2.5 Partitioned Matrices

Very often we will have to consider certain groups of rows and columns of a matrix $\data{A}(n\times p)$. In the case of two groups, we have

$$\data{A}=\left(\begin{array}{ll} \data{A}_{11} & \data{A}_{12}\\ \data{A}_{21} & \data{A}_{22} \end{array}\right)$$

where $\data{A}_{ij} (n_i \times p_j) ,\ i,j=1,2,\ n_1+n_2=n$ and $p_1+p_2=p$.

If $\data{B} (n \times p)$ is partitioned accordingly, we have:

$$\begin{eqnarray*} \data{A}+\data{B} & = & \left(\begin{array}{ll} \data{A}_{11}... ...p}_{21} + \data{A}_{22}\data{B}^{\top}_{22} \end{array}\right). \end{eqnarray*}$$

An important particular case is the square matrix $\data{A}( p \times p)$, partitioned such that $\data{A}_{11}$ and $\data{A}_{22}$ are both square matrices (i.e., $n_j=p_j,j=1,2$). It can be verified that when $\data{A}$ is non-singular ( $\data{A}\data{A}^{-1}=\data{I}_p$):

$$\data{A}^{-1}=\left(\begin{array}{ll} \data{A}^{11} & \data{A}^{12}\\ \data{A}^{21} & \data{A}^{22} \end{array}\right)$$ (2.26)

where

$$\left\{\begin{array}{lclcl} \data{A}^{11}&=& (\data{A}_{11}-... ...})^{-1} \data{A}_{12}\data{A}^{-1}_{22}&&. \end{array}\right.$$

An alternative expression can be obtained by reversing the positions of $\data{A}_{11}$ and $\data{A}_{22}$ in the original matrix.

The following results will be useful if $\data{A}_{11}$ is non-singular:

$$\vert\data{A}\vert=\vert\data{A}_{11}\vert\vert\data{A}_{22}... ...}\vert =\vert\data{A}_{11}\vert\vert\data{A}_{22\cdot 1}\vert.$$ (2.27)

If $\data{A}_{22}$ is non-singular, we have that:

$$\vert\data{A}\vert=\vert\data{A}_{22}\vert\vert\data{A}_{11}... ...}\vert =\vert\data{A}_{22}\vert\vert\data{A}_{11\cdot 2}\vert.$$ (2.28)

A useful formula is derived from the alternative expressions for the inverse and the determinant. For instance let

$$\data{B} =\left(\begin{array}{ll} 1 & b^{\top}\\ a & \data{A} \end{array}\right)$$

where $a$ and $b$ are $(p \times 1)$ vectors and $\data{A}$ is non-singular. We then have:

$$\vert\data{B}\vert=\vert \data{A}-ab^{\top} \vert= \vert \data{A}\vert\vert 1-b^{\top}\data{A}^{-1}a \vert$$ (2.29)

and equating the two expressions for $\data{B}^{22}$, we obtain the following:

$$(\data{A}-ab^{\top})^{-1}=\data{A}^{-1}+\frac{\data{A}^{-1}ab^{\top}\data{A}^{-1}}{1-b^{\top}\data{A}^{-1}a}.$$ (2.30)

EXAMPLE 2.9   Let's consider the matrix

$${\data A}=\left(\begin{array}{cc}1&2\\ 2&2\end{array}\right).$$

We can use formula (2.26) to calculate the inverse of a partitioned matrix, i.e., ${\data A}^{11}=-1, {\data A}^{12}={\data A}^{21}=1, {\data A}^{22}=-1/2$. The inverse of ${\data A}$ is

$${\data A}^{-1}=\left(\begin{array}{cc}-1&1\\ 1&-0.5\end{array}\right).$$

It is also easy to calculate the determinant of ${\data A}$:

$$\vert{\data A}\vert=\vert 1\vert\vert 2-4\vert=-2.$$

Let $\data{A}(n\times p)$ and $\data{B}(p \times n)$ be any two matrices and suppose that $n\geq p$. From (2.27) and (2.28) we can conclude that

$$\left\vert \begin{array}{cc} -\lambda\data{I}_n&-\data{A}\\ ... ...ata{I}_p\vert = \vert\data{A}\data{B}-\lambda \data{I}_n\vert.$$ (2.31)

Since both determinants on the right-hand side of (2.31) are polynomials in $\lambda$, we find that the $n$ eigenvalues of $\data{A}\data{B}$ yield the $p$ eigenvalues of $\data{B}\data{A}$ plus the eigenvalue $0$, $n-p$ times.

The relationship between the eigenvectors is described in the next theorem.

THEOREM 2.6   For $\data{A}(n\times p)$ and $\data{B}(p \times n)$, the non-zero eigenvalues of $\data{A}\data{B}$ and $\data{B}\data{A}$ are the same and have the same multiplicity. If $x$ is an eigenvector of $\data{A}\data{B}$ for an eigenvalue $\lambda\neq 0$, then $y=\data{B}x$ is an eigenvector of $\data{B}\data{A}$.

COROLLARY 2.2   For $\data{A}(n\times p)$, $\data{B} (q\times n)$, $a (p\times 1)$, and $b (q\times 1)$ we have

$$\mathop{\rm {rank}}(\data{A}ab^{\top}\data{B})\leq 1.$$

The non-zero eigenvalue, if it exists, equals $b^{\top}\data{B}\data{A}a$ (with eigenvector $\data{A}a$).

PROOF:
Theorem 2.6 asserts that the eigenvalues of $\data{A}ab^{\top}\data{B}$ are the same as those of $b^{\top}\data{B}\data{A}a$. Note that the matrix $b^{\top}\data{B}\data{A}a$ is a scalar and hence it is its own eigenvalue $\lambda_1$.

Applying $\data{A}ab^{\top}\data{B}$ to $\data{A}a$ yields

$$(\data{A}ab^{\top}\data{B})(\data{A}a) = (\data{A}a) (b^{\top}\data{B}\data{A}a) = \lambda_1 \data{A}a.$$

${\Box}$