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Which of the following matrices have Jordan canonical form of equal to the $3\times 3$ matrix

$$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

a)$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

b)$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

c)$ \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

d)$ \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

Here characteristic equation of the matrix is $x^3$.Hence the 3 eigenvalues of the matrix are zero. Do we want to find the eigenvalues of all the matrices in the options?Is there any other way?

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    $\begingroup$ A friendly note, you can check out this page here to learn how to properly format mathematics on this site. In doing so, your matrices can look like $\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}$ $\endgroup$
    – JMoravitz
    May 6, 2015 at 16:36
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    $\begingroup$ @jgon great edit. As a matter of efficiency, however, note that \pmatrix{a&b\\c&d} will work; I find this to be quicker than the whole "begin/end" deal. $\endgroup$ May 6, 2015 at 16:39
  • $\begingroup$ @Omnomnomnom Wow thanks so much for that, I had no idea, that will help so much. $\endgroup$
    – jgon
    May 6, 2015 at 16:40

2 Answers 2

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Finding the eigenvalues of any of these matrices is not particularly difficult; they are all upper-triangular.

Hint: Suppose that $A$ is a $3 \times 3$ matrix with zero as its only (complex) eigenvalue. Note that $A$ has the desired J-C form if and only if $A \neq 0$ but $A^2 = 0$ (why?).

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The given Jordan canonical form implies, minimal polynomial of corresponding matrix should be $x^2=0.$ Hence if matrix $A$ is having the property that $A \neq 0$ and $A^2=0 ,$ it will have desired Jordan canonical form.

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