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Put the matrix $$\begin{bmatrix} 3 & -4\\ 1 & -1\end{bmatrix}$$ in Jordan Canonical Form. Moreover, find the appropriate transition matrix to the basis in which the original matrix assumes its Jordan form.

I'm having a lot of trouble with this. I know that the eigenvalue has multiplicity two and is $\lambda = 1$. I can find the first eigenvector, which is:

\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix} I'm having trouble finding the second since both eigenvalues tell us the same thing. But I'm not nearly as concerned about the eigenvectors as I am about what to do after.

If anyone could explain thoroughly the next steps involved (not necessarily the answer but how to obtain it), I would be forever grateful. This homework is in 2 days and it may determine my grade letter.

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  • $\begingroup$ Yes, I know the only eigenvalues are one. The problem is finding the eigenvectors corresponding to 1 and more importantly how to proceed. $\endgroup$ – Arbitrationer Nov 23 '14 at 5:43
  • $\begingroup$ You don't seem to understand that when vadim wrote, "the eigenspace corresponding to $\lambda=1$ is one-dimensional", he was telling you there is only the one eigenvector that you have found (and scalar multiples of that eigenvector, but they don't help). What you need is a generalized eigenvector; a vector $w$ such that $(A-\lambda I)w=v$, where $v$ is the eigenvector you have found. Do you not have a text, or some notes to follow? Anyway, "generalized eigenvector" is the keyphrase: go look it up, and then post an answer when you have worked out how to solve the problem. $\endgroup$ – Gerry Myerson Nov 23 '14 at 6:16
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To put a matrix in Jordan normal form requires to know three matrices such that $A=PJP^{-1}$ i.e: a matrix $P^{-1}$ that transform to the canonical basis $(\mathbf{i},\mathbf{j},\mathbf{k})$ to a new basis in which the matrix $J$ represents the transformation of a vector $\mathbf{v}$ such that the transformed vector $\mathbf{v'}$ is the same as we find when we transform $\mathbf{v}$ with $A$ in the canonical basis. Last the matrix $P$ returns this result to the canonical basis.

As noted in OP the matrix $A$ has eigenvalues $\lambda_1=\lambda_2=1$ and a single eigenvector $$ \mathbf{u_1}=\left[ \begin{array}{cccc} 2\\ 1 \end {array} \right] $$

So the main problem is to find another vector that completes the new basis.

To find such a vector notes that all vectors $\mathbf{x}$ such that $(A-\lambda I)\mathbf{x}=0$ are transformed in the eigenspace generated by the eigenvector $\mathbf{u_1}$, so we want a vector $\mathbf{u_2}$ such that $(A-\lambda I)\mathbf{u_2} \ne 0$, and the way to do this is to find a vector such that : $(A-\lambda I)\mathbf{u_2} = \mathbf{u_1}$. (Note that this equation is the same as $(A-\lambda I)^2\mathbf{u_2}= 0$).

Solving in our case we find: $$ \left[ \begin{array}{cccc} 2&-4\\ 1&-2 \end {array} \right] \left[ \begin{array}{cccc} x\\ y \end {array} \right]= \left[ \begin{array}{cccc} 2\\ 1 \end {array} \right] $$

so that the components $x$ and $y$ of the searched vector must satisfies $x-2y=1$, and we can find the vector $$ \mathbf{u_2}= \left[ \begin{array}{cccc} 1\\ 0 \end {array} \right] $$ So the matrix $P$ we are searching is $$ P=[\mathbf{u_1},\mathbf{u_2}]= \left[ \begin{array}{cccc} 2&1\\ 1&0 \end {array} \right] $$ And the inverse is: $$ P^{-1}= \left[ \begin{array}{cccc} 0&1\\ 1&-2 \end {array} \right] $$ The matrix $J$ is a typical Jordan block, with the eigenvalues as diagonal elements and an entry $1$ up-right them: $$ J= \left[ \begin{array}{cccc} 1&1\\ 0&1 \end {array} \right] $$ and we can easily verify that $A=PJP^{-1}$.

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The characteristic polynomial $\det (A - \lambda I) = (\lambda -1)^2.$ when the dimension of the null space(1) of an eigenvalue($\lambda = 1$) is less than the algebraic multiplicity(2), you need to find generalized eigenvectors. in this instance, you need to solve $(A - I_2) \left( \begin{array}{l} x \cr y\end{array} \right) = \left( \begin{array}{l} 2 \cr 1 \end{array} \right).$ this gives you $x = 1, y = 0$

with respect to the basis $\{ \left( \begin{array}{l} 2 \cr 1 \end{array} \right), \left( \begin{array}{l} 1 \cr 0 \end{array} \right) \}$ you transformation is represented by the Jordan canonical form $\left( \begin{array}{ll} 1 & 1 \cr 0 & 1 \end{array} \right).$

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  • $\begingroup$ Might have been better to let OP do some of the work. $\endgroup$ – Gerry Myerson Nov 23 '14 at 23:27
  • $\begingroup$ @myerson, perhaps i should have not done all the work. $\endgroup$ – abel Nov 23 '14 at 23:57

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