# Finding eigenvalues for matrix when eigenvectors are known.

## Question:

We have matrix $A$ defined as:

$$A=\begin{bmatrix} -29 & 39 & -69 \\ -41 & 41 & -81 \\ -11 & 1 & -11 \end{bmatrix}$$

Eigenvectors for matrix $A$ are $$v_1= \begin{bmatrix}1 \\ 2 \\ 1 \end{bmatrix} v_2=\begin{bmatrix}-1 \\ 1 \\ 1 \end{bmatrix} v_3=\begin{bmatrix}-3 \\ -2 \\ 1 \end{bmatrix}$$ If you would actually calculate eigenvalues and then eigenvectors for matrix $A$ you wouldn't get $v_1$, $v_2$ and $v_3$ as shown here. I guess the idea is to calculate eigenvalues if we were to make assumption that eigenvectors were these.

## Attempt to solve:

First i've calculated characteristic polynomial for matrix A without solving eigenvalues from it. Characteristic polynomial is defined as:

$$P_a(\lambda)=\det(A-\lambda I)$$ $$P_a(\lambda)=\det\left(\begin{bmatrix}-29 & 39 & -69 \\ -41 & 41 & -81 \\ -11 & 1 & -11 \end{bmatrix}\times \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\lambda \right)$$ $$P_a(\lambda)=\det \left(\begin{bmatrix}-29-\lambda & 39 & -69 \\ -41 & 41-\lambda & -81 \\ -11 & 1 & -11-\lambda \end{bmatrix}\right)$$ $$P_a(\lambda)= -\lambda^3+\lambda^2+400\lambda -400$$

Eigenvectors could be solved from equation. Calculating one vector per one $\lambda$

$$(A-\lambda I)v=0$$ $$Av-\lambda I v =0$$ $$\lambda I = Av$$ Now writing the equation in matrix form:

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix} \lambda \\ \lambda \\ \lambda \end{bmatrix}=\begin{bmatrix} -29 & 39 & -69 \\ -41 & 41 & -81 \\ -11 & 1 & -11\end{bmatrix} \times v_1,v_2 \ldots v_3$$

But now how do you solve one $\lambda$ for one eigenvector $v_n$ where $n \in [1,2,3]$ and at this point I am starting to doubt if this even correct way to begin with ?

If someone could give hint towards correct solution to this problem that would be greatly appreciated.

Thanks,

Tuki

• Write $Av_i$ as a multiple of $v_i$. – John Griffin Oct 15 '17 at 17:37

Given the eigenvector $v$. To find the corresponding eigenvalue $\lambda$, recall that it should satisfy

$$Av=\lambda v$$

Since $v$ is known, we can compute $Av$. Let's call it $w$.

In the event that $w=0$, since $v \neq 0$, $\lambda$ must be $0$.

If $w \neq 0$. Then we have $w=\lambda v$. $w$ has a non-zero component, let say it is at index $i$.

Let me denote $v_{[i]}$ as the $i$-th component of $v$. For example, for vector $v= \begin{bmatrix} 4 \\ 5 \\ 6\end{bmatrix}$, $v_{[1]}=4, v_{[2]}=5, v_{[3]}=6$

$w_{[i]} = \lambda v_{[i]}$

Hence $\lambda = \frac{w_{[i]}}{v_{[i]}}$.

• $$Av_1=\begin{bmatrix}-20 \\ -40 \\ -20 \end{bmatrix}$$. We could call this $w_1$ and then we could calculate value for $\lambda_1$ $$\lambda_1 = w_1*v_1^{-1}$$ $v_1^{-1}$ since i have no clue how you would divide matrix we can multiply instead ? $$\lambda_1=\begin{bmatrix} -20 \\ -40 \\ -20 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}^{-1}$$ This would be dot product of two vectors ? But again this doesn't make sense since we cannot calculate inverse $v_1^{-1}$ ? – Tuki Oct 15 '17 at 18:04
• I edited the answer and introduce notation $v_{[i]}$ to make things clearer. In this case $w_{1[1]}=-20 \neq 0$. $v_{1[1]}=1$, can you compute $\lambda_1$? – Siong Thye Goh Oct 15 '17 at 18:10
• I think it would be $$\lambda_1 = \frac{-20}{1}=-20$$ but how did you come up with these values ? $w_{1[1]}=-20$ and $v_{1[1]}=1$ ? probably not – Tuki Oct 15 '17 at 18:15
• Your answer is correct. I look at the vector $w_1$, and see what is the first non-zero index, the first entry $-20$ is non-zero. Hence $w_{1[1]} = -20$, then i use the first entry for $v_1$ as well. – Siong Thye Goh Oct 15 '17 at 18:22
• To my understanding $w_1$ is vector and $v_1$ is also vector. then we can of course index all entries but the thing that doesn't make sense to me that we have 3 eigenvectors and we are using only one ? $\lambda_2$ wouldn't be with index 2 from $w_{1[2]}$ and $v_{1[2]}$ right ? $$\lambda_2 = \frac{-40}{2}=-20$$ ? – Tuki Oct 15 '17 at 18:30