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i would be very thankful if someone could help me on this question, i know how to do the first bit but the last two questions confuse me a little. thanks in advance

$M = \left( \begin{smallmatrix} 8&40&-30\\ 25&98&-75\\ 35&140&-107 \end{smallmatrix} \right)$

Let $v = [1, 2, 3]^T$

Show that $v$ is an eigenvector for the matrix M and determine the associated eigenvalue, say $\lambda$.

Determine the dimension of the eigenspace, $E_\lambda$, of M.

It is known that 3 is also an eigenvalue of M. Quoting any general result you need, determine whether or not M is diagonalisable.

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The title is not sentence. –  lhf Jul 8 '12 at 12:12
1  
What are your thoughts? Please include them in the question. The simplest way for answering all 3 questions would be to simply find out the eigenvalues (They are 3,-2,-2), the corresponding eigen vectors and then find out whether all of them are linearly independent. Diagonalisable is a little tricky since you don't have 3 different eigen values, so you'll have to work it out. Do you know how? –  Inquest Jul 8 '12 at 12:20
    
Do you know what dimension is? Do you know what Nullity is? Do you know when an endomorphism (linear transformation on some vectorspace V, your matrix is just representation of such transformation) is diagonalisable? In general, it's good to know what you already know. If you don't understand some of these terms, let me know, we'll try to make them clear ;) –  Lissa Jul 8 '12 at 15:57

3 Answers 3

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$$M =\begin{pmatrix} 8&40&-30\\ 25&98&-75\\ 35&140&-107 \end{pmatrix}\begin{pmatrix} 1\\2\\3\end{pmatrix}=\begin{pmatrix} -2\\-4\\-6\end{pmatrix}=(-2)\begin{pmatrix}1\\2\\3\end{pmatrix}$$

Also $$M-(-2)I=\begin{pmatrix} 10&40&-30\\ 25&100&-75\\ 35&140&-105 \end{pmatrix}=10\cdot 25\cdot 35\begin{pmatrix} 1&4&-3\\ 1&4&-3\\ 1&4&-3 \end{pmatrix}$$ So $\,\dim\ker(M+2I)=1\,$ and since the eigenvalue $\,-2\,$ has algebraic multiplicity $\,2\,$ then the matrix is not diagonalizable.

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The dimension of the eigenspace $E_{\lambda}$ just means the dimension of the nullspace of $M-\lambda I$. You know how to find the dimension of the nullspace of a matrix?

For the last part, you need the answers to the first parts. Some facts that may be useful:
1. The sum of the eigenvalues is equal to the trace.
2. Eigenvectors belonging to distinct eigenvalues are linearly independent.

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In case you knew Jordan forms and you're entitled to work over the complex field, you can use the following fact (which perhaps you can yourself prove!) The number of Jordan blocks corresponding to an eigen value is the geometric multiplicity of the eigen value (i.e., the dimension of the nullspace of $M-\lambda I$) To get the Jordan form you need the second answer. The Jordan form turns out to be a diagonal matrix with diagonal entries $3,2$ and $2$.

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Your answer and that of DonAntonio can't both be right. –  Gerry Myerson Jul 9 '12 at 6:22

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