I am having trouble understanding this proof that every operator has an upper-triangular matrix.

$\lambda=$ is an eigenvalue of $T$, for $T \in L(V)$ where $V$ is a vector space on $F^n$, they say :

suppose $U = \mathcal{R}(T-\lambda I)$, then $\dim(U)<\dim(V)$ because $U$ is not surjective.

$Tu=(T-\lambda I)u + \lambda u$ shows $T$ is invariant under $U$, because both terms are in U.

So above they are showing that since $T$ is an operator on $U$, $T$ has an upper triangular matrix with respect to some basis of $U$, $u_1,...u_m$ because in this is a claim they are trying to prove by induction.

Because $T$ has an upper triangular matrix, this means $Tu_j=(T|_u)(u_j)\in \operatorname{span}(u_1...u_j)$

I think I get all of this so far. Here is what I don't understand:

Extend $u_1...u_m$ to a basis of $V$, $u_1,...u_m,v_1,...,v_n$. For each $k$, $Tv_k=(T-\lambda I)v_k + \lambda v_k$

$(T-\lambda I)v_k \in U = \operatorname{span}(u_1,...,u_m) => Tv_k \in \operatorname{span}(u1,...,u_m,v_1,...,v_k)$

I guess $(T-\lambda I)v_k \in U $ because it equals zero, and zero is in $U?$ So why does it mean $Tv_k \in \operatorname{span}(u1,...,u_m,v_1,...,v_k)?$ $\lambda v_k$ is an eigenvector/eigenvalue, but is it true that there is only 1 independent eigenvector per eigenvalue? This one eigenvalue works on $v_1...v_k?$

Given Proof

  • 5
    $\begingroup$ The page copied above starting with 5.27 is from my book Linear Algebra Done Right (third edition). When that much material is copied, the original source should be cited. The chapter titled Eigenvalues, Eigenvectors, and Invariant Subspaces, which includes the page above, is freely available on the book's website: linear.axler.net $\endgroup$ Jun 21, 2018 at 1:01
  • $\begingroup$ Is the only reason why we have to choose an eigenvalue ${\lambda}$ and define the operator as T - ${\lambda}$I because we want dim(U) < dim(V)? Otherwise any other choice of a scalar k (thus defining the operator as T - kI) would work also for the rest of the proof right? $\endgroup$ Oct 30, 2023 at 2:56

1 Answer 1


By definition, $\mathrm{range}(A)=\{A.v ; v\in V\}$ for any $A \in \mathcal L(E)$. So $U=\mathrm{range}(T-\lambda I)=\{\left(T-\lambda I\right)v ; v\in V\}$. $v_k$ being a vector of $V$ like any other vectors of this vector space, you have $\left(T-\lambda I\right)v_k \in \mathrm{range}(T-\lambda I)=U$.

The rest follows.

The $v_k$ are not supposed in that proof to be related to the eigenvalue $\lambda$.

  • $\begingroup$ In this proof, $U $ is made of every vector that is not an eigenvector for the eigenvalue $\lambda$. Isn't whats left vectors which correspond to the eigenvalue? So how can the $v_k$ not be related to the eigenvalue? Also if $v_k \not\in span(u_1...u_m)$ then $v_k \in null(T-\lambda I)$ isn't it? Since $range(T-\lambda I)$ is every vector that $(T-\lambda I)$ does not send to $0$? So isnt the null space made of eigenvectors? $\endgroup$
    – Frank
    Jun 21, 2018 at 2:27
  • $\begingroup$ LIn this proof... No. I suggest that you read and read again the definition of $U$ to convince yourself of what is this object. Then the $v_k$ are used to extend a basis of $U$ to a basis of $V$. Again, there is no related to eigenvalue. A vector that is not in $U$ is NOT an eigenvector. $\endgroup$ Jun 21, 2018 at 4:42

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