Discover Linear Algebra

Example 34.4.1. Determining Jordan normal form.

Consider the matrix

from Example 30.2.1. This matrix has two eigenvalues: $\lambda_1 = 3$ of multiplicity $m_1 = 4\text{,}$ and $\lambda_2 = -1$ of multiplicity $m_2 = 3\text{.}$ So we expect the Jordan normal form of $A$ to have “macro” blocks

where the first block $3 I + N_1$ is a $4 \times 4$ block-diagonal matrix of Jordan blocks for $\lambda_1 = 3\text{,}$ and the second block $(-1) I + N_2$ is a $3 \times 3$ block-diagonal matrix of Jordan blocks for $\lambda_2 = -1\text{.}$

Analyze $A - \lambda_1 I$ first. We calculate the ranks of powers of $A - \lambda_1 I\text{,}$ subtracting off the multiplicity of the other eigenvalue $\lambda_2\text{:}$

We can stop here, since only the zero matrix has rank zero. So we know the $4 \times 4$ nilpotent part of the $\lambda_1$ block has already become zero at the second power. (In general, we know it will at worst become zero at exponent equal to the algebraic multiplicity for that eigenvalue.) From the first rank, we know that there are $m_1 - 1 = 3$ Jordan blocks for $\lambda_1 = 3\text{,}$ and from the second rank, we know that the largest is of size $2 \times 2\text{.}$ From the first rank again, we can conclude that there is only one $2 \times 2$ block, so there must be two more $1 \times 1$ blocks, from which we get the partial Jordan form

Now analyze $A - \lambda_2 I\text{.}$ Again, we calculate the ranks of powers of $A - \lambda_2 I\text{,}$ subtracting off the multiplicity of the other eigenvalue $\lambda_1\text{:}$

From the first rank, we know that there is only $m_2 - 2 = 1$ Jordan block for $\lambda_2 = -1\text{.}$ The third rank corroborates this, since it tells us that the largest block has size $3 \times 3$ (which is equal to the overall size $m_2 = 3$ of the blocks for $\lambda_2 = -1$). And the second rank corroborates this as well, since it tells us that there is only one block of that largest size $3 \times 3\text{.}$ With this information, we now have the exact Jordan form for $A\text{:}$

Example 34.4.2. Calculating a transition matrix for Jordan form.

In Example 30.2.1, we determined a matrix $Q$ that puts the following $7 \times 7$ matrix $A$ in triangular-block form:

(Note that we called the transition matrix $P$ in that example).

Now, let's put the nilpotent part of each block of $\inv{Q} A Q$ into nilpotent triangular form. Since the blocks are small, we will determine cyclic space bases by inspection, using our knowledge of the theory from Section 33.6, instead of using either of the procedures from Subsection 33.4.3. Break each of the two blocks of $\inv{Q} A Q$ into its scalar and nilpotent parts:

Start with $N_1\text{.}$ Let $N_1'$ represent the nilpotent triangular form for $N_1\text{.}$ Since $N_1^2 = \zerovec\text{,}$ the largest block in $N_1'$ has size $2 \times 2\text{.}$ We have $\rank N_1 = 1\text{,}$ so there is only one $2\times 2$ block, and from $\nullity N_1 = 3$ (nullity is size minus rank), we see that $N_1'$ will have three blocks. Therefore, we need one two-dimensional $N_1$-cyclic subspace of $\R^4\text{,}$ and two one-dimensional subspaces. Since the fourth column of $N_1$ is nonzero, we can use $\uvec{e}_4$ to generate the two-dimensional space. We can use any vectors in the null space of $N_1$ that are linearly independent with that fourth column to generate the one-dimensional spaces. For example, one can easily see by inspection that $\uvec{e}_1,\uvec{e}_2$ both lie in the null space of $N_1\text{.}$ Thus, set

Using these cyclic bases, we get

For $N_2\text{,}$ we have

Since $N_2$ is $3\times 3$ and the first power of $N_2$ that is zero is the $\ird[3]$ power, $N_2$ should be similar to a matrix in elementary nilpotent form. The third column of $N_2^2$ is nonzero, so we can set

and use this cyclic basis to get

We can also apply $R_1,R_2$ to $A_1,A_2\text{:}$

Recall that the matrix $Q$ puts $A$ into triangular-block form. We see above that the matrices $R_1,R_2$ break each block of $\inv{Q} A Q$ into Jordan blocks. We can combine these by forming the block-diagonal matrix

and setting $P = Q R\text{.}$ Then,

the Jordan normal form of $A$ as desired.