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Suppose we have the matrix $$A=\begin{bmatrix}4 & -3 \\ 2 & -1 \end{bmatrix} $$ which has eigenvalues $\lambda_1=1$ and $\lambda_2 = 2$. Then $$A-\lambda_1 I=\begin{bmatrix}3 & -3 \\ 2 & -2 \end{bmatrix}\quad \text{and} \quad A-\lambda_2 I=\begin{bmatrix}2 & -3 \\ 2 & -3 \end{bmatrix}.$$ Why do vectors in the column space for $A-\lambda_2 I$ act as eigenvectors corresponding to $\lambda_1$, and vice versa? (E.g. $(2,2)^T$ is an eigenvector for $\lambda_1$.)

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  • $\begingroup$ Observe that $(A-\lambda_1I)/(\lambda_2-\lambda_1)$ is a projection onto the eigenspace of $\lambda_2$ and $(A-\lambda_2)/(\lambda_1-\lambda_2)$ projects onto the eigenspace of $\lambda_1$. $\endgroup$ – amd Oct 26 '16 at 8:36
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Note that by Cayley-Hamilton theorem

$$(A-\lambda_1 I)(A-\lambda_2 I)= A^2 -(\lambda_1 + \lambda_2) A+\lambda_1 \lambda_2 I=0$$

taking $\vec u_2, \vec v_2$ as the column vectors of $A-\lambda_2 I$

We must have $(A-\lambda_1 I)\vec u_2= \vec 0$ and $(A-\lambda_1 I) \vec v_2 = \vec 0$

So the column vectors of $A-\lambda_2$ are in the null space of $A-\lambda_1 I$, so they are eigenvectors of $A$ corresponding to $\lambda_1$

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  • $\begingroup$ Do we need to work over $\mathbb{C}$ to be able to factor the characteristic polynomial as $(A-\lambda_1 I)(A-\lambda_2 I)$? And it seems that this occurs for a matrix of any size, as long as we have (only) two eigenvalues? $\endgroup$ – Alain Oct 26 '16 at 7:12
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    $\begingroup$ @pikachau: You don't have to work over $\mathbb{C}$. It works whenever $A \in M_n(\mathbb{F})$ is diagonalizable and has two distinct eigenvalues (which, by definition, are in $\mathbb{F}$). When $n = 2$, the fact that $A$ has two distinct eigenvalues guarantees that it is diagonalizable and so you don't need to check that. If $n > 2$, then it doesn't and even if you find that it has two distinct eigenvalues, the generalization of the method might fail. $\endgroup$ – levap Oct 26 '16 at 7:29
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Assuming that $A\in\mathbb R^{2\times2}$ has distinct eigenvalues $\lambda_1$ and $\lambda_2$, we can write any vector as a linear combination $a\mathbf v_1+b\mathbf v_2$ of eigenvectors corresponding to these eigenvalues. Since $A\mathbf v_2=\lambda_2\mathbf v_2$, we then have $$(A-\lambda_2I)(a\mathbf v_1+b\mathbf v_2) = a(\lambda_1-\lambda_2)\mathbf v_1,$$ that is, the image (column space) of $A-\lambda_2I$ is spanned by the eigenvector $\mathbf v_1$. Similarly, $(A-\lambda_1I)(a\mathbf v_1+b\mathbf v_2)=b(\lambda_2-\lambda_1)\mathbf v_2$.

Observe that this suggests a decomposition of $A$ into a sum of projections $P_1$ and $P_2$ onto its eigenspaces, with $$P_1={A-\lambda_2I\over\lambda_1-\lambda_2}\text{ and }P_2={A-\lambda_1I\over\lambda_2-\lambda_1}.$$

N.B.: This is essentially the same as Nick Liu’s answer.

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View $A$ as a linear operator from $\Bbb R^2$ to $\Bbb R^2$,

Note $\Bbb R^2=E_{\lambda_1}\oplus E_{\lambda_1}$, where $E_{\lambda_i}$ denotes the eigenspace of $\lambda_i$.

and $E_{\lambda_1}=N(A-\lambda_1)=A-\text{col}(A-\lambda_1I)=\text{col}(A-\lambda_2I$), and similar case for $\lambda_2$.
Therefore any $v\in\text{col}(A-\lambda_2I)$ is an eigenvector of $\lambda_1$, and similar case for $\lambda_2$.

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  • $\begingroup$ Sorry, I don't really understand the notation.. $A$ is a function and $col(A-\lambda_1 I)$ is a vector space thats being subtracted from $A$? $\endgroup$ – Alain Oct 26 '16 at 7:21
  • $\begingroup$ This is column space. $\endgroup$ – Nick Oct 26 '16 at 8:25
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I am not quite satisfied with this answer, so I hope others can give better ones.

$A$ has distinct eigenvalues so it is diagonalizable with an eigenbasis $\{v_1,v_2\}$ corresponding to $\lambda_1$ and $\lambda_2$.

If you apply $A-\lambda_2 I$ on some vector $v=c_1v_1+c_2v_2$, then $$(A-\lambda_2 I)(c_1v_1+c_2 v_2) = (A-\lambda_2 I)(c_1 v_1) = c_1(\lambda_1-\lambda_2)v_1 \in \operatorname{span}\{v_1\}.$$

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