvectors are complex).

If $x$ is a right-eigenvector with $Ax = \lambda x$ and $y$ is a left-eigenvector with $A^Ty = \mu y$ with $\lambda \neq \mu$, then it necessarily holds that $x^Ty = 0$. I suspect that in your terminology, we might say that left and right eigenvectors are "biorthogonal, but not necessarily biorthonormal" (that said, this is the first time I remember seeing the term "biorthonormal").

A matrix can be completely decomposed in terms of left and right eigenvectors if and only if it is diagonalizable, which is to say that each eigenvalue has equal algebraic and geometric multiplicity. In this case, if $P$ is a matrix whose columns form a basis consisting of right-eigenvectors of $A$ (i.e. an $A$-eigenbasis), then the matrix $\Lambda := P^{-1}AP$ is diagonal with diagonal entries $\lambda_1,\dots,\lambda_n$ equal to the diagonal entries of $A$. Moreover, every matrix $P$ for which $P^{-1}AP$ is diagonal has columns consisting of an $A$-eigenbasis. If we let $Q = [P^{-1}]^T = [P^T]^{-1} = P^{-T}$, then we can "decompose" $A$ into the form $$ A = PDP^{-1} = PDQ^T = \sum_{j=1}^n \lambda_j \, p_jq_j^T, $$ where $p_1,\dots,p_n$ are the columns of $P$ and $q_1,\dots,q_n$ are the columns of $Q$. Notably, the columns of $Q$ are left-eigenvectors of $A$, as can be seen from the fact that $$ D = D^T = [P^{-1}AP]^T = P^{T}A^TP^{-T} = Q^{-1}A^TQ, $$ which means that the columns of $Q$ form an $A^T$-eigenbasis. The vectors $q_1,\dots,q_n$ (obtained in this particular fashion from $P$) form a "biorthonormal pair" with the vectors $p_1,\dots,p_n$ (i.e. $q_i^Tp_j$ is $0$ when $i \neq j$ and $1$ when $i = j$), as can be seen from the fact that $Q^TP = I$.