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Let B be a 10 x 10 matrix and let $\lambda$ be a scalar. Suppose it is known that

null$(B-\lambda I)=5$,

null$(B-\lambda I)^2=8$

null$(B-\lambda I)^3=9$

Based on this information I am required to find all possible jordan forms of B. I'm not quite sure how to approach this, as I don't know how many eigenvalues there are. I think $\lambda$ has an algebraic multiplicity of 4, but I'm not sure if that will help.

Exodia
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  • $\lambda$ can't have an algebraic multiplicity of 4 because it has a geometric multiplicity of 5, and the algebraic multiplicity is always greater than or equal to the geometric multiplicity. – symplectomorphic Jul 07 '16 at 03:13
  • Is the second equation supposed to be $\mathrm{null}(B-\lambda I)^2$? – stewbasic Jul 07 '16 at 03:14
  • Do you mean those numbers as the dimensions of the null spaces? How could $null(B-\lambda I)^3$ be 8 and 9? – Christian Jul 07 '16 at 03:14
  • Yeah sorry its meant to be ^2 my bad – Exodia Jul 07 '16 at 03:24
  • @symplectomorphic I think $\lambda$ can have an algebraic multiplicity of 4, it just means that B isn't diagonlisable – Exodia Jul 07 '16 at 03:26
  • @Exodia: no. An operator is diagonalizable when the algebraic multiplicity equals the geometric multiplicity for each eigenvalue, but for any operator, diagonalizable or not, the algebraic multiplicity is at least the geometry multiplicity, for each eigenvalue. – symplectomorphic Jul 07 '16 at 04:31

1 Answers1

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Since $dim null(B-\lambda I) = 5$, there must be exactly 5 eigenvectors with eigenvalue $\lambda$. This does not, of course, take into account generalized eigenvectors, which we will do next. This means we will be working with 5 Jordan chains corresponding to $\lambda$

Since the null space increases in dimension by 3 when looking at $dim null(B-\lambda I)^2$, there are 3 Jordan blocks of size 2 or greater corresponding to eigenvalue $\lambda$, and we also can conclude that there are exactly 2 Jordan blocks of size 1 corresponding to \lambda, as those Jordan chains must have been terminated at length 1, or else the dimension of $null(B- \lambda I)^2$ would be greater. We have 3 Jordan chains remaining for this eigenvalue.

Similarly, since $dimnull(B-\lambda I)^3 = 9$, and the previous dimension was 8, there must be one Jordan block of size 3 or greater, meaning that 2 of the Jordan blocks corresponding to $\lambda$ must be of size exactly 2, as the increase in dimension of 1 means that 2 chains must exactly of length 2. If one of the chains I claim have been terminated were longer, then the dimension of this null space would be greater.

So what we know so far, for blocks corresponding to $\lambda$

2 Blocks of size exactly 1

2 Blocks of size exaclty 2

1 Block of size 3 or greater

If the block of size 3 or greater is a block of size 4, which is possible, as we know nothing about $dim null (B- \lambda I)^4$, then this completes 1 possible Jordan form, since we now have 10 entries on the diagonal. (2*1) + (2*2) + (1*4) = 10.

However, this block may indeed be of size 3, and if that is the case, there must be a second eigenvalue of the transformation, as all 5 of the Jordan chains have been terminated, which will have Jordan block of size 1, as it is the only possible size left. (9 of the diagonal slots are already filled)

So there are two possible Jordan forms, not including forms found by rearranging the Jordan blocks:

Form 1:

2 Blocks of size 1 for $\lambda$

2 blocks of size 2 for $\lambda$

1 Block of size 4 for $\lambda$

$\begin{bmatrix} \lambda & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \lambda & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \lambda & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \lambda & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda \end{bmatrix}$

Form 2:

2 Blocks of size 1 for $\lambda$

2 blocks of size 2 for $\lambda$

1 Block of size 3 for $\lambda$

1 Block of size 1 for some eigenvalue $\lambda_2$

$\begin{bmatrix} \lambda & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & \lambda & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \lambda & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & \lambda & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \lambda & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \lambda_2 \end{bmatrix}$

Christian
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