Given an irreducible non-symmetric, non-negative real matrix $A$ with eigenvalues $\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq \cdots$ and $k$ of its largest right eigenvectors:

\begin{eqnarray} % A v_1 &= \lambda_1 v_1 \\ A v_2 &= \lambda_2 v_2 \\ \vdots & \vdots \\ A v_k &= \lambda_k v_k \\ % \end{eqnarray}

What (if anything) can we say about it's $k$ largest left eigenvectors? \begin{eqnarray} % w_1 A &= \lambda_1 w_1 \\ w_2 A &= \lambda_2 w_2 \\ \vdots & \vdots \\ w_k A &= \lambda_k w_k \\ % \end{eqnarray}


2 Answers 2


If $wA = \lambda w$, we have, using transposition, $A^T w^T = \lambda w^T$.

Assuming $A$ is diagonalizable, we have $A = T^{-1} D T$ for an invertible matrix $T$ and a diagonal matrix $D$, so $A^T = T^T D^T (T^{-1})^T = T^T D (T^T)^{-1}$.

Therefore, $\lambda w^T = A^T w^T = T^T D (T^T)^{-1} w^T$, or equivalently, $\lambda (T^T)^{-1} w^T = (T^T)^{-1}A^T w^T = D (T^T)^{-1} w^T$. But $\lambda v = D v$ iff $T^{-1} v$ is an eigenvector of $A$ (simple calculation), so $T^{-1} (T^T)^{-1} w^T = (T^T T)^{-1} w^T$ is an eigenvector of $A$.

Thus, $(T^T T)^{-1} w^T = v$ for some right eigenvector $v$, or equivalently, $w = ((T^T T) v)^T = v^T T^T T$. This should generalize to the non-diagonalizable case using arguments of the Jordan form.

  • $\begingroup$ Thanks for the answer Johannes. As I see it, doesn't the term $T^T T$ require to know all of the right eigenvectors $v$? Can you calculate a subset of the $w$'s from a subset set of the $v$'s? In the question as stated, I only know a small subset of the $v$'s ($k$ largest). $\endgroup$
    – Hooked
    Apr 19, 2012 at 16:22
  • $\begingroup$ It can probably be done by doing something smart with block matrices; just play around a bit. $\endgroup$ Apr 19, 2012 at 16:37
  • $\begingroup$ Actually, scratch that. It could well be that there is no good solution if you have too little information about the eigenvectors; maybe there is some numeric trickery that helps approximate $T^T T$? $\endgroup$ Apr 19, 2012 at 16:39
  • $\begingroup$ If you come across it, I'd be interested. Either way, your answer, and more importantly, your quick proof was very useful. $\endgroup$
    – Hooked
    Apr 19, 2012 at 18:08

Your left eigenvectors are characterized by the null space of $V$, as explaned in the following.

The matrix form of your known equations is $$AV = VD$$ where $D = diag(\lambda_i)$ for $1 \le i \le k$ and the columns of $V$ are the known eigenvectors.

Writing in terms also of the column matrix of unknown vectors, call it $W$, we have $$A\pmatrix{V & W} = \pmatrix{V & W}\pmatrix{D & \mathbf{0} \\ \mathbf{0} & D_w}$$

Since $\pmatrix{V & W}$ is invertible (it is a basis) it diagonalizes A, and we may write $$\pmatrix{V & W}^{-1}A\pmatrix{V & W} = \pmatrix{D & \mathbf{0} \\ \mathbf{0} & D_w}$$

Denote $\pmatrix{V & W}^{-1}=\pmatrix{\tilde{V}^T \\ \tilde{W}^T}$ to obtain (using $n$ as the dimension of $A$) $$\pmatrix{\tilde{V}^T \\ \tilde{W}^T}\pmatrix{V & W} = \pmatrix{\mathbf{I}_k & \mathbf{0} \\ \mathbf{0} & \mathbf{I}_{n-k}} =\pmatrix{\tilde{V}^T V & \tilde{V}^TW \\ \tilde{W}^TV & \tilde{W}^T W}\tag{1}$$ In terms of the $k$ known right eigenvectors $V$: $$\pmatrix{\tilde{V}^T \\ \tilde{W}^T} V = \pmatrix{\mathbf{I}_k \\ \mathbf{0}}$$ From this we see that $n-k$ of the left eigenvectors $\tilde{W}^T$ are contained in the perpendicular space to $V$ since $\tilde{W}^TV = \mathbf{0}$.


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