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[I'm sorry, I don't know how to type a matrix using codes...]

I have a 4x4 matrix A,

[2 -1 -3  2; 
 4 -4 -4  4; 
-4  4  4 -4; 
 6  5  7 -6]

and its Jordan normal form

J=
[0 0 0  0;
 0 0 1  0;
 0 0 0  0;
 0 0 0 -4]

I have to find a P such that $P^{-1}AP=J$ but I can't for this one.

Please help, I am hopeless with this one now... thanks a lot in advance!


The method I know is as follows:

①Find the eigenvalues

②Use the eigenvalues to find corresponding eigenvectors.

③If there are eigenvalues with multiplicity more than 1, then

i)Find the eigenvector of the eigenvalue $\lambda$ by solving for $x$ in $(A-\lambda I)x=0$.

ii)Then, solve for $x'$ for $(A-\lambda I)x'=x$.

iii) If multiplicity more than 3, then carry on as $(A-\lambda I)x''=x'$.

④After all the eigenvectors are found by above, then the column matrix of all the vectors, $[x_1 x_1' x_1'' x_1''' ...x_2 x_2'... ]$ where each $x_i$ is the eigenvector for distinct eigenvalues $\lambda_i$.

Then that matrix is P.

Well, at least that's how it goes for 2x2 and 3x3 matrices in my textbook. But The trouble with this one in particular, is that, for $\lambda=0$, I don't get an eigenvalue of the form $c_1x$ where $x$ is the vector, but $c_1x_1+c_2x_2$ thus a linear combination of 2 vectors.

I don't know how to treat that and fit it into the matrix to form a column vector matrix which will be a part of the solution $P$...

Are there any other methods??? Or, what should I do with the eigenvector in this case?

What if I get more combinations...say $c_1x_1+c_2x_2+c_3x_3+....$ for my eigenvector? Or...have I calculated the eigenvector for 0 wrongly?

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Notice that if we let $P=[X_1,X_2,X_3,X_4]$ then $AP=PJ$ implies \begin{equation}AX_1=0, \end{equation} \begin{equation}AX_2=0, \end{equation} \begin{equation}AX_3=X_2, \end{equation} \begin{equation}AX_4=-4X_4. \end{equation}


What this tells you: There are two linearly independent vectors in the eigenspace of $A$ associated with the eigenvalue 0. Furthermore, one of these vectors are also in the range of $A$. You can also make use of: $X_3$ must be non-zero, and $A^2X_3=AX_2=0$. The remaining eigenspace, associated with -4, you can determine in the usual way.

I hope this helps.

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  • $\begingroup$ Thanks for answering! Ah, okay...I'll try that method and see how it goes, will come back if I have further issues or get stuck! $\endgroup$ – Melba1993 May 13 '15 at 14:54

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