Let $A \in \mathbb{C}^{n\times n}$ upper triangular matrix with complex entries. The eigenvalues will then be the diagonal entries of $A$. Given an eigenvalue $\lambda$ of $A$, is there a unique $\mathbf{v} \in \mathbb{R}^{n}$ that satisfies $$ A\mathbf{v} = \mathbf{v}\lambda $$ ? I wrote a a small algorithm that will compute the eigenvalues and eigenvectors of a real-valued upper triangular matrix. Comparing my results with MATLAB’s eigs function, I always get the correct eigenvalues and eigenvectors.

Now for a complex-valued upper triangular matrix, my results for the eigenvectors do not match those given by eigs, but upon further inspection, my results always satisfy $$ A\mathbf{v} = \mathbf{v}\lambda $$

which leads me to conclude that the eigenvector associated with a particular $\lambda$ in this case is not unique.


1 Answer 1


In general, any multiple of an eigenvector is also an eigenvector.

Let $A$ be any matrix and $v$ an eigenvector with eigenvalue $\lambda$. Write $w = \mu v$. We show that $w$ is an eigenvector for the eigenvalue $\lambda$ as follows. $$ Aw = A(\mu v) = \mu Av = \mu \lambda v = \lambda w $$

In fact, if the multiplicity of the eigenvalue $\lambda$ is $n>0$ then the number of linearly independent eigenvectors can be anything between $1$ and $n$. For example, tlet $I$ be the $(n\times n)$-identity matrix. Of course, every vector is an eigenvector of $I$ and they all correspond to eigenvalue $1$. So if $n \geq 2$ then $I$ will have multiple linearly independent eigenvectors all corresponding to eigenvalue $1$.


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