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This is a follow up question to, Eigenvectors of a matrix and its diagonalization.

Do the eigenspaces corresponding to the same eigenvalues of similar matrices describe the same subspaces? For example, let's say I have two similar matrices $A,D\in \mathbb{R}^3$ and that $D$ is diagonal. Suppose $\lambda_1$ is a unique eigenvalue of both and that $V_{\lambda_1}$ is an eigenspace of $A$ describing a line in $\mathbb{R}^3$. Does the eigenspace of $D$ corresponding to $\lambda_1$ describe the same line?

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No, in fact, if $A=C^{-1}BC^{}$ and $w$ is an eigenvector of $A$, then $Cw$ is an eigenvector of $B$ with same eigenvalue. In general $w$ and $Cw$ are not in the same space. –  juanrapha Jun 26 '12 at 7:29
    
@JuanSimões Could you give an example? –  Robert S. Barnes Jun 26 '12 at 7:30
    
Take a diagonal 2x2 matrix that is not the identity, and $C$ to be a rotation of $\theta$. Then the eigenspaces will be rotated of the same angle. –  juanrapha Jun 26 '12 at 7:34
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Robert, you're not even trying. Let $$A=\pmatrix{1&2\cr0&3\cr},\qquad D=\pmatrix{1&0\cr0&3\cr}$$ They are similar, and 3 is an eigenvalue of both. Are the eigenvectors the same? –  Gerry Myerson Jun 26 '12 at 9:43
    
See also Eigenvectors of $P^{-1}AP$ –  Jonas Meyer Jun 26 '12 at 16:05
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You have to distinguish between the "underlying" linear transformation and the representation.

Recall that similar matrices $A$ and $B$ may be thought of as representing the same linear transformation with respect to different basis. What this means, explicitly, is that we have an automorphism (a bijective map that respects sums and scalar multiplications) $P\colon\mathbb{R}^n\to\mathbb{R}^n$, with the property that for every vector $\mathbf{v}$, $P(A\mathbf{v})) = B(P(\mathbf{v}))$.

You can think of $P$ as a "dictionary" that from $\beta$-speak to $\gamma$-speak, where $\beta$ and $\gamma$ are two distinct basis, in the sense that if $[\mathbf{v}]_{\beta}$ is the coordinate vector of $\mathbf{v}$ with respect to $\beta$, and $[\mathbf{v}]_{\gamma}$ is the coordinate vector of $\mathbf{v}$ with respect to $\gamma$, then $P[\mathbf{v}]_{\beta}=[\mathbf{v}]_{\gamma}$.

In that sense, if you look at the underlying vectors (not at the coordinate vectors), then the eigenvectors of similar matrices are the same vector: after all, $A$ and $B$ represent the same linear transformation, it's just that $A$ "speaks" $\beta$-language and $B$ "speaks" $\gamma$-language. It's just like saying that The Odyssey in English is the same as The Odyssey in Russian: it's the same story, with the same characters, doing the same things.

On the other hand, if you look at the coordinate vectors, so that you view each of $A$ and $B$ as simply operating on $\mathbb{R}^n$ with the standard basis, then the eigenspaces need not be the same; for instance, the matrices $$A=\left(\begin{array}{cc}1&1\\1&1 \end{array}\right)\quad\text{and}\quad B=\left(\begin{array}{cc}2&0\\0&0\end{array}\right)$$ are similar, via $P^{-1}AP = B$ with $$P=\left(\begin{array}{rr}1&1\\ 1 & -1\end{array}\right),$$ but the eigenspaces of $A$ are $\{(a,a)\mid a\in\mathbb{R}\}$ and $\{(b,-b)\mid b\in\mathbb{R}\}$, while the eigenspaces of $B$ are $\{(a,0)\mid a\in\mathbb{R}\}$ and $\{(0,b)\mid b\in\mathbb{R}\}$. Following the analogy from above, it's because on their face, The Odyssey in English is not the same as The Odyssey in Russian: different alphabets, different words, different sentence structure, etc.

So it depends how you want to view the matrices. As operators on $\mathbb{R}^n$, no, they don't have to have the same eigenspaces. As coordinate matrices of a particular linear transformation with respect to different bases, then yes, they have the same eigenspaces... but they describe them differently: one uses coordinate vectors with respect to one basis, the other matrix uses coordinate vectors with respect to a different matrix.

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You got the point of what was bugging me. But I don't quite get this last sentence: "As coordinate matrices of a particular linear transformation with respect to different bases, then yes, they have the same eigenspaces... but they describe them differently". The point about the eigenvectors being in some sense "the same" vector according to different bases was what was bugging me, but since the resulting eigenspaces in your example describe different lines I'm not really sure what the nature of that "sameness" is. Is it like the abstract nature of transformations in $Hom(V,W)$? –  Robert S. Barnes Jun 27 '12 at 8:16
    
This may be similar to my problem conceptualizing the nature of the objects in $Hom(V,W)$. math.stackexchange.com/q/158542/9381 –  Robert S. Barnes Jun 27 '12 at 8:18
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@RobertS. Barnes: Say $\beta$ is the standard basis, and $\gamma$ is another basis, and $P$ is the change-of-basis matrix from $\beta$ to $\gamma$. If we interpret $A$ as the coordinate matrix of the transformation $(\mathbb{R}^n,\beta)\to(\mathbb{R}^n,\beta)$ given by $\mathbf{v}\mapsto A\mathbf{v}$, and $B=PAP^{-1}$, then we can view $B$ as acting on $\mathbb{R}^n$ directly by matrix multiplication, or we interpret $B$ as the coordinate matrix relative to $\gamma$, by the rule $\mathbf{v}\mapsto [B[\mathbf{v}]_{\gamma}]_{\beta}$. If you work out that formula in terms of $A$, $B$, $P$ (cont) –  Arturo Magidin Jun 27 '12 at 16:04
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@RobertS.Barnes: (cont) you have that $[\mathbf{v}]_{\gamma}=P\mathbf{v}$, and $[B[\mathbf{v}]_{\gamma}]_{\beta} = P^{-1}(B[\mathbf{v}]_{\gamma})$. So this map sends $\mathbf{v}$ to $P^{-1}BP\mathbf{v}$. But $P^{-1}BP = A$, so viewing $B$ as "speaking" in terms of the basis $\gamma$, the map it defines is identical to the map defined by $A$. It's only when you try to interpret $B$ as speaking in $\beta$ (instead of $\gamma$, which is what $B$ "actually" speaks) that it looks like a different linear transformation than $A$. (cont) –  Arturo Magidin Jun 27 '12 at 16:07
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@RobertS.Barnes: This problem only arises when you are working directly on the "standard vector space" $\mathbf{R}^n$. If you are working on a arbitrary vector space $V$ with a linear transformation $T$ and basis $\beta$ and $\gamma$, what we have is that $A=[T]_{\beta}^{\beta}$ and $B=[T]_{\gamma}^{\gamma}$. But the underlying linear transformation, $T$, is the same, so they have the same eigenvectors; it's just that in order to use $A$ you first must translate the vectors into $\beta$-speak, and if you want to use $B$ instead you must translate them into $\gamma$-speak. –  Arturo Magidin Jun 27 '12 at 16:09
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