Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So my professor gave me this question:

$A=\begin{pmatrix} 0 & 2 & 5\\ -5 & 5 & 10\\ 2 & -2 & -4 \\ \end{pmatrix}$

I had to calculate $\forall 0 < i$ $kerA^{i}$ and $ImA^{i}$

So after calculating I reached those results:

$kerA$={$\begin{pmatrix} 1 \\ 5 \\ -2 \\ \end{pmatrix}$}

and $\forall 1<i$

$kerA^{i}$={$\begin{pmatrix} 1 \\ -1 \\ 0 \\ \end{pmatrix}$,$\begin{pmatrix} -3 \\ 0 \\ 1 \\ \end{pmatrix}$}

regarding the image it is just the span of the columns of $A^{i}$

Then he asked us to calculate $\forall 0 < i$ $ker(A-I)^{i}$ and $Im(A-I)^{i}$

So after calculating I reached those results:

$\forall 0<i$

$kerA^{i}=$ {$\begin{pmatrix} 0 \\ -5/2 \\ 1 \\ \end{pmatrix}$}

regarding the image it is just the span of the columns of $(A-I)^{i}$

Then he asked us to show the Jordan form of $A$

How can I conclude that from what I proved. As far as I know the diagonal of jordan form contains the eigenvalues of $A$ (each eigenvalue repeat in the diagonal as the number of Algebraic multiplicities of the eigenvale ) and nothing more but how can I conclude from what I proved the eigenvalues of $A$.

Any help would be appreciated

Thanks in advanced !!

share|improve this question
    
There seems to be a typo. Judging from your computed kernel it should be $-4$ in the matrix. –  Julian Kuelshammer Jun 23 '13 at 8:58
    
@JulianKuelshammer fixed, you were right !! –  wantToLearn Jun 23 '13 at 9:06
add comment

1 Answer

up vote 2 down vote accepted

For a general method to compute the normal form in such case: For each eigenvector, start with the power such that the kernel of $A-\lambda I$ does not change anymore (call that $m_\lambda$. Take a basis of a complement of $\operatorname{ker}(A-\lambda I)^{m_\lambda-1}$ in $\operatorname{ker}(A-\lambda I)^{m_\lambda}$. Now apply $(A-\lambda I)$ to that basis. The elements will lie in $\operatorname{ker}(A-\lambda I)^{m_\lambda -1}$. Now take a basis of the complement of the span of these vectors and $\operatorname{ker}(A-\lambda I)^{m_\lambda-2}$ in that space. Continue inductively. You will then get a basis for your whole space. If you change your matrix to that basis, you will get a matrix in Jordan normal form.

In your example (assuming your calculations are correct) take for example $b_0=(1,-1,0)^T$ and $Ab_0$ (for the eigenvalue $0$ and $b_1=(0,-5/2,1)^T$ for the eigenvalue $1$. Then your matrix represented in Jordan normal form will look like $$\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix},$$ the first column corresponding to $Ab_0$, the second to $b_0$ and the third to $b_1$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.