# Doubt on finding generalized eigenvectors of a matrix

I have a system with $$A$$ matrix given as $$A=\begin{pmatrix} 4 & 1 & -2 \\ 1 & 0 & 2 \\ 1 & -1 & 3 \end{pmatrix}$$ and i've asked to find the Jordan canonical representation of the given matrix. The first step is to find the eigen-vectors, in this case the eigen- values are $$\lambda_1=1,\lambda_2=3,\lambda_3=3$$ which necessitates to find a generalized eigen-vector for the eigen-value $$3$$ . I'm able to find it by following general definition i.e $$(A-\lambda I)^m V=0$$ and $$(A-\lambda I)^{m-1} V \neq 0$$ But in the book they have followed some other method. First they found $$\lambda_1 I-A$$ i.e $$\lambda_1 I-A=\begin{pmatrix} -3 & -1 & 2 \\ -1 & 1 & -2 \\ -1 & 1 & -2 \end{pmatrix}$$ Then to find the eigen-vectors they calculated the row co-factors of $$\lambda_1I-A$$ , the co-factors along the first row gives null solution so they found co-factors of the second row i.e $$v_1=\begin{pmatrix} 0 & 8 & 4 \end{pmatrix}^{T}$$

My first doubt is: Why rowcofactors gives the eigen-vectors ? What is the logic behind it ?

Now to find the generalized eigen-vector first they found $$\lambda_2 I-A=\begin{pmatrix} \lambda_2-4 & -1 & 2 \\ -1 & \lambda_2 & -2 \\ -1 & 1 & \lambda_2-3 \end{pmatrix}$$ Again the row-cofactor, here the first row works out , so \begin{align}v_2&=\begin{pmatrix} \lambda_2(\lambda_2-3)+2& \lambda_2-3+2 & -1+\lambda_2 \end{pmatrix}^{T}\bigg{|}_{\lambda_2=3} \\&=\begin{pmatrix} 2 & 2 & 2 \end{pmatrix}^{T}\end{align}

And to complete the basis another eigenvector is found by taking derivative of $$v_2$$ i.e \begin{align}v_3 &=\begin{pmatrix} \frac{d}{d\lambda_2} \left\{ \lambda_2(\lambda_2-3)+2\right\} \\ \frac{d}{d\lambda_2}\left\{ \lambda_2-3+2 \right\} \\ \frac{d}{d\lambda_2}\left\{-1+ \lambda_2 \right\} \end{pmatrix}\Bigg{|}_{\lambda_2=3}&=\begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}\end{align}

My second doubt is: Why they are taking derivatives ? how taking derivatives are related with repeated roots

• What book is this from? – Moo Feb 12 at 5:48
• It's from a book called 'control system engineering' by I.J.Nagrath and M.Gopal – Siddhartha Ganguly Feb 12 at 5:50
• Hint: Compute the determinant of $\lambda I-A$ via cofactor expansion along a row. – amd Feb 12 at 6:33
• – J. W. Tanner Feb 12 at 6:37
• @amd if i find det $(\lambda_1 I-A)$ via co-factor expansion along all three rows i'm getting zero, does it tell me anything about eigen vectors ? sorry i can't understand. I can say that det $(\lambda_1 I-A)$ has rank=2 and nullity=1 , so the column space consists of linearly dependent vector..but how eigenvectors are related with this ? – Siddhartha Ganguly Feb 12 at 8:43

I’d not seen these techniques before, but now that I understand how they work, I’ll be adding them to my repertoire.

The first computation is pretty easy to understand. I’ll work here with the adjugate $$\operatorname{adj}(M)$$ of the square matrix $$M$$ (often called the “adjoint” in older sources). This is simply the transpose of the matrix of cofactors. A fundamental property of the adjugate is that $$M\operatorname{adj}(M) = \operatorname{adj}(M) M = \det(M)I.\tag1$$ Now, by construction $$\det(\lambda I-A)=0$$ when $$\lambda$$ is an eigenvalue of $$A$$, so in that case equation (1) says that every column of $$\operatorname{adj}(\lambda I-A)$$ is an element of the null space of $$\lambda I-A$$, but this means that every nonzero column of this adjugate matrix is an eigenvector of $$A$$ with eigenvalue $$\lambda$$.

The computation that involves differentiation to find generalized eigenvectors is a bit trickier. It looks like it’s covered in a 1967 paper by Dzoković. I don’t have access to the entire paper, but the one page of the preview on the Springer site gives us most of what we need. Following Dzoković, let $$f(\lambda)=\det(\lambda I-A)=(\lambda-\lambda_1)^{r_1}(\lambda-\lambda_2)^{r_2}\cdots$$ be the characteristic polynomial of $$A$$ and $$B(\lambda)=\operatorname{adj}(\lambda I-A)$$, so that by (1), $$(\lambda I-A)B(\lambda) = f(\lambda)I, \tag2$$ which again says that every nonzero column of $$B(\lambda_i)$$ is an eigenvector of $$A$$. For a zero column of $$B(\lambda_1)$$, say the first, Dzoković then goes on to show that if $$k_1\lt r_1$$ is the highest power of $$(\lambda-\lambda_1)$$ that divides $$B_1(\lambda_1)$$, then differentiating it $$k_1$$ times and setting $$\lambda=\lambda_1$$ produces $$(\lambda_1I-A)B_1^{(k_1)}(\lambda_1)=0$$, i.e., differentiating $$B_1(\lambda)$$ $$k_1$$ times generates an eigenvector of $$A$$. He then continues to differentiate: $$(\lambda_1I-A)B_1^{(k_1)}(\lambda_1)=0 \\ (\lambda_1 I-A)B_1^{(k_1+1)}(\lambda_1)+(k_1+1)B_1^{(k_1)}(\lambda_1)=0 \\ \vdots \\ (\lambda_1 I-A)B_1^{(r_1-1)}(\lambda_1)+(r_1-1)B_1^{(r_1-2)}(\lambda_1) = 0.$$ Setting $$x_i = \frac1{(k_1+i-1)!}B_1^{(k+i-1)}(\lambda_1),$$ this becomes $$(A-\lambda_1 I)x_1=0 \\ (A-\lambda_1 I)x_2=x_1 \\ \vdots \\ (A-\lambda_1 I)x_{r_1-k_1+1} = x_{r_1-k_1},$$ but this is none other than the definition of a chain of generalized eigenvectors for $$\lambda_1$$.

• Thanks , now it makes sense. – Siddhartha Ganguly Feb 14 at 8:16