Finding out Eigenvalues and Eigenvectors

Question

Question: $A=\begin{pmatrix}1&1&2\\0&1&3\\0&0&2\end{pmatrix}$

I am trying to find out the Eigenvalues and Eigenvector for this question. During my working out (I use the cover-up method- First row/First column, Second row/Second column etc.), I've gone until $(λ-1)[(λ-1)(λ-2)]+(1)[(0)(λ-2)]-(2)[(0)(λ-1)]$. Is the working out right so far? Also not sure what $1[(0)\cdot(λ-2)]$ equals to?

Mainly the eigenvalues and I'm having issues with.

Answer 1

From here

$$|A-\lambda I|=\begin{vmatrix} 1-\lambda & 1 & 2\\ 0 & 1-\lambda & 3\\ 0 & 0 & 2-\lambda\end{vmatrix}=(1-\lambda)^2(2-\lambda)=0$$

we can see that for a up triangular matrix eigenvalues are the diagonal entries.

Now for each $\lambda$ solve $(A-\lambda I)x=0$ to find the corresponding eigenvectors.

Answer 2

$$A~=~\begin{pmatrix}1&1&2\\0&1&3\\0&0&2\end{pmatrix}$$

By plugging in the eigenvalues and compute the determinate we get

$$A~=~\begin{vmatrix}1-\lambda&1&2\\0&1-\lambda&3\\0&0&2-\lambda\end{vmatrix}=(1-\lambda)\cdot\begin{vmatrix}1-\lambda&3\\0&2-\lambda\end{vmatrix}+1\cdot\begin{vmatrix}3&0\\2-\lambda&0\end{vmatrix}+2\cdot\begin{vmatrix}0&1-\lambda\\0&0\end{vmatrix}=(1-\lambda)^2(2-\lambda)+1\cdot(0)+2\cdot(0)$$

Where two of the three summands vanish because a product with $0$ is always $0$ aswell and so only $(1-\lambda)^2(2-\lambda)$ remains which you have to set equal to zero. This term is already in the form where you can just see the roots, which are $\lambda_{1,2}=1,\lambda_3=2$.

Answer 3

Hint: All Eigenvalues of a upper-tringular matrix are the items on it's diagonal so they are $1,2$.

Answer 4

If $A$ is a triangular matrix then the determinant of $A$ is $det(A) = \prod_{i=1}^{n} a_{ii}$

It follows that

$$ det(A-\lambda I) = \prod_{i=1}^{n} a_{ii} - \lambda$$

then if

$$ A = \begin{bmatrix} 1& 1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 2\end{bmatrix} $$

$$det(A- \lambda I) = \prod_{i=1}^{n} a_{ii} - \lambda = (1-\lambda)(1-\lambda)(2-\lambda) $$

Then we know

$$\lambda_{1},\lambda_{2} = 1, \lambda_{3} = 2 $$

To determine the eigenvectors we solve the linear equation $(A-\lambda_{j}I)x =\mathbb{0} $ for $ j=1,2,3$ using $\lambda_{3} =2$ we have

$$(A-I)x = \begin{bmatrix} -x_{1}+ x_{2} + 2x_{3} \\ -x_{2} + 3x_{3} \\ 0 x_{3} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} $$

$$ -x_{2} + x_{3} = 0 \implies x_{2} = 3x_{3}$$ $$ -x_{1} + x_{2} +2x_{3} \implies -x_{1} + 3x_{3} +2x_{2} =0 \implies x_{1} = 5x_{3} $$

$$ u_{3} = \begin{bmatrix} 1 \\ \frac{3}{5} \\ \frac{1}{5}\end{bmatrix}$$ $\lambda_{1} = 1 $

$$ \| u_{3}\| = \sqrt{1+ (\frac{3}{5})^{2} + (\frac{1}{5})^{2}} $$ $$ \| u_{3}\| = \sqrt{1+ \frac{9}{25} + \frac{1}{25}} $$ $$ \| u_{3}\| = \sqrt{1+ \frac{10}{25}} $$ $$ \| u_{3}\| = \sqrt{ \frac{35}{25}} $$ $$ \| u_{3}\| = \frac{\sqrt{35}}{5} $$ $$ v_{3} = \frac{u_{3}}{\| u_{3}\|} = [\frac{1}{\frac{\sqrt{35}}{5}} , \frac{\frac{3}{5}}{\frac{\sqrt{35}}{5}}, \frac{\frac{1}{5}}{\frac{\sqrt{35}}{5}} ]$$

now $\lambda_{1} = 1 $ $$(A-I)x = \begin{bmatrix} x_{2} + 2x_{3} \\ x_{2} + 3x_{3} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} $$ $$ x_{3} = 0 \implies x_{2} + 3x_{3} = x_{2} + 0 = 0 \implies x_{2} = 0 $$ our assumption was $x_{1} =1$ $$u_{1} = \begin{bmatrix}1 \\ 0 \\0 \end{bmatrix} $$ $$u_{2} = \begin{bmatrix}-1 \\ 0 \\0 \end{bmatrix} $$ testing this in computer

A = np.array([[1,1,2],[0,1,3],[0,0,2]])
vals, vecs = LA.eig(A)
test = np.sqrt(35)/5
test1 = [1/test, (3/5)/test, (1/5)/test]

Out[2]: 
array([[ 1.00000000e+00, -1.00000000e+00,  8.45154255e-01],
       [ 0.00000000e+00,  2.22044605e-16,  5.07092553e-01],
       [ 0.00000000e+00,  0.00000000e+00,  1.69030851e-01]])

test1
Out[3]: [0.8451542547285166, 0.50709255283711, 0.16903085094570333]

Answer 5

There is actually just one genuine eigenvector for the eigenvalue $1.$ There are "generalized" eigenvectors, which lead to the Jordan Normal Form. For the middle column of my matrix $R,$ I chose from among the vectors $v$ for which $(A-I)^2 v = 0$ but $(A-I) v \neq 0.$ Then the first column, a genuine eigenvector, is $u =(A-I) v $

$$ R = \left( \begin{array}{ccc} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right) $$

The Jordan form is $ R^{-1} A R = J$ and has a special shape.

$$ \left( \begin{array}{ccc} 1 & 0 & -5 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right)= \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array} \right) $$

For applications such as finding $e^{At}$ it is the reverse identity that is important, $RJR^{-1} = A,$

$$ \left( \begin{array}{ccc} 1 & 0 & 5 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 & -5 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{array} \right) $$