I have matrix $A=\begin{pmatrix}3 &1 \\ -1 &1 \end{pmatrix}$. I have found that the eigenvalue is $2$ and the eigenvector is $\begin{pmatrix}1\\ -1\end{pmatrix}$. How do I find $T$ so that I can calculate $T^{-1}AT=J$? How to form $T$ when I have only one eigenvector?

  • $\begingroup$ Do the same as here. $\endgroup$ – Dietrich Burde May 20 '16 at 9:36
  • $\begingroup$ @DietrichBurde Your link doesn't say anything about generalized eigenvectors. $\endgroup$ – Abstraction May 20 '16 at 9:42
  • $\begingroup$ Then you could look here. $\endgroup$ – Dietrich Burde May 20 '16 at 9:44

You need to find a generalized eigenvector since your eigenvalue's geometric multiplicity is only one. So you need a vector $\;\vec v=\binom xy\;$ such that

$$A\vec v=2\vec v+(-1,\,1)^t\iff \binom{3x+y}{-x+y}=\binom{2x}{2y}+\binom{\!-1}1=\binom{2x-1}{2y+1}\implies$$

$$\implies\begin{cases}x+y=-1\\{}\\-x-y=1\end{cases}\implies\vec v=\binom{-2}1\;,\;\;\text{for example .}$$

Now form the matrix from the eigenvector and the generalized eigenvector

$$P=\begin{pmatrix}\!-1&\!-2\\1&1\end{pmatrix}\implies |P|=1\;\;\;\text{and}\;\;\; P^{-1}=\begin{pmatrix}1&2\\\!-1&\!-1\end{pmatrix}\implies$$

and now you can easily check that


Note: Inadvertently I took $\;\binom{-1}1\;$ as eigenvector. Of course, any non-zero scalar multiple of this is an eigenvector.

  • $\begingroup$ You got $2x-y$ where it should be $2x-1$ (first line). $\endgroup$ – Abstraction May 20 '16 at 10:12
  • $\begingroup$ @Abstraction Thank you very much, you're right. Editing now. Interestingly enough, the given generalized eigenvalue remains the same... :) $\endgroup$ – DonAntonio May 20 '16 at 10:38

A general strategy for generalised eigenspaces:

If we have an eigenvalue $\lambda$, with multiplicity $m$, one proceeds as follows to find a basis for the generalised eigenspace associated to $\lambda$: we have a sequence of subspaces: $$0\varsubsetneq \ker(A-\lambda I)\subset\ker(A-\lambda I)^2\subset\dots\subset\ker(A-\lambda I)^k\subset\dotsm$$ This sequence eventually stabilises, i.ee, there exists integer $r$ such that the inclusions $\ker(A-\lambda I)^{k-1}\subset\ker(A-\lambda I)^k$ are strict up to $r$, then $\ker(A-\lambda I)^r=\ker(A-\lambda I)^{r+1}=\dotsm$.

Start with a maximal system of linearly independent vectors in $\ker(A-\lambda I)^{r}\smallsetminus\ker(A-\lambda I)^{r-1}$, map them by $A-\lambda I$ to $\ker(A-\lambda I)^{r-1}\smallsetminus\ker(A-\lambda I)^{r-2}$, and complete in a maximal system of linearly independent vectors in $\ker(A-\lambda I)^{r-1}\smallsetminus\ker(A-\lambda I)^{r-2}$, and so on. The union of all these sets of vectors is a basis of the generalised eigenspace associated to $\lambda$.


$A-2I=\begin{bmatrix}1&1\\-1&-1\end{bmatrix},\enspace (A-2I)^2=0$. Then $Au_2=(1,-1)$.

Let $u_2$ a non-zero vector which is not an eigenvector, say $(0,1)$. The $(A-2I)u_2=(1,-1)$ is an eigenvector $u_1$. $(u_1,u_2)$ is the required basis.

Indeed $Au_2=2u_2+u_1$, by definition, and $Au_1=2u_1$. Hence the matrix of $A$ in this basis is $$ B=\begin{bmatrix}2&1\\0&2\end{bmatrix}.$$


You know that one column in $\bf T$ should be $(1,-1)^T$ So assume we have for unknowns $a,b$:

$${\bf T} = \left[\begin{array}{rr}1&a\\-1&b\end{array}\right]$$

You will also need to decide how you want $\bf J$, assuming it is the Jordan Normal Form which the $\bf J$ makes me believe, then it will have the form $${\bf J}=\left[\begin{array}{rr}\lambda_1&1\\0&\lambda_1\end{array}\right]$$

Now we can rewrite the equation as $\bf TJ = AT$ to get rid of the matrix inversion.

This should now be a quite easy equation system to find $a$ and $b$ since you already know both $\bf A$ and $\bf J$, right? Difference is that here you will get one equation for each element in the matrix.


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