Spectral Decomposition calculating the Eigenvalue

Question

I currently have this problem i can't seem to figure out. A symmetric matrix

       .. 6 2/5 ..
 S =  [..  ..   ..]
       ..  ..   ..

It has the Eigenvalues 9, 18 and c is not equel to 9 and 18 with eigenspace:

               a           
 E9 = Span([   b  ])     
            -2 1/2    ,                     

               2           
 E18 = Span([  1  ])     
               2      ,

               1           
 Ec = Span([  -2  ])     
               0

Now i wonder why is a = 2 and b = 1 How do you proof these values?

and a second question, how do you calculate the Eigenvalue of c with Spectral Decomposition ?

Thank you so much in advantage

Answer 1

The eigenspaces of a symmetric matrix associated with distinct eigenvalues are always orthogonal. Since all the eigenvalues of your matrix are distinct, this means we must have

$$ \left< \begin{pmatrix} a \\ b \\ -2.5 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \right> = 2a + b - 5 = 0, \\ \left< \begin{pmatrix} a \\ b \\ -2.5 \end{pmatrix}, \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} \right> = a - 2b = 0. $$

Hence, $5 = 2a + b - 2(a - 2b) = 5b$ so $b=1$ and then $a = 2$. Since we know the eigenvectors of $S$, we must have

$$ \begin{pmatrix} \frac{4}{\sqrt{45}} & \frac{2}{3} & \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{45}} & \frac{1}{3} & -\frac{2}{\sqrt{5}} \\ -\frac{5}{\sqrt{45}} & \frac{2}{3} & 0 \end{pmatrix} \begin{pmatrix} 9 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & c \end{pmatrix} \begin{pmatrix} \frac{4}{\sqrt{45}} & \frac{2}{\sqrt{45}} & -\frac{5}{\sqrt{45}} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} & 0 \end{pmatrix} = \begin{pmatrix} * & \frac{32}{5} & * \\ * & * & * \\ * & * & * \end{pmatrix}. $$

Computing the $(1,2)$ entry of the product, we get

$$ \begin{pmatrix} \frac{4}{\sqrt{45}} & \frac{2}{3} & \frac{1}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} \frac{18}{\sqrt{45}} \\ \frac{18}{3} \\ -\frac{2c}{\sqrt{5}} \end{pmatrix} = \frac{72}{45} + \frac{36}{9} - \frac{2c}{5} = \frac{32}{5}. $$

Hence, $c = -2$.

Answer 2

Hint:

Let the three eigenvectors, corresponding to eigenvalues 9, 18 and $c$, are written $\mathbf{v}_1$, $\mathbf{v}_2$ and $\mathbf{v}_3$ and let $V$ be the matrix that contains the eigenvectors as its columns.

Let the original matrix with the single known element be $A$ and let $\Lambda$ be a diagonal matrix with the eigenvalues as its diagonal elements, i.e.

$$ \Lambda = \begin{pmatrix} 9 & & \\ & 18 & \\ & & c \end{pmatrix} $$

The relation between $A$ and its eigenvectors/eigenvalues can be written in the matrix equation

$$ A V = V \Lambda $$

You can use this to set up a system of equations to solve for the values you need.