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I have the matrix $A=\begin{pmatrix}2&2&1\\-1&0&1\\4&1&-1\end{pmatrix}$, I want to write it in Jordan-Normal Form. I have $x_1=3,x_2=x_3=-1$ and calculated eigenvectors $v_1=\begin{pmatrix}1\\0\\1\end{pmatrix},v_2=\begin{pmatrix}1\\-4\\5\end{pmatrix},v_3=\begin{pmatrix}0\\0\\0\end{pmatrix}$. But, the matrix $Z=\begin{pmatrix}1&1&0\\0&-4&0\\1&5&0\end{pmatrix}$ is not invertible since $\text{det}(Z)=0$. Does this mean the matrix cannot be written in JNF or do I need to find different eigenvectors?

I have tried to find different eigenvectors, but keep arriving at the same problem, any suggestions?

Thanks

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  • $\begingroup$ Note that the null vector is by definition no eigenvector. The null vector can’t be part of a basis either. That’s the problem you’re encountering. $\endgroup$ – k.stm Mar 27 '13 at 17:12
  • $\begingroup$ The zero vector is usually excluded from the definition of "eigenvector", since it is such for any transformation and any value..! $\endgroup$ – DonAntonio Mar 27 '13 at 17:12
  • $\begingroup$ Ah yes, I have forgotten this, but the only reason I chose $0$ was that I was struggling to find to linearly independent eigenvectors for $x_2$ and $x_3$ $\endgroup$ – hello123 Mar 27 '13 at 17:14
  • $\begingroup$ Technically you don't count $(0,0,0)$ as an eigenvector. What has happened here is that the eigenspace $W_{-1}$ only has one eigenvector. This should give you a clue as to how to reduce it to Jordan normal form as the whole reason we need Jordan blocks is because we can't find as many eigenvectors as we can eigenvalues (counting multiplicity). Otherwise we'd just be diagonalising it. $\endgroup$ – muzzlator Mar 27 '13 at 17:14
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    $\begingroup$ It's been a long day... $\endgroup$ – hello123 Mar 27 '13 at 17:16
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You need a generalized eigenvector for the third eigenvalue and it looks like that went wrong somehow, so lets fix it.

We have $\left(A - \lambda_2 I\right)v_3 = v_2$

From this, we get:

$\begin{pmatrix}3&2&1&1\\-1&1&1&-4\\4&1&0&5\end{pmatrix}$

The RREF yields:

$\displaystyle \begin{pmatrix}1&0&-\frac{1}{5}&\frac{9}{5}\\0&1&\frac{4}{5}&-\frac{11}{5}\\0&0&0&0\end{pmatrix}$

This gives us a generalized eigenvector of: $\displaystyle \left(\frac{9}{5}, -\frac{11}{5}, 0 \right)$.

To write the Jordan Normal Form, we form:

$\displaystyle A = S\cdot J\cdot S^{-1} = \begin{pmatrix} 1 & \frac{9}{5} & 1\\ -4 & -\frac{11}{5} & 0 \\ 5 & 0 & 1\end{pmatrix} \cdot \begin{pmatrix} -1 & 1 & 0\\ 0 & -1 & 0\\ 0 & 0 & 3 \end{pmatrix} \cdot \begin{pmatrix} -\frac{11}{80} & -\frac{9}{80} & \frac{11}{80} \\ \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{11}{16} & \frac{9}{16} & \frac{5}{16}\end{pmatrix}$.

Notice the structure of the Jordan block. Also, notice what the columns of $S$ and $J$ are made of? Clear?

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  • $\begingroup$ Very nicely done, once again $\ddot\smile\quad \checkmark$ $\endgroup$ – amWhy Apr 12 '13 at 0:14
  • $\begingroup$ Yes, there's a limit to how many posts one can read, so I tend to read the posts of those who tend to take care with their answers, whose approach I appreciate, and whose contributions are valuable (but occasionally go unnoticed!) $\endgroup$ – amWhy Apr 12 '13 at 0:19
  • $\begingroup$ Interesting question you link to...I missed it until now. $\endgroup$ – amWhy Apr 12 '13 at 0:22
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As already stated, the zero vector is usually excluded from the definition or eigenvector.

Besides this, solving the linear homogeneous sytem form the eigenvalue $\,-1\,$ we get that the solution space is of dimension 1 and from here that there's only one single linearly independent eigenvector corresponding to this value.

Since the eigenvalue has algebraic multiplicity ($=(x+1)^2\,$ is its factor in the charac. polynomial) of$\,2\,$ but geometric multiplicity (=the corresponding eigenspace's dimension) is $\,1\,$ , your matrix isn't diagonalizable.

Knowing the eigenvectors and that the matrix isn't diagonalizable, the JCF is immediate now:

$$\begin{pmatrix}3&0&0\\0&\!\!\!\!-1&1\\0&0&\!\!\!\!-1\end{pmatrix}$$

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