# Thread: eigen value geometric multiplicity problem

1. ## eigen value geometric multiplicity problem

find the jordan form of

A=\left(\begin{array}{cccc}
4 & 1 & 1 & 1\\
-1 & 2 & -1 & -1\\
6 & 1 & -1 & 1\\
-6 & -1 & 4 & 2\end{array}\right)

?

my latex around the matrices is not working and it meesed up the whole thing
so i removed it

|tI-A|=\left|\begin{array}{cccc}
t-4 & -1 & -1 & -1\\
1 & t-2 & 1 & 1\\
-6 & -1 & t+1 & -1\\
6 & 1 & -4 & t-2\end{array}\right|[/TEX]

=(t+3)\left|\begin{array}{cccc}
1 & 1 & 1 & 1\\
1 & t-2 & 1 & 1\\
-6 & -1 & t+1 & -1\\
6 & 1 & -4 & t-2\end{array}\right|=(t-3)^{2}(t-2)

i did R_{1}+R_{2}+R_{3}+R_{4}->R_{1} and took out (t+3) and got the polinomial above
so i need to have one block jordan of eigen value 2
so when i calculate
dim ker(A-2I)

A-2I=\left(\begin{array}{cccc}
2 & 1 & 1 & 1\\
-1 & 0 & -1 & -1\\
6 & 1 & -3 & 1\\
-6 & -1 & 4 & 0\end{array}\right)

2R_{2}+R_{1}->R_{1}

=\left(\begin{array}{cccc}
0 & 1 & -1 & -1\\
-1 & 0 & -1 & -1\\
6 & 1 & -3 & 1\\
-6 & -1 & 4 & 0\end{array}\right)

6R_{2}+R_{3}->R_{3}

=\left(\begin{array}{cccc}
0 & 1 & -1 & -1\\
-1 & 0 & -1 & -1\\
0 & 1 & -9 & -5\\
-6 & -1 & 4 & 0\end{array}\right)

-6R_{2}+R_{4}>R_{4}

=\left(\begin{array}{cccc}
0 & 1 & -1 & -1\\
-1 & 0 & -1 & -1\\
0 & 1 & -9 & -5\\
0 & -1 & 10 & 6\end{array}\right)

R_{3}+R_{4}>R_{4}

=\left(\begin{array}{cccc}
0 & 1 & -1 & -1\\
-1 & 0 & -1 & -1\\
0 & 1 & -9 & -5\\
0 & 0 & 1 & 1\end{array}\right)
R_{4}+R_{2}>R_{2}
R_{4}+R_{1}>R_{1}

=\left(\begin{array}{cccc}
0 & 1 & 0 & 0\\
-1 & 0 & 0 & 0\\
0 & 1 & -9 & -5\\
0 & 0 & 1 & 1\end{array}\right)

-R_{1}+R_{3}>R_{3}

=\left(\begin{array}{cccc}
0 & 1 & 0 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & -9 & -5\\
0 & 0 & 1 & 1\end{array}\right)

9R_{4}+R_{3}>R_{3}

\left(\begin{array}{cccc}
0 & 1 & 0 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 4\\
0 & 0 & 1 & 1\end{array}\right)=0

so from here if my vector is (x,y,z,t) i get
4t=0 ->t=0
z+t=0 -> z=0
-x=0 -> x=0
y=0
so i get
that the geometric multiplicity of t=2 is zero
and it should be 1
where is my mistake
?

2. ## Re: eigen value geometric multiplicity problem

The eigenvalues of $\displaystyle A$ are $\displaystyle \lambda_1=3$ (triple) and $\displaystyle \lambda_2=-2$ (simple), so $\displaystyle \dim \ker (A+2I)=1$ and $\displaystyle \dim \ker (A-3I)=4-\textrm{rank}(A-3I)=\ldots=4-2=2$ . Hence the Jordan form of $\displaystyle A$ is:

$\displaystyle J=\begin{bmatrix}{3}&{1}&{0}&\;\; 0\\0&{3}&{0} & \;\;0\\{0}&{0}&{3}&\;\;0\\0&0&0&-2\end{bmatrix}$

3. ## Re: eigen value geometric multiplicity problem

Originally Posted by transgalactic
my latex around the matrices is not working and it meesed up the whole thing so i removed it
Look at the horrible code you have written. Only with a miracle as the multiplication of the loaves and fishes you could get a correct expression.