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I calculated the eigenvalues of the following matrix to be $a$ and $-b$.

$J = \left( \begin{array}{ccc} a & 0 \\ 0 & -b \end{array} \right)$

But when I use the formula $(J - \lambda I)v = 0$ with either $a$ or $-b$ as eigenvalue to calculate the eigenvectors I get an indeterminate system of equations as they cancel with the $a$ and $-b$ in the matrix.

Yet Wolfram Alpha says the eigenvectors are $(1, 0)$ and $(0, 1)$?

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There's two cases:

  • if $a=-b$ then the given matrix is $aI_2$ and then any non zero vector $v$ is an eigenvector for $J$
  • if $a\ne-b$ then the eigenvectors associated to $a$ are $\lambda (1,0)^T$ where $\lambda\ne0$ and the eigenvectors associated to $-b$ are $\lambda(0,1)^T$ where $\lambda\ne0$.
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if $a=b=0$

Then you have

$$\begin{pmatrix}a&0\\0&-b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$

This means that every basis $(v_1,v_2)$ of $\Bbb{R}^2$ is a basis of eigenvectors. One can verify this by seeing that the dimension of the kernel of $A$ is $2$ and the fact that multiplying the zero matrix by $v_1$ and $v_2$ we get the vector $(0,0)^T$.

If $a=-b$ Then we have

$$(J-aI)=\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}=\begin{pmatrix}0&0\\0&-b+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$

You can do the same reasoning as above to conclude that every basis of $\Bbb{R}^2$ is a basis of eigenvectors.

If $a\neq -b$ we have:

$$(J-aI)v_1=0 \iff\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}v_1=0\implies v_1 = k\cdot \begin{pmatrix}1\\0\end{pmatrix},\quad k\in \Bbb{R},$$

because $$\begin{pmatrix}0&0\\0&-b-a\end{pmatrix} \begin{pmatrix}k\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}.$$

Do you get the second eigenvector?

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  • $\begingroup$ assuming $a\neq b$ otherwise there are two independent solutions $\endgroup$
    – Belgi
    Oct 14 '14 at 20:57
  • $\begingroup$ @Belgi If $a=b$ then $(J-aI) = \begin{pmatrix}0&0\\0&-2a\end{pmatrix}$ and you still have $v = (1,0)^T$ as eigenvector. If $a = -b$ then you have the zero matrix and the eigenvectors are whatever basis of $\Bbb{R}^2$ hence also $e_1, e_2$. What do you mean exactly? $\endgroup$
    – Bman72
    Oct 14 '14 at 21:01
  • $\begingroup$ If $a=b$ we get the zero matrix so the implication that $v=(1,0)^T$ in the first line is not correct $\endgroup$
    – Belgi
    Oct 14 '14 at 21:03
  • $\begingroup$ @Belgi if $a=b$ then $J=\begin{pmatrix} b&0\\0&-b\end{pmatrix}$. $b$ is an eigenvalue and so is $-b$. The eigenvector corresponding to $b$ is the vector $v_1$ such that $(J-bI)v_1=0$ that is $\begin{pmatrix}0&0\\0&-2b\end{pmatrix}v_1=0 \implies $v_1 = \begin{pmatrix}1\\0\end{pmatrix}$ $\endgroup$
    – Bman72
    Oct 14 '14 at 21:06
  • $\begingroup$ Oh sorry, but then I would argue that if $a=-b$ the implication is not correct $\endgroup$
    – Belgi
    Oct 14 '14 at 21:07
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One way to answer this question is to remember the definition of an eigenvector.

An eigenvector $v$ of a matrix $A$ is a vector which satisfied $Av=\lambda v$ for some scalar $\lambda$.

So we can guess and check. My guesses would be $\left(\begin{array}{c} 1\\0\end{array}\right)$ and $\left(\begin{array}{c} 0\\1\end{array}\right)$. And a simple check would verify that these are indeed eigenvectors.

Guess and check is a very reasonable way to solve a math problem.

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