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I'm trying to find the Jordan Normal Form of the following matrix: $\pmatrix{2 & 0 & 1 & 1 \\0 & 2 & 1 & 1\\0 & 0 & 2 & 1\\0 & 0 & 0 & 2\\}$.
Now since its upper triangular I know that the only eigenvalue is 2 with algebraic multiplicity equal to $4$. Then by calculating $\text{rank}(A-2I)=2$, I can find the geometric multiplicity which is $\text{gm}=4-2=2$. Hence I will have two Jordan Block with dimensions summing to $4$.
Which of the following should I choose: $J_3(2) \oplus J_1(2)$ or $J_2(2) \oplus J_2(2)$?

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One way is to find the minimal polynomial of the matrix. The multiplicity of $2$ as a root of that polynomial is the size of the largest block of that eigenvalue in the Jordan Normal Form. That will give you the size of the second block as well.

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  • $\begingroup$ Just to confirm. I know that the minimal polynomial must divide the characteristic polynomial. Also the minimal polynomial equals $m(x)=(x-\lambda_1)^{a_1} \cdots (x-\lambda_r)^{a_r}$, where $a_i$ is the size of the largest $\lambda_i$-block in the JCF. So my minimal polynomial is here $m(x)=(x-2)^{2}$ (as if I picked for $a_i=3$ that would not divide the characteristic polynomial). Hence, my JCF is $J_2(2) \oplus J_2(2)$. Correct? $\endgroup$ – Nick Aug 18 '18 at 14:23
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    $\begingroup$ No, the minimal polynomial here is actually $m(x)=(x-2)^3$. It does divide the characteristic polynomial, as the characteristic polynomial is $p(x)=(x-2)^4=m(x)(x-2)$. So the Jordan form is $J_3(2) \oplus J_1(2)$. $\endgroup$ – Mark Aug 18 '18 at 14:34
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You can look at the behaviour of $A-2I $. Namely, just looking at the first row, $$ (A-2I)^2=\begin {bmatrix}0 & 0 & 1 & 1 \\0 & 0 & 1 & 1\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end {bmatrix}^2=\begin {bmatrix}0 & 0 & 0 &1 \\ \ \\ \ \\ \ \end {bmatrix}. $$ The fact that this is nonzero tells you that you have a block of size at least 3, since $J_2(0)^2=0$.

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You have already shown that the dimension of $2-Eigenspace$ is $2$. And you also know that the only eigen value is $2$.So if you have atleast $2$ many blocks of order $1$,then that dimension would be atleast $3$.S o you are left with only $2$ more possibilities.

$1$.you have a single block of size $4$, which would imply that dimension of $2-eigenspace$ is $1$.

$2$.you have a single block of order $1$ and a block of order $3$, which is one of your correct guesses i.e. $J_3(2)\oplus J_1(2)$.

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