I was wondering if there is any relation between the space span by all eigenvector of a matrix $A$ and its column space.
Also, is there any condition on $A$ that ensures that these two spaces are the same?
Linear Algebra – Relation Between Eigenspace and Column Space of a Matrix
eigenvalues-eigenvectorslinear algebramatrices
Related Solutions
Suppose that $v$ is an eigenvector of $A$ with eigenvalue $\lambda.$ Then $v$ is also an eigenvector of $A^{-1}$, but with eigenvalue $1/\lambda:$
$$v = 1.v = (A^{-1}A)v = A^{-1}(\lambda v) = \lambda (A^{-1}v) \quad \Rightarrow \quad A^{-1}v = \lambda^{-1} v.$$
Now let $\{ \lambda_i \}_{i=1}^n$ be the spectrum of $A$, and let the $\lambda_i$ be ordered: $\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n.$ What can you say about the spectrum of $A^{-1}$?
Lemma 1: Given an $m\times n$ matrix $A,$ the null space of $A^T$ is the orthogonal complement of the column space of $A.$
Proof: Write $A=[c_1\:\cdots\:c_n]$ where the $c_j$ are the columns of $A,$ and note that for any $m$-dimensional vector $x$ we have $$A^Tx=\left[\begin{array}{c}c_1^T\\\vdots\\c_n^T\end{array}\right]x=\left[\begin{array}{c}c_1^Tx\\\vdots\\c_n^Tx\end{array}\right]=\left[\begin{array}{c}c_1\cdot x\\\vdots\\c_n\cdot x\end{array}\right].$$ Since the column space of $A$ is spanned by $c_1,...,c_n$, then $x$ is in the orthogonal complement to the column space of $A$ if and only if $x$ is orthogonal to each $c_j$ if and only if each $c_j\cdot x=0$ if and only if $A^Tx$ is the $n$-dimensional zero vector if and only if $x$ is in the null-space of $A^T.$ $\Box$
Lemma 2: Let $V,W$ be subspaces of some finite-dimensional space $X$. $V$ and $W$ have the same orthogonal complement if and only if $V=W$.
Proof: If $V=W$, then their orthogonal complements are trivially the same.
Suppose $V,W$ have the same orthogonal complement. Take $v\in V$. Since $X$ is the direct sum of $W$ and its orthogonal complement, and since $x\in V\subseteq X$, then there exist unique $w,w'$ such that $v=w+w',$ $w\in W$, and $w'$ in the orthogonal complement of $W$. Since $V,W$ have the same orthogonal complement, then $w'$ is orthogonal to $v,$ and so $$0=v\cdot w'=(w+w')\cdot w'=w\cdot w'+w'\cdot w'.\tag{$\star$}$$ Since $w'$ is in the orthogonal complement of $W$ and $w\in W$, then $w\cdot w'=0$, so it follows by $(\star)$ that $$w'\cdot w'=0.$$ Now, no non-zero vector is self-orthogonal, so $w'$ must be the zero vector, whence $v=w\in W$, and so $V\subseteq W$. By symmetrical arguments, we likewise have $W\subseteq V$, so $V=W$. $\Box$
Proposition: Given matrices $A,B$ of the same dimensions, $A$ and $B$ have the same column space if and only if $A^T$ and $B^T$ have the same reduced row echelon form.
Proof: Let $rref(M)$ indicate the reduced row echelon form of a matrix $M$. Recall that we can obtain $rref(M)$ by Gauss-Jordan elimination, which involves multiplication on the left by some finite collection of elementary matrices--that is, for any $M$, there exist elementary matrices $E_1,\cdots,E_n$ of appropriate dimension such that $rref(M)=E_n\cdots E_1M.$ This collection of elementary matrices is not unique, but that isn't important. Note, though, that elementary matrices are invertible, so it follows that the null spaces of $rref(M)$ and $M$ are the same.
Thus, $A^T$ and $B^T$ have the same reduced row echelon form if and only if they have the same null space. By Lemma 1, $A^T$ and $B^T$ have the same null space if and only if the column spaces of $A$ and $B$ have the same orthogonal complement. By Lemma 2, the column spaces of $A$ and $B$ have the same orthogonal complement if and only if the column spaces of $A$ and $B$ are the same. $\Box$
Upshot: The Proposition lets us get around needing to know what the column spaces of two matrices are, and simply determine whether they have the same column space by converting their transposes to reduced row echelon form.
Best Answer
To add a bit to Gerry's answer...
If you look at a matrix as a linear operator, $T(v)=Av$ then the column space is just the range of that linear operator. Eigenvectors for non-zero eigenvalues will be members of the range (if $Av=\lambda v$, then $A(\lambda^{-1}v) = v$).
So the span of the eigenvectors with non-zero eigenvalues, is contained in the column space.
(...and the span of the eignevectors with eigenvalue zero is the null space.)