Applications of the Jordan Canonical Form

For B an n × n matrix with positive eigenvalues, the Jordan canonical form can be used to find a matrix A for which A ² = B. The procedure is as follows:

Find the Jordan canonical blocks for the matrix B. For a single Jordan block, this will be of the form λI + N, where λ > 0 and N is nilpotent; that is, there is a positive integer k for which N ^k is the zero matrix.

Find the Maclaurin series for λ + x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0001.tif"/>

In the Maclaurin series for λ + x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0002.tif"/> , replace λ by λI and replace x by N.

1.Since N is nilpotent, there is a positive integer k for which N ^k is the zero matrix, so the series has only finitely many terms.

For P the matrix whose columns are the generalized eigenvectors, the matrix can be converted back to the standard basis via

P λ I + N P − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0003.tif"/>

Example

We find the square root of A = − 1 − 2 − 1 2 4 − 1 6 3 6 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0004.tif"/>

The only eigenvalue is λ = 3, and a basis for the eigenspace is316 v ^ 1 = − 1 2 0 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0005.tif"/>

To find the generalized eigenvectors, we solve A − 3 I v ^ 2 = v ^ 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0006.tif"/> , and find that v ^ 2 = 1 − 1 − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0007.tif"/>

We also solve A − 3 I v ^ 3 = v ^ 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0008.tif"/> , and find that v ^ 3 = − 2 / 3 2 / 3 1 / 3 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0009.tif"/>

Thus, P = − 1 1 − 2 / 3 2 − 1 2 / 3 0 − 1 1 / 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0010.tif"/>

and P − 1 A P = − 1 1 − 2 3 2 − 1 2 3 0 − 1 1 3 − 1 − 1 − 2 − 1 2 4 − 1 6 3 6 − 1 1 − 2 3 2 − 1 2 3 0 − 1 1 3 = 3 1 0 0 3 1 0 0 3 = 3 I + N https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0011.tif"/>

where N = 0 1 0 0 0 1 0 0 0 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0012.tif"/>

We note that N 2 = 0 0 1 0 0 0 0 0 0     and     N 3 = 0 0 0 0 0 0 0 0 0 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0013.tif"/>

The first three terms of the Maclaurin series of 3 + x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0014.tif"/> are 3 + 3 6 x − 3 72 x 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0015.tif"/>

so we form the sum317 3 I + 3 6 N − 3 72 N 2 = 3 1 1 / 6 − 1 / 72 0 1 1 / 6 0 0 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0016.tif"/>

One can check that it is indeed the case that 3 1 1 / 6 − 1 / 72 0 1 1 / 6 0 0 1 2 = 3 1 0 0 3 1 0 0 3 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0017.tif"/>

Thus, we have found A https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0018.tif"/> with respect to the basis v ^ 1 , v ^ 2 , v ^ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0019.tif"/> . To find A https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0020.tif"/> with respect to the standard basis, we compute P 3 1 1 / 6 − 1 / 72 0 1 1 / 6 0 0 1 P − 1 = 3 / 4 − 3 3 / 8 − 5 3 / 24 3 / 2 5 3 / 4 − 3 / 12 3 3 / 2 3 3 / 2 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0021.tif"/>

One can check it is indeed the case that 3 / 4 − 3 3 / 8 − 5 3 / 24 3 / 2 5 3 / 4 − 3 / 12 3 3 / 2 3 3 / 2 2 = − 1 − 2 − 1 2 4 − 1 6 3 6 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0022.tif"/>

One can extend this idea to finding a function of A as long as the function has a Taylor series. Note that applying the Taylor series to the original matrix will normally not be beneficial if the matrix is not nilpotent.

Example

In this example, we find the square root of a matrix that has two eigenvalues.

Let A = 3 0 0 0 0 1 4 1 2 1 − 1 0 3 1 0 1 0 − 1 1 0 − 2 − 1 0 0 2 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0023.tif"/>

The Jordan canonical form of A is 3 0 0 0 0 1 3 0 0 0 0 0 3 0 0 0 0 0 2 0 0 0 0 1 2 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0024.tif"/>

We consider the Jordan blocks separately.

Let318 B = 3 1 0 0 3 0 0 0 3 = 3 1 0 0 3 0 0 0 3 + 0 1 0 0 0 0 0 0 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0025.tif"/>

and C = 2 1 0 2 = 2 0 0 2 + 0 1 0 0 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0026.tif"/>

For the matrix B, λ I = 3 1 0 0 0 1 0 0 0 1 = 3 0 0 0 3 0 0 0 3     and     N = 0 1 0 0 0 0 0 0 0 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0027.tif"/>

The Taylor expansion of 3 + x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0028.tif"/> to three terms is 3 + 3 x 6 − 3 x 2 72 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0029.tif"/>

Thus we compute 3 0 0 0 3 0 0 0 3 + 1 6 3 0 0 0 3 0 0 0 3 0 1 0 0 0 0 0 0 0     − 1 72 3 0 0 0 3 0 0 0 3 0 1 0 0 0 0 0 0 0 2 = 3 3 6 0 0 3 0 0 0 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0030.tif"/>

and note that 3 3 6 0 0 3 0 0 0 3 2 = 3 1 0 0 3 0 0 0 3 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0031.tif"/>

We follow the same procedure for C = 2 1 0 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0032.tif"/> .

The Taylor expansion of 2 + x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0033.tif"/> to two terms is

2 + 2 x 4 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0034.tif"/> so with λ I = 2 0 0 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0035.tif"/> and N = 0 1 0 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0036.tif"/> , λ I + 1 4 λ I N = 2 0 0 2 + 1 4 2 0 0 2 0 1 0 0 = 2 2 4 0 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0037.tif"/>

and note that319 2 2 4 0 2 2 = 2 1 0 2 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0038.tif"/>

Now, if we let P = 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 ,     then     P − 1 = 1 0 − 1 1 − 1 0 1 0 0 − 1 − 1 0 1 − 1 2 0 − 1 0 1 0 0 0 1 − 1 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0039.tif"/>

So P − 1 = 1 0 − 1 1 − 1 0 1 0 0 − 1 − 1 0 1 − 1 2 0 − 1 0 1 0 0 0 1 − 1 1 3 3 6 0 0 0 0 3 0 0 0 0 0 3 0 0 0 0 0 2 2 4 0 0 0 0 2 1 0 − 1 1 − 1 0 1 0 0 − 1 − 1 0 1 − 1 2 0 − 1 0 1 0 0 0 1 − 1 1 3 3 6 2 − 3 3 − 2 2 − 7 3 6 0 7 3 6 0 0 − 3 6 0 3 − 2 5 2 4 − 2 4 5 2 4 − 3 0 7 3 6 − 2 2 4 3 2 4 2 4 − 3 6 0 3 6 0 0 5 3 6 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0040.tif"/>

and320 3 3 6 2 − 3 3 − 2 2 − 7 3 6 0 7 3 6 0 0 − 3 6 0 3 − 2 5 2 4 − 2 4 5 2 4 − 3 0 7 3 6 − 2 2 4 3 2 4 2 4 − 3 6 0 3 6 0 0 5 3 6 2 = 3 1 − 1 1 − 2 0 4 0 0 − 1 0 1 3 − 1 0 0 2 1 1 0 0 1 0 0 2 = A . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315152240/31a6f155-b0b7-46e3-a5d0-139021db1359/content/TNFUK-CH011_eqn_0041.tif"/>