Discover Linear Algebra

Example 33.5.1.

Suppose \(A\) is an \(19 \times 19\) matrix that is similar to the triangular-block nilpotent matrix

where the elementary nilpotent blocks \(N_1\) and \(N_2\) are each \(5 \times 5\text{,}\) blocks \(N_3\) and \(N_4\) are each \(3 \times 3\text{,}\) block \(N_5\) is \(2 \times 2\text{,}\) and block \(N_6\) is \(1 \times 1\text{.}\)

We will carry out our analysis as if we don't actually know the form matrix \(N\text{,}\) but at the same time we will use our “secret” knowledge of \(N\) to “know” the required properties of \(A\text{.}\)

We should first compute the powers of \(A\) to determine the degree of nilpotency. We should find \(A^5 = \zerovec\) while \(A^4 \neq \zerovec\text{.}\) If we didn't already know the form \(N\text{,}\) this computation would tell us that the largest block in \(N\) is size \(5 \times 5\text{,}\) because any larger block would not yet be zero at the \(\nth[5]\) power, and smaller blocks become zero before the \(\nth[5]\) power. After this we should compute \(\rank A^4\text{,}\) which in this case we would find to be \(2\text{.}\) When we back off the exponent by one like this, blocks that are smaller than \(4 \times 4\) will still all be taken to zero blocks, but a single \(1\) in the bottom left corner of each of the \(5 \times 5\) blocks will have reappeared. So \(\rank A^4 = 2\) tells us that we have two such “corner ones” in \(N^4\text{,}\) one in each of two blocks of size \(5 \times 5\text{.}\)

We continue with rank computations of lower powers. Backing off the exponent by one again, in this case we should find \(\rank A^3 = 4\text{.}\) Our expectation is that in \(N^3\text{,}\) each of the two (now known) \(5 \times 5\) blocks would have grown in rank from \(1\) to \(2\) as they each “re-grow” a subdiagonal of two \(1\)s in the lower left, for a total rank contribution of \(4\) from these two blocks. Since that is our total rank for \(A^3\text{,}\) this tells us that there are no blocks of size \(4 \times 4\text{,}\) since if there were they would have “reappeared” at the \(\ird\) power with a “corner one” in the lower left of each, causing a larger jump in rank than expected.

However, calculating \(\rank A^2\) should yield \(8\) in this case, a larger rank than expected. The two \(5 \times 5\) blocks in \(N\) each gain one in rank, as their subdiagonals of \(1\)s march higher up from the lower left toward the main diagonal. So their total contribution to the rank increases from \(4\) to \(6\) as we move from \(A^3\) to \(A^2\text{.}\) Since the calculated rank is \(2\) larger than the expected rank of \(6\) from our two known \(5 \times 5\) blocks, we conclude that two new blocks have reappeared with a solitary \(1\) in their respective lower-left corners. Since these blocks reappeared at the second power, they must have size \(3 \times 3\text{.}\) So now we know about two \(5 \times 5\) blocks and two \(3 \times 3\) blocks.

Finally, with four known blocks, we would expect the rank to increase from \(8\) to \(12\) as we move from \(A^2\) to \(A^1\text{,}\) an increase by \(1\) for each known block. But a calculation should reveal \(\rank A = 13\text{,}\) which means that one new block has reappeared. And since it appeared at exponent one, it must have size \(2 \times 2\text{.}\)

From above, we have \(\rank A = 13\text{,}\) so

This means that \(N\) should have six blocks (which we secretly know it does). And since we already have five blocks (two \(5 \times 5\) blocks, two \(3 \times 3\) blocks, and one \(2 \times 2\) block), this means that there must be one remaining unknown block. It can't be size \(2 \times 2\text{,}\) as the \(\rank A^1\) calculation should have revealed all of the blocks of that size. So there must be one remaining \(1 \times 1\) block, and we now have enough information to know the form matrix \(N\) precisely.

Example 33.5.3. A \(9 \times 9\) “bottom-up” example.

Consider the matrix

To carry out Procedure 33.4.2, we first need to determine the degree of nilpotency of \(A\text{.}\) In Discovery 33.2, we found this degree to be \(k = 4\text{.}\) So the largest block(s) in the triangular-block nilpotent form for \(A\) will have size \(4\text{,}\) and we need to look for the corresponding \(4\)-dimensional \(A\)-cyclic subspaces of \(\R^9\text{.}\)

Example 33.5.4. A \(5 \times 5\) “top-down” example.

Consider the \(5 \times 5\) matrix

Compute the powers of \(A\text{:}\)

So \(A\) is nilpotent of degree \(3\text{.}\)

The standard basis for \(\R^5\) has five vectors in it, so we begin with five \(A\)-cyclic spaces:

But recall that for any matrix \(B\text{,}\) the product \(B \uvec{e}_j\) is precisely the \(\nth[j]\) column of \(B\text{.}\) So, by examining when the columns of powers of \(A\) first become zero, we can use Theorem 32.6.3 to obtain a cyclic basis for each of these spaces:

We don't have any dimension \(1\) cyclic spaces, so let's move on to considering the collection of final vectors from each basis:

Examining the columns of \(A^2\text{,}\) notice that the first three vectors are all equal to each other, and the fifth vector is the negative of the those first three. Let's begin with the dependence relation

Factoring out the \(A^2\text{,}\) this turns into

We form the new cyclic space generated by \(\uvec{f}_1 = \uvec{e}_1 - \uvec{e}_2\text{,}\) and call it \(U_1\text{.}\) The above relation says that \(A^2 \uvec{f}_1 = \zerovec\text{,}\) so we have

Our dependence relation involved the final vectors from \(W_1\) and \(W_2\text{.}\) These two spaces have equal dimension, so we can use the new space \(U_1\) to replace either of \(W_1\) or \(W_2\text{.}\) If we decide to replace \(W_1\text{,}\) our new collection of cyclic subspaces is

We will work more quickly through this process now.

• From the dependence relation
  \begin{equation*} A^2\uvec{e}_2 - A^2\uvec{e}_3 = \zerovec \end{equation*}
  obtain new cyclic vector \(\uvec{f}_2 = \uvec{e}_2 - \uvec{e}_3\text{,}\) and use it to replace \(W_2\text{:}\)
  \begin{gather*} U_1 = \Span \{ \uvec{f}_1, A\uvec{f}_1 \}, \quad U_2 = \Span \{ \uvec{f}_2, A\uvec{f}_2 \}, \quad W_3 = \Span \{ \uvec{e}_3, A\uvec{e}_3, A^2\uvec{e}_3 \},\\ W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}
• From the dependence relation
  \begin{equation*} A^2\uvec{e}_3 + A^2\uvec{e}_5 = \zerovec \end{equation*}
  obtain new cyclic vector \(\uvec{f}_3 = \uvec{e}_3 + \uvec{e}_5\text{,}\) and use it to replace \(W_3\text{:}\)
  \begin{gather*} U_1 = \Span \{ \uvec{f}_1, A\uvec{f}_1 \}, \quad U_2 = \Span \{ \uvec{f}_2, A\uvec{f}_2 \}, \quad U_3 = \Span \{ \uvec{f}_3, A\uvec{f}_3 \},\\ W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}
• From the dependence relation
  \begin{equation*} A\uvec{f}_2 - A\uvec{e}_4 = \zerovec \end{equation*}
  obtain new cyclic vector \(\uvec{f}_4 = \uvec{f}_2 - \uvec{e}_4\text{,}\) and use it to replace \(U_2\text{.}\) However, \(\uvec{f}_4\) only generates a cyclic space of dimension \(1\text{,}\) and is the negative of \(A^2 \uvec{e}_5\text{.}\) So discard \(\uvec{f}_4\text{,}\) bringing us down to four cyclic spaces:
  \begin{gather*} U_1 = \Span \{ \uvec{f}_1, A\uvec{f}_1 \}, \quad U_3 = \Span \{ \uvec{f}_3, A\uvec{f}_3 \},\\ W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{gather*}
• From the dependence relation
  \begin{equation*} A \uvec{f}_1 + A \uvec{f}_3 + 2 A \uvec{e}_4 = \zerovec \end{equation*}
  obtain new cyclic vector \(\uvec{f}_5 = \uvec{f}_1 + \uvec{f}_3 + 2 \uvec{e}_4 \text{,}\) and use it to replace \(U_1\text{.}\) However, \(\uvec{f}_5\) only generates a cyclic space of dimension \(1\text{,}\) and is equal to \(A^2 \uvec{e}_5 - A \uvec{e}_4 \text{.}\) So discard \(\uvec{f}_5\text{,}\) bringing us down to three cyclic spaces:
  \begin{equation*} U_3 = \Span \{ \uvec{f}_3, A\uvec{f}_3 \}, \quad W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{equation*}
• From the dependence relation
  \begin{equation*} A \uvec{f}_3 - A \uvec{e}_4 - A^2 \uvec{e}_5 = \zerovec \end{equation*}
  obtain new cyclic vector \(\uvec{f}_6 = \uvec{f}_3 - \uvec{e}_4 - A \uvec{e}_5 \text{,}\) and use it to replace \(U_3\text{.}\) However, \(\uvec{f}_6\) only generates a cyclic space of dimension \(1\text{,}\) and is equal to \(A^2 \uvec{e}_5 \text{.}\) So discard \(\uvec{f}_5\text{,}\) bringing us down to two cyclic spaces:
  \begin{equation*} W_4 = \Span \{ \uvec{e}_4, A\uvec{e}_4 \}, \quad W_5 = \Span \{ \uvec{e}_5, A\uvec{e}_5, A^2\uvec{e}_5 \}. \end{equation*}

Since \(A\) does not have maximum degree of nilpotency, we know it is not similar to an elementary nilpotent form matrix. In other words, it is not possible for us to reduce down to a single \(A\)-invariant space, and the fact that \(A \uvec{e}_4\) and \(A^2 \uvec{e}_5\) are independent confirm that we are finished at this point. All that is left is to form the transition matrix

where we have ordered the bases for our \(A\)-cyclic spaces in the columns of \(P\) according to their dimensions to ensure that the elementary nilpotent blocks appear in order of decreasing size in the form matrix \(\inv{P} A P\text{:}\)